THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES Abstract. Basic notions of calculus are reviewed and geometric applications discussed in order to build up to Green’s Theorem in the plane and its extensions in three dimensions.
Outline of Contents 1 Derivative of one variable functions 1.1 Definition of derivative (one variable). 1.2 Geometric Interpretation 1.3 Basic laws of continuous functions with derivatives in an interval. (Finite Covering, Bounded, Max/Min, Intermediate Value, Law of the Mean)
2. Di↵erential of one variable functions 2.1 Approximating y by a linear function of x for small x 2.1.1 In general, y is a function of both x, x 2.1.2 Definition of the total di↵erential at a point x0 2.1.3 The di↵erential of the independent variable, dx = x 2.1.4 The total di↵erential and the derivative. If f has a total di↵erential at x0 0 then it can be written dy = f (x0 ) (x) + ✏1 x 2.2 Existence of the total di↵erential; Fundamental lemma for one variable (If f has a continuous derivative on (a, b) then the total di↵erential exists on (a, b) and 0 we write dy = f (x)dx. 2.2.1 Chain Rule for di↵erentials. If z = g(y), y = f (x) and the total di↵erentials df (x) dzdy dz exist at x0 , y0 = f (x0 ), then dx = dydx = dg(y) dy dx 2.2.2 Chain rule from 1st principles (definition of derivative). Example.
3 Di↵erential of two variable functions 3.1 Definition of the total di↵erential at a point (x0 , y0 ) 3.2 If f (x, y) has a total di↵erential at (x0 , y0 ) then it can be written dz = fx (x0 , y0 ) x + fy (x0 , y0 ) y 3.3 Existence of the total di↵erential: Fundamental lemma for two variables (If f (x, y) has continuous partial derivatives fx , fy in domain D then the total di↵erential exists in D and we write dz = fx dx + fy dy) 3.3.1 Chain Rule for di↵erentials I. If z = f (x, y), x = x(t), y = y(t), then dy dz dx dx = zx dt + zy dt . In particular, still have dz = zx dx + zy dy 3.3.2 Chain Rule for di↵erentials II. If z = f (x, y), x = x(u, v), y = y(u, v), then zu = zx xu + zy yu and zv = zx xu + zy yv . In particular, still have dz = zx dx + zy dy Date: January 28, 2014 . 1
2 THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES
3.3.3 Practical Application:
Geometric Applications 3.4 Di↵erentials and Linearization Rb 3.5 Curves in Space [Tangent Line, arc length s = a sec ↵ dx] 3.6 Surfaces in Space Tangent Plane 3.6.1 Tangent line to a curve defined as the intersection of two surfaces 3.6.2 Alternate derivation of the equation of the tangent plane to surface at a point [Ford and Ford] 3.6.3 Tangent plane when surface represented F (x, y, z) = 0 3.7 Surface Area 3.7.1 The area of a rectangle projected onto another plane RR 3.7.2 Surface area S = Rxy sec ↵ dxdy 3.7.3 Surface area (parametric form)
4 Two-dimensional Line Integral 4.1 Line Integrals of Functions 4.2 Line Integrals of Vector Fields 4.3 Green’s Theorem in the plane relating line integral on a closed curve with ordinary double integral
5 Three-dimensional Line Integral
6 Surface Integrals 6.1 Parameterized Surfaces 6.2 Orientable Surfaces 6.3 Surface Integrals of Vector Fields 6.3.1 Definition 6.3.2 Evaluation 6.4 Stokes’ Theorem - A higher dimensional version of Green’s Theorem 6.5 Divergence Theorem or Gauss’ Theorem - Another higher dimensional version of Green’s Theorem
7 Review and Discussion 7.1 Three theorems and proofs 7.2 A special vector field, F⇤ 7.2.1 A family of velocity vector fields 7.2.2The F⇤ vector field 7.2.3 F⇤ is not (globally) conservative 7.2.4 The angle function 7.2.5 Extensions to complex variables
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 3
Introduction These notes take the reader from first principles through the theorems of Green, Gauss and Stokes in which interrelationships between line, surface and ordinary integrals are established. The presentation is meant as a summary with discussions and proofs based on presentations made in the references. There is no claim to originality in this paper and readers are encouraged to read the references for more complete detail and mathematical scope. What is intended is a useful summary. As in most mathematical presentations, credit for the approach is diverse and clouded. Each individual mathematician adds his or her insights hoping to make the topic at hand reachable for the future reader. Although Mathematics is seen by some as rigid, there is an art in presentation and at best a presentation can shed light in a helpful way on the rigorous topics it is describing. Di↵erent presentations of the subject matter reach di↵erent audiences in di↵erent ways. In that vein, the references cited were helpful approaches to the subject for this reader. Hopefully, this presentation will shed a little light for others. 1. Derivative of one variable functions 1.1. Definition. Let f be a function of one independent variable x and write y = f (x) where y is a variable whose value depends on x. We say that f is di↵erentiable at a point x0 if lim x!0 f (x0 + x)x f (x0 ) = lim x!0 xy exists, and we 0 0 write f (x0 ) to represent the derivative of f at x0 where f (x0 ) = lim x!0 xy . 1.2. Geometric Interpretation. If we look at two points on the graph of f , (x0 , y0 ) and (x0 + x, y0 + y), and form the secant joining these two points, it 0 + y y0 has slope xy = xy00 + x x0 . So, geometrically speaking, f (x0 ) = limit of the slope of the secant from (x0 , y0 ) to (x0 + x, y0 + y) as x ! 0. The limiting secant is 0 called the tangent at x0 , a line with slope f (x0 ) and containing the point (x0 , y0 ). That is the tangent line: 0 y y0 = f (x0 ) x x0 0 If f (x0 ) > 0 then there is a neighborhood about x0 where f (x) < f (x0 ) for x < x0 , f (x) > f (x0 ) for x > x0 , for x in the neighborhood. That is, f is increasing in the neighborhood. 0 Proof: Let f (x0 ) = c > 0. Then lim x!0 f (x0 + x)x f (x0 ) = c > 0. So if we are close enough to x0 (that is, for x sufficiently small) we know the ratio is positive, so numerator and denominator have the same sign. If x > 0 we have f (x0 + x) > f (x0 ) and if x < 0 we have f (x0 + x) < f (x0 ). A similar argument applies to the case of a negative derivative at x0 showing that f is decreasing in a neighborhood of x0 . 1.3. Basic Laws of Continuous Functions. This section is taken from [1] where frames are used to define continuity in a most helpful way. Definition (Frame): A closed rectangle F with sides parallel to the axes with (1) x1 < x x2 , x1 < x2 and (2) y1 y y2 and containing at least one point of the graph of y = f (x) will be called a frame for the function f (x) if for each x satisfying (1), the corresponding y satisfies (2). A point (x, y) is enclosed by a frame if the
4 THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES
point lies in the frame but is not on a vertical side. Frames are overlapping if their bases contain common interior points. They are called adjacent if their bases abut. Definition (Continuity): f (x) is continuous at x0 if and only if for every h > 0, there is a frame enclosing (x0 , y0 ) whose height is no greater than h. [This is equivalent to the ✏, - definition.] Axiom: Every set of real numbers which possesses an upper bound also possesses a least upper bound. Theorem: (Finite Covering) If f (x) is continuous on [a, b], then, for any h > 0, a finite number of adjacent, non-overlapping frames can be found, each of height no more than h, which cover the graph in the interval. Proof: Starting from a suppose we can cover the graph for some way with frames with height less than or equal to h. The frame enclosing (a, f (a)) (with the part to the left of a cut o↵) suffices for the base a x x1 . The frame enclosing (x1 , f (x1 )) (with the part to the left of x1 cut o↵) extends the cover to a new point x2 , etc. Consider all the intervals a x b1 b which can be covered as required (finite, non-overlapping) by the theorem. That is, we can cover these intervals from a to b1 by finite, non-overlapping frames of height no greater than h. The set of numbers b1 have a least upper bound (by the axiom), say u b. By continuity of f there is a frame enclosing (u, f (u)) with base b0 x b00 where b0 < u < b00 . The graph for the interval a x b0 can be covered, and one additional frame covers the graph to b00 > u. This shows that u cannot be less than b and u = b means the covering can be extended to b. Corollary: Taking h = A/(b a) we conclude that if f (x) is continuous in [a, b] and A > 0, then a finite number of non-overlapping frames can be found which cover the graph in the interval and whose total area is no greeter than A. Theorem (Continuous in closed interval means Bounded): If f cont. in [a, b], then f bounded in [a, b]. Proof: Consider any finite covering of the interval. The finite collection of horizontal sides has a minimum (y = y1 ) and maximum (y = y2 ) value. Corollary: If f is continuous on the closed interval [a, b] then there are points in [a, b] where f takes on a maximum and minimum value. Proof: The function is bounded above and below by the theorem and so it must have a least upper bound,M , and a greatest lower bound, m. We need to show these values are taken on by the function. Suppose M is not taken on by the function. Then the function 1/(M f (x)) is continuous on [a, b] and has an upper bound K. Since the function is positive (M is an upper bound), K > 0. Thus, 1/(M f (x)) K, 1/K (M f (x)), so f (x) M 1/K, contrary to M being a least upper bound of the values of f . Thus M is attained. Similarly, m is attained.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 5
Theorem (Intermediate Value) If f is continuous in a closed interval [a, b] and y0 is a value between its least value m and its greatest value M , then f (x) = y0 for at least one point in the interval. Proof: Suppose f (x) y0 is nowhere zero. Then 1/(f (x) y0 ) is continuous in the interval and so bounded, |f (x) y0 | > 1/B. But this is contrary to the continuity of f . That is, we know by the continuity of f in [a, b] that we can find a finite covering with all frames having height less than 1/B. Since m < y0 < M we know the frame containing M must be entirely above y = y0 and the frame containing m must be entirely below this line. Hence in our finite collection of frames we have some (at least one) wholly above the line y = y0 and some (at least one) wholly below this line. At some point, two of these frames must be adjacent to each other but having no point in common, contrary to the construction of the cover. Theorem (Derivative zero at interior max/min) If the maximum or minimum value of a function in an interval occurs at an interior point c of the interval, and 0 if the derivative exists at c, then f (c) = 0. 0
Proof: If f (c) 6= 0, then the function is increasing or decreasing at c so it cannot have a maximum or minimum there. 0
Theorem (Rolle) Suppose f is continuous in [a, b] and f (a) = f (b) = 0. If f (x) 0 exists in (a, b) then there must be at least one point c 2 (a, b) with f (c) = 0. Proof: Know f has maximum and minimum values in the interval. If f > 0 anywhere, it has a maximum in the interval. If f < 0 anywhere it has a minimum inside the interval. If f = 0 everywhere then any point in the interval yields a maximum or minimum value. In all cases (by the last theorem) we have an interior 0 maximum or minimum and therefore a point c where f (c) = 0. Theorem (Extension of Rolle’s Theorem): Let f (x) and g(x) be continuous in [a, b] and be equal at the endpoints, f (a) = g(a), f (b) = g(b). Then if both functions are di↵erentiable in (a, b), there is at least one point in (a, b) at which the derivatives are equal. Proof: Apply Rolle’s Theorem to the function f
g.
Theorem (Law of the Mean) Let f (x) be continuous in [a, b] and di↵erentiable in 0 (a, b). Then there is at least one point c 2 (a, b) at which f (b) f (a) = (b a)f (c). Proof: In the preceding theorem let g(x) = mx + k so chosen that it is equal to f (x) at the endpoints. Then at some interior point c the derivatives are equal, i.e., 0 f (c) = m, where m = [f (b) f (a)]/[b a]. Note: In applying this theorem we usually do not require the precise location of the point c, only that it exists.
6 THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES
2. Differentials in one variable 2.1. Approximating y by a linear function of x for small x. If f is a ”complicated” function of x it may be ”complicated” to compute y for a given x. 0 But if f is di↵erentiable at x0 , then we know |(f (x0 + x) f (x0 ))/ x f (x0 )| is 0 close to zero for sufficiently small x and so too is |(f (x0 + x) f (x0 )) f (x0 ) x|. 0 This means that f (x0 ) x is close to (f (x0 + x) f (x0 )) = (y0 + y) y0 = y 0 0 for x sufficiently small. Since f (x0 ) is a fixed quantity, we have f (x0 ) x as a linear function in x which approximates y as x ! 0. Computing this linear approximation might be a lot less complicated than the actual value of y. 2.1.1. In the discussion above there is no mention of the particular function f with which we might be dealing. We used only its di↵erentiability to arrive at an estimate for its change in value given a certain change in the value of the independent variable. This change in the functional value (f (x0 + x) f (x0 )) = y, might, in general, be a complicated function of both x0 and x made up of multiple powers of these terms. 2.1.2. Definition of the Total Di↵erential. We look at (f (x0 + x) f (x0 )) = y. This expression, when expanded in terms of x0 and x, consists of: some linear expression in x and some higher order x terms. [Assume here we have an expression for f (x) - such as a Taylor series - so that all terms are in powers of x.] When x is small the higher order terms are very small compared with the ”principal part” of y, that being the expression involving x0 and linear in x. Example: f (x) = x3 . y = f (x + x) f (x) = (x + x)3 x3 = 3x2 x + 3( x)2 + ( x)3 . As x ! 0 we have y ⇠ 3x2 x. 3 5 Example: f (x) = sin x = x x3! + x5! · · · and y = sin(x + x) sin(x). Thus y = [(x +
x)
x]
(1/3!)[3x
[(x + 2
x)3 /3! 2
=
x
=
x[1
3x2 /3! + 5x4 /5!
=
x[1
x2 /2! + x4 /4!
=
x cos x + ( x)2 [· · ·] + · · ·
y ⇠ cos x x as
x3 /3!] + [(x + 3
x)5 /5!
x + 3x( x) + ( x) ] + (1/5!)[5x · · ·] + ( x)2 [· · ·] + · · ·
4
x5 /5!] 3
···
x + 10x ( x)2 + 15x2 ( x)3 + · · ·] + etc.
· · ·] + ( x)2 [· · ·] + · · ·
x ! 0.
Now we formalize this idea. For y = f (x) we say that the total di↵erential dy(x, x) = dy exists at x if we can write y = f (x + x) f (x) as y = ax x + ✏1 x where ax is independent of x and ✏1 ! 0 as x ! 0. In such cases, we define the total di↵erential to be (1)
dy = ax x
where ax is evaluated at x. 2.1.3. The di↵erential of the independent variable. Consider the special case f (x) = x, the identity function where y = x. We can write y = x = 1 · x + 0 · x, so we can identify the change in the independent variable with its total di↵erential: x = dx. In particular we can write (2) (3)
dy = ax x = ax dx dy/dx = ax
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 7
2.1.4. The total di↵erential and the derivative. For f to have a total di↵erential at x0 means y = ax0 x + ✏1 x, where ax0 is independent of x and ✏1 ! 0 as x ! 0. Therefore, if f has a total di↵erential at x0 then y/ x = ax0 + ✏1 so as 0 x ! 0 we get lim x!0 y/ x = f (x0 ) = ax0 . 2.2. Existence of the Total Di↵erential: Fundamental Lemma for one variable. If y = f (x) has a derivative in (a, b), then by the Law of the Mean we 0 know that for any fixed x0 2 (a, b): y = f (x0 + x) f (x0 ) = f (x⇤0 ) x for ⇤ some x0 2 (x0 , x0 + x). Now, if the derivative is also continuous in (a, b), then 0 0 0 0 0 f (x⇤0 ) ! f (x0 ) as x ! 0. Therefore, y = f (x⇤0 ) x = f (x0 ) x + (f (x⇤0 ) 0 0 0 0 f (x0 )) x = f (x0 ) x + ✏ x where ✏ = f (x⇤0 ) f (x0 ) ! 0 as x ! 0. That is, the total di↵erential exists at x0 . So we have: 0
Lemma 1. f (x) continuous on (a, b) ) total di↵erential exists on (a, b) 0
0
0
We write dy = f (x) x or dy = f (x)dx or dy/dx = f (x). 2.2.1. Chain Rule for Di↵erentals. If z = g(y) and y = f (x) are defined and the 0 0 total di↵erentials exist at x0 and y0 = f (x0 ), then dz = g (y0 ) y = g (y0 )dy and 0 0 dy = f (x0 ) x = f (x0 )dx, so 0
(4)
0
dz = g (y0 )f (x0 )dx dz/dx]x0 = dz/dy]y0 · dy/dx]x0 dz dz dy = dx dy dx dg(y) df (x) = dy dx
(5) (6)
2.2.2. Chain Rule from first principles. In the last section we obtained the chain rule as a direct application of the Fundamental Lemma (1). We can also look at z x from first principles: z g(f (x + x)) g(f (x)) = x x g(f (x + x)) g(f (x)) f (x + x) f (x) = f (x + x) f (x) x g(y + y) g(y) f (x + x) f (x) = y x so as x, y ! 0, z/ x ! dz/dx and on the right hand side, the product 0 0 goes to g (y) · f (x). That is dz/dx = dg(y)/dy · df (x)/dx = dz/dy · dy/dx. Example: z = g(y) = y 3 ; y = f (x) = x2 . Then from the Fundamental Lemma dz = 3y 2 dy; dy = 2xdx so dz = 3y 2 (2x)dx = 6x5 dx. From the definition of di↵erential we get dz = a x + ✏ x and see that a = 6x5 , that is: z = z(x +
x)
= 6x5 x +
z(x) = (x +
x)6
x6
x[15x4 x + 20x3 ( x)2 + 15x2 ( x)3 + 6x( x)4 + ( x)5 ]
= 6x5 x + ✏ x
8 THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES
where ✏ ! 0 as
x ! 0. 3. Differential of two variable functions
3.1. Definition of the total di↵erential at a point (x0 , y0 ). We can extend the definition of di↵erential to a function of two (or more) variables. For z = f (x, y), the total di↵erential dx(x, y, x, y) = dz exists at (x, y) if we can write z = f (x + x, y + y) f (x, y) = a x + b y + ✏1 x + ✏2 y where a, b are independent of x, y and ✏1 , ✏2 ! 0 as both x, y ! 0 and we define (7)
dz = a x + b y
where a, b are evaluated at (x, y). As before, for independent variables x, y we can write x = dx, y = dy. 3.2. If f (x, y) has a total di↵erential at (x0 , y0 ) then it can be written dz = fx (x0 , y0 ) x+fy (x0 , y0 ) y. Proof: Taking y = 0; have z = a x+✏1 x = (a+ ✏1 ) x so z/ x ! a as x ! 0. That is, lim x!0 [z(x0 + x, y0 ) z(x0 , y0 )]/ x = a. So z/ x = fx (x0 , y0 ) = a. Similarly, if x = 0, get b = fy (x0 , y0 ). 3.3. Existence of the total di↵erential: Fundamental lemma for two variables. The proof of the Fundamental Lemma is the bulk of the work in proving the chain rule. Lemma 2. If f (x, y) has continuous partial derivatives fx , fy in domain D then the total di↵erential exists in D and we write dz = fx dx + fy dy. Proof. Fix a point (x, y) 2 D. First, suppose only x changes ( y = 0). Then z = f (x + x, y) f (x, y). Since y is fixed, we have z as a function of x, with continuous partial derivative fx defined in D, so, by the Law of the Mean, f (x + x, y) f (x, y) = fx (x1 , y) x for some x1 2 (x, x + x). Now by the continuity of fx , fx (x1 , y) ! fx (x, y) as x ! 0. Let ✏1 = fx (x1 , y) fx (x, y), so we know ✏1 ! 0 as x ! 0. Then, z = fx (x1 , y) x = (fx (x, y) + ✏1 ) x = fx (x, y) x + ✏1 x. In this final sum, the first coefficient of x is independent of x, y and the second goes to zero as x and y go to zero. In particular, for fixed y, z = fx (x, y) x + ✏1 x where ✏1 ! 0 as x ! 0. Similarly, for fixed x, z = fy (x, y) y + ✏2 y where ✏2 ! 0 as y ! 0. Now suppose both variables change. Then z = f (x + x, y + y) f (x, y) = [f (x + x, y) f (x, y)] + [f (x + x, y + y) f (x + x, y)] where in the first summand only x changes and in the second only y changes. Thus, as above, the first summand goes to fx (x, y) x + ✏1 x and the second to fy (x, y) y + ✏2 y and both ✏1 , ✏2 ! 0 as both x, y ! 0. Note: with ✏2 = fy (x + x, y1 ) fy (x, y) and fy continuous, (that is, each of its component functions is continuous) we have ✏2 ! 0 as both x and y ! 0. Thus we have shown that z = fx (x, y) x + fy (x, y) y + ✏1 x + ✏2 y where both ✏1 , ✏2 ! 0 as both x, y ! 0 and so, by definition, the di↵erential dz exists at each point of D. ⇤ We write dz = fx x + fy y or dz = fx dx + fy dy.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 9
We will assume now that functions we discuss are defined and have continuous partial derivatives in a domain D, so that, by Lemma (??) we know the di↵erentials exist.
3.3.1. Chain Rule for di↵erentials I. If z = f (x, y), x = x(t), y = y(t), then dy dz dx dx = zx dt + zy dt . In particular, still have dz = zx dx + zy dy. Suppose z = f (x, y) where x = g(t) and y = h(t), so we write z(t) = [g(t), h(t)] and consider z a function of t. Then we can consider z as a function of a single variable and ask what is dz/dt? If we fix t we define the particular values x = g(t), y = h(t), z = f (x, y). For a given change in t, t, x, y are determined: x = g(t + x) g(t) and y = h(t + t) h(t) which both go to zero as t ! 0. (Continuity of f means continuity of its component functions.) Also, z = f (x + x, y + y) f (x, y) is determined. By the Fundamental Lemma, z = f (x + x, y + y) f (x, y) = fx (x, y) x + fy (x, y) y + ✏1 x + ✏2 y so
(8)
z x y x y = fx + fy + ✏1 + ✏2 t t t t t z dx dy dx dy lim = fx + fy +0· +0· t!0 t dt dt dt dt dx dy = zx + zy dt dt dz dx dy = zx + zy dt dt dt
Notice that x, y, z as functions of t, have di↵erentials dx = g0 (t) t = dx/dt t, dy = h0 (t) t = dy/dt t, dz = z0 (t) t = dz/dt t. Multiplying both sides of equation (8) by t we get
(9)
dz dx dy t = zx ( t) + zy ( t) dt dt dt dz = zx dx + zy dy
as above when dx = x and dy = y were arbitrary increments of the independent variables. That is, this relationship holds whether x, y independent and dz the corresponding di↵erential or x, y, z all functions of an independent t and dx, dy, dz are the di↵erentials with respect to t. Now we will extend this result even further.
3.3.2. Chain Rule for di↵erentials II. If z = f (x, y), x = x(u, v), y = y(u, v), then zu = zx xu + zy yu and zv = zx xu + zy yv . In particular, still have dz = zx dx + zy dy. Proof: We will show zu = zx xu + zy yu . The second result is proven similarly. Again, we assume our functions are continuously di↵erentiable as needed so the
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 10
di↵erentials exist. z = zx x + zy y + ✏ 1 x + ✏ 2 y x = xu u + xv v + ✏ 3 u + ✏ 4 v y = yu u + y v v + ✏ 5 u + ✏ 6 v z = zx [xu u + xv v + ✏3 u + ✏4 v] + zy [yu u + yv v + ✏5 u + ✏6 ] + ✏1 [xu u + xv v + ✏3 u + ✏4 v] + ✏2 [yu u + yv v + ✏5 u + ✏6 v] z= +
u[zx xu + zy yu + zx ✏3 + zy ✏5 + xu ✏1 + yu ✏2 + ✏1 ✏3 + ✏2 ✏5 ] v[zx xv + zy yv + zx ✏4 + zy ✏6 + xv ✏1 + yv ✏2 + ✏1 ✏4 + ✏2 ✏6 ]
where all ✏i ! 0 as x, y, u, v ! 0. Now fix v ( v = 0). We have z = u[zx xu + zy yu + zx ✏3 + zy ✏5 + xu ✏1 + yu ✏2 + ✏1 ✏3 + ✏2 ✏5 ] and so z = [zx xu + zy yu + zx ✏3 + zy ✏5 + xu ✏1 + yu ✏2 + ✏1 ✏3 + ✏2 ✏5 ] u so that, as z, u ! 0 we get (10)
(11)
z = zu = [zx xu + zy yu ] u
Similarly, z = zv = [zx xv + zy yv ] v We consider u, v as independent variables and x, y, z dependent on u, v. The di↵erentials are:
(12)
dx = xu u + xv v dy = yu u + yv v dz = zu u + zv v , so using equations (11) and (12), = (zx xu + zy yu ) u + (zx xv + zy yv ) v = zx (xu u + xv v) + zy (yu u + yv v) dz = zx dx + zy dy so our familiar equation still holds when x, y, z are all functions of independent variables u, v and dx, dy, dz are all di↵erentials with respect to u, v. Conclude: The di↵erential formula dz = zx dx + zy dy + zw dw + · · · which holds when z = z(x, y, w, ...) and dx = x, dy = y, .... remains true when x, y, w, ... (hence z) are functions of other independent variables and dx, dy, ... are corresponding di↵erentials. In fact, any equation in di↵erentials treats all variables on an equal basis. That is, if an equation in di↵erentials is correct for one choice of independent and dependent variables, it remains true for any other such choice. If dz = 2dx 3dy then dx = (1/2)dz + (3/2)dy, etc.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 11
3.3.3. Practical Application. To compute partial derivatives one can first compute di↵erentials (treating all variables as functions of a hypothetical single variable 2 say t) so that all rules of ordinary di↵erential calculus apply. Example: z = x y 1 . Then dz =
y2xdx (x2 1)dy y2
=
2x y dx
x2 1 y 2 dy
so zx = 2x/y and zy = (1
x2 )/y 2 .
Geometric Applications 3.4. Di↵erentials and Linearization. Taking di↵erentials of an equation (or system of equations) corresponds to: a) replacement of equations in x, y, etc. by linear equations in dx, dy, etc. b) replacement of curves and surfaces by tangent lines and tangent planes.
3.5. Curves in Space [Tangent Line, arc length]. We will call a (parameterized) curve in space a set of points C = {(x, y, z) : x = f (t), y = g(t), z = h(t)} where the component functions are assumed to have continuous derivatives of all orders. Consider the secant line through the points P = (x(t1 ), y(t1 ), z(t1 )) and Q = (x(t1 + t), y(t1 + t), z(t1 + t)) = (x1 + x, y1 + y, z1 + z) on the curve. The tangent line at P = (x1 , y1 , z1 ) is the limit of the secant lines as t ! 0. Thus, the direction of the tangent line to C at the point P is 0 0 0 lim t!0 ( x/ t, y/ t, z/ t) = (f (t1 ), g (t1 ), h (t1 )) and the tangent line to C through P is the set (13)
T = {(x, y, z) : (x1 , y1 , z1 ) + (t
0
0
0
t1 )(f (t1 ), g (t1 ), h (t1 ))}
or (14)
T = {(x, y, z) :
x x1 y y1 z z1 = 0 = 0 =t 0 f (t1 ) g (t1 ) h (t1 )
t1 }
as long as the derivatives are non-zero. 0 0 0 That is, (x x1 ) = f (t1 )(t t1 ), (y y1 ) = g (t1 )(t t1 ), (z z1 ) = h (t1 )(t t1 ); x, y, z on T . Now identify (x x1 ) = dx; y y1 = dy, z z1 = dz, t t1 = dt to get the equivalent expression: 0 0 0 dx = f (t1 )dt, dy = g (t1 )dt, dz = h (t1 )dt, the di↵erentials with respect to t at t1 . So, if r¯ = (x, y, z); and d¯ r = (dx.dy.dz) then we have r¯]P = (xP , yP , zP ) is 0 0 0 a point on the curve and d¯ r]P = (dxP , dyP , dzP ) = (f (tP ), g (tP ), h (tP ))dt is tangent vector at the point P . That is, (15)
0 0 0 d¯ r ]P = (f (tP ), g (tP ), h (tP )) dt
is a vector in the direction of the tangent at P . Let r¯(t) = (x(t), y(t), z(t)) = (x, y, z) be a point on curve C = (x(t), y(t), z(t)), where x = x(t), y = y(t), z = z(t).
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 12
d¯ r r¯(t + t) r¯(t) = lim = (dx/dt, dy/dt, dz, dt) t!0 dt t 0 0 0 d¯ r = (f (t1 ), g (t1 ), h (t1 ))dt = (dx, dy, dz) ⇠ (x x1 , y
y1 , z
z1 )
where (x, y, z) is on tangent line through (x1 , y1 , z1 ) on C p |d¯ r| = dx2 + dy 2 + dz 2 = ds = di↵erential arc length along curve (16)
|d¯ r/dt| = |d¯ r|/dt = ds/dt
the speed along curve (arc length per unit time) - assuming s increasing with t. So we have,
(17)
d¯ r d¯ r dt d¯ r/dt d¯ r/dt = = = = Tˆ ds dt ds ds/dt |d¯ r/dt|
where Tˆ is the unit vector in the direction of increasing s. If the parameter t is itself the arc length s along the curve then d¯ r dx dy dz = ( , , ) = Tˆ ds ds ds ds
(18)
We referred to ds above as the di↵erential arc length along the curve. The arc length can be approximated by the tangent segment. That is, let P be a point on a curve y = f (x) in the x, y-plane. Let ↵ be the angle the tangent at P makes with the x-axis. Then dT cos ↵ = dx = x and ds ⇠ dT = sec ↵ dx. So the arc length Rb 0 from a point a to a point b on the curve is s = a sec ↵ dx where tan ↵ = f (x) so p sec ↵ = 1 + (f 0 (x))2 . That is. (19)
s=
Z
b
a
q
1 + (f 0 (x))2 dx
It can be shown (using ds2 = dx2 + dy 2 ) that if the curve is given parametrically by C = (x(t), y(t)) then
(20)
s=
Z
b a
q
(x0 (t))2 + (y 0 (t))2 dt
and for a curve in space (ds2 = dx2 + dy 2 + dz 2 )
(21)
s=
Z
b a
q (x0 (t))2 + (y 0 (t))2 + (z 0 (t))2 dt
with suitable continuity and di↵erentiability constraints on the component functions.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 13
3.6. Surfaces in Space [Tangent Plane]. A function F (x, y, z) = 0 defines a surface in space. The point P = (x0 , y0 , z0 ) is on the surface if F (x0 , y0 , z0 ) = 0. Let C = (f (t), g(t), h(t)) be a curve on F which contains P on F . Then F (f (t), g(t), h(t)) = 0 for all t and, say, at t = t0 we have (f (t0 ), g(t0 ), h(t0 )) = (x0 , y0 , z0 ) = P . Take the di↵erential at P : dF ]P = Fx ]P dx + Fy ]P dy + Fz ]P dz = 0 where dx = 0 0 0 f (t0 )dt, dy = g (t0 )dt, dz = h (t0 )dt. As before identify the di↵erentials by dx = x x0 , dy = y y0 , dz = z z0 where (x, y, z) is any point on the tangent to C at P to get the equation of the tangent plane at P (22)
dF ]P = Fx ]P (x
x0 ) + Fy ]P (y
y0 ) + Fz ]P (z
z0 ) = 0
Since this equation is satisfied by every point on the tangent line to C at P , the tangent line is included in the tangent plane (i.e. the set of all points satisfying the equation includes the points of the tangent line). Since this equation is independent of the particular curve we used to construct it, we conclude that for any curve in F which contains P , the tangent line to that curve at P will be included in the tangent plane. That is, the tangent plane at P to the surface defined by F contains the tangent lines of all the curves in F that include the point P . 3.6.1. Tangent line to a curve defined by the intersection of two surfaces. Given surfaces defined by the two equations F (x, y, z) = 0 and G(x, y, z) = 0, let P be a point on both of the surfaces (i.e. P is included in the curve of intersection of the two surfaces.) Let dF ]P = 0 and dG]P = 0 be the equations of the tangent planes to F and G at P . As a point on a curve (of intersection) through P which lies in both surfaces F and G, the tangent line must lie in both tangent planes as we saw in the last section. As a line common to two planes, the tangent line at P must be the line of intersection of the two planes. If d¯ r is the direction of the tangent at P , then since the tangent line is in both planes, we must have d¯ r · nF = d¯ r · nG = 0 where nF , nG are normals to the planes dF ]P = 0, dG]P = 0, respectively. Notice in equation (22) that (Fx ]P , Fy ]P , Fz ]P ) · (x x0 , y y0 , z z0 ) = 0 for any (x, y, z) in the plane. This means (Fx ]P , Fy ]P , Fz ]P ) is normal to the plane. We denote this vector by rF ]P and call it the gradient of F at P . In particular, rF ]P · d¯ r = 0, rG]P · d¯ r = 0, that is d¯ r is perpendicular to both normals so we have an alternative equation for the tangent plane as a vector equation: (23) d¯ r ⇥ (rF ]P ⇥ rG]P ) = 0¯ which says d¯ r is parallel to the direction perpendicular to both rF ]P , rG]P . The tangent line through P = (x0 , y0 , z0 ) has direction rF ]P ⇥ rG]P = (Fx , Fy , Fz )]P ⇥ (Gx , Gy , Gz )]P =(
(F, G) (F, G) (F, G) , , ) (y, z) (z, x) (x, y)
and therefore can be written as (24)
x
x0 (F,G) (y,z)
=
y
y0 (F,G) (z,x)
=
z
z0 (F,G) (x,y)
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 14
where the denominators are evaluated at P . 3.6.2. Alternate derivation of the equation of the tangent plane to surface at a point [Ford and Ford]. Suppose a surface is given as z = f (x, y) and P = (x0 , y0 , z0 ) where z0 = f (x0 , y0 )) is on the surface. The tangent plane containing P has the form z z0 = A(x x0 ) + B(y y0 ), for some constants A, B. Take the partial derivative with respect to x (at (x0 , y0 )) on both sides, holding y fixed at y0 , z ](x ,y ) = A x 0 0 Take the partial derivative with respect to y (at (x0 , y0 )) on both sides, holding x fixed at x0 , (25)
z ](x ,y ) = B y 0 0
(26)
So the tangent plane to the surface at P is: (27)
z
z0 = zx ](x0 ,y0 ) (x
x0 ) + zy ](x0 ,y0 ) (y
y0 )
Now if we write the equation in the form F (x, y, z) = 0, we have F (x, y, z) = f (x, y) z = 0, so Fx = zx , Fy = zy , Fz = 1 so this equation can be written (28)
0 = Fx ]P (x
x0 ) + Fy ]P (y
y0 ) + Fz ]P (z
z0 )
as in equation (22) above. Notice here that the curves z = f (x0 , y) and z = f (x, y0 ) intersect at P . The tangent plane contains the tangent lines at P of both these curves. In particular, these two lines define the tangent plane. That is, by taking the normal, n ¯ , as the cross product of the direction vectors of these two lines, and P as a point on ¯ )·n the plane, we can write the equation of the plane as (¯ r OP ¯ = 0 where r¯ is the direction vector to the general point (x, y, z) on the plane. The tangent along z = f (x0 , y) has slope zy ](x0 ,y0 ) = B in the x = x0 plane, so the tangent line is (x, y, z) = (x0 , y0 , z0 ) + t(0, 1, B). The tangent along z = f (x, y0 ) has slope zx ](x0 ,y0 ) = A in the y = y0 plane, so the tangent line is (x, y, z) = (x0 , y0 , z0 ) + t(1, 0, A). Thus n ¯ = (0, 1, B) ⇥ (1, 0, A) = (A, B, 1) and the plane is given by ((x, y, z) (x0 , y0 , z0 )) · (A, B, 1) = 0, as before. 3.6.3. Tangent plane when surface represented F (x, y, z) = 0. If the surface is represented as F (x, y, z) = 0, we can solve using di↵erentials to get the tangent plane at P . dF ]P = Fx ]P dx + Fy ]P dy + Fz ]P dz = 0 dz = Thus, zx ](x0 ,y0 ) =
F x ]P Fz ]P
z
as before.
Fx ] P dx Fz ] P
and zy ](x0 ,y0 ) =
Fy ] P dy Fz ] P F y ]P F z ]P
and the tangent plane is
Fx ] P Fy ] P (x x0 ) (y y0 Fz ] P Fz ] P 0 = Fx ]P (x x0 ) + Fy ]P (y y0 ) + Fz ]P (z
z0 =
z0 )
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 15
3.7. Surface Area. 3.7.1. The area of a rectangle projected onto another plane. Given a plane ax + (a,b,1) by + z = 0 with unit normals n ˆ = ± p1+a , the upper normal is the one with 2 +b2
(a,b,1) the positive z-component: n ˆ = p1+a . 2 +b2 Let A and B be two planes whose upper normals make an angle of ↵. The area of a rectangle on A is cos ↵ times the area of its projection onto B. Proof:Take A to be the xy-plane with rectangle (0, 0, 0), (a, 0, 0), (0, b, 0), (a, b, 0). Take B to be the plane, through the origin, z = zx ](0,0) x + zy ](0,0) y. The upper normals are (0, 0, 1) and ( zx , zy , 1). Note, if zx = zy = 0 the z = 0 which means B is also the xy-plane and sec ↵ = sec(0) = 1 so result holds. So we assume henceforth that they are not both zero. The area of the rectangle on A is ab. Using the equation of B, we see the corners of the rectangle on A project onto the following points of B: (0, 0, 0) to (0, 0, 0); (a, 0, 0) to (a, 0, azx ); (0, b, 0) to (0, b, bzy ); (a, b, 0) to (a, b, azx +bzy ). Thus the rectangle projects to a parallelogram on B and the area of a parallelogram with adjoining sides u ¯ = (a, 0, azx ) and q
v¯ = (0, b, bzy ) is |¯ u ⇥ v¯| = |( abzx , abzy , ab)| = ab 1 + zx2 + zy2 . q ( zx , zy ,1)·(0,0,1) p 2 2 Since cos ↵ = = 1/ 1 + zx2 + zy2 , the area of the rectangle on 1+zx +zy
A is cos ↵ times the area of its projection onto B. If the area of the rectangle on A is a and the area of its projection onto B is b, then we have a = cos ↵ b (29)
b = sec ↵ a
Note 1: If the original plane B did not pass through the origin, it could be moved vertically (i.e. in the direction of the normal to A) without changing the projected area. Note 2: More general shapes can be shown to have the same ratio in areas by reducing them to a sum (limit) involving rectangles. Note 3: If d is the (positive) area of a plane section with unit normals n ˆ U (upper 0 0 ˆ normal) and n ˆ L (lower normal), the angle used in dxdy = cos d is = \(ˆ nU , k), ˆ that is, the angle between the upper normal and k. RR 3.7.2. Surface area S = Rxy sec ↵ dxdy. Given a surface S : z = f (x, y) or z f (x, y) = 0, the tangent plane at a point P = (x1 , y1 , f (x1 , y1 )) on S is z z1 = fx ](x1 ,y1 ) (x x1 ) + fy ](x1 ,y1 ) (y y1 ) or fx (x x1 ) fy (y y1 ) + (z z1 ) = 0 with partial derivatives evaluated at (x1 , y1 ). The upper normal to this surface is ( fx , fy , 1) and the unit upper normal is (30)
q
( fx , fy , 1) n ˆ=q 1 + fx2 + fy2
q ˆ we have cos ↵ = n For ↵ = \(ˆ n, k) ˆ · kˆ = 1/ 1 + fx2 + fy2 , and so sec ↵ =
1 + fx2 + fy2 .
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 16
Let dxdy be a di↵erential area on the xy-plane and project this rectangle onto the tangent plane at P on the surface S. Let d be the projected area, a di↵erential parallelogram on the tangent plane at P . We have dxdy = cos ↵ d . Using these di↵erential areas to approximate (locally) the area of the surface, we get the total surface area by integrating over the domain Rxy on the xy-plane. That is ZZ surface area of S = d (31)
surface area of S =
ZZ
Rxy
sec ↵ dxdy Rxy
3.7.3. Surface area (parametric form). Suppose we are given the surface in parametric form: x = x(u, v), y = y(u, v), z = z(u, v) where (u, v) are defined in a domain Ruv in the u, v-plane. We can divide the domain into a grid with lines parallel to the u, v-axes at intervals of u, v. Then we can consider an area element A = u v as an element of the total area of Ruv . The linear grid on Ruv translates to a curvilinear grid on the surface in x, y, zspace. That is, for a constant value of v = v0 , we get a curve on the surface of S, Cv0 (u) = {x = x(u, v0 ), y = y(u, v0 ), z = z(u, v0 )}, parameterized by u, and similarly for a constant value of u. Let P be a point on the surface and for ease of discussion assume P lies at a corner of the curvilinear grid, the image of (u0 , v0 ). That is P = (x(u0 , v0 ), y(u0 , v0 ), z(u0 , v0 )). Let T¯v0 (u0 ) be the tangent to the curve Cv0 (u) at P . We have T¯v0 (u0 ) = (xu , yu , zu )]u0 ,v0 . Similarly we have the tangent to the curve Cu0 (v) at P is T¯u0 (v0 ) = (xv , yv , zv )]u0 ,v0 . Since all tangents at a point are in the tangent plane, these two tangents can be used to determine the normal to the tangent plane: (32)
n ¯ = T¯v0 (u0 ) ⇥ T¯u0 (v0 )
We now show that we can approximate the area d , the image of the region u v on the surface S as the area of the parallelogram with sides T¯v0 (u0 ) u and T¯u0 (v0 ) v. If we think of T¯v0 (u0 ) as the velocity vector along the curve Cv0 (u) at P , then |T¯v0 (u0 )| is the speed at P which we can write as dsv0 /du where sv0 is the distance measured along the curve in time u. Thus, the distance moved in the duration u is approximated by sv0 = |T¯v0 (u0 )| u = |T¯v0 (u0 ) u| which is the length of the vector T¯v0 (u0 ) u. The same holds in the other direction so we can approximate the sides of the approximating image parallelogram at P with T¯v0 (u0 ) u and T¯u0 (v0 ) v, and the area of the approximating parallelogram is |T¯v0 (u0 ) u ⇥ T¯u0 (v0 ) v| = |T¯v0 (u0 ) ⇥ T¯u0 (v0 )| u v. As the number of subdivisions of Ruv increase, that is, as u, v ! 0, the sum of these approximating parallelograms approximate the surface area of S. We have, Z ZZ (33) surface area of S = d = |T¯v0 (u0 ) ⇥ T¯u0 (v0 )| dudv S
Ruv
where d , the di↵erential element of surface are on S is equal, in the limit, to |T¯v0 (u0 ) ⇥ T¯u0 (v0 )| times the di↵erential element dudv in the region Ruv .
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 17
We have seen that T¯v0 (u0 ) = (xu , yu , zu )]u0 ,v0 and T¯u0 (v0 ) = (xv , yv , zv )]u0 ,v0 , so we can write |T¯v0 (u0 ) ⇥ T¯u0 (v0 )| = |(xu , yu , zu )]u0 ,v0 ⇥ (xv , yv , zv )]u0 ,v0 |
= |(yu zv zu yv , zu xv xu zv , xu yv yu xv )| p = (yu zv zu yv )2 + (zu xv xu zv )2 + (xu yv p = EG F 2
(34) where
yu x v ) 2
E = (xu , yu , zu ) · (xu , yu , zu ) G = (xv , yv , zv ) · (xv , yv , zv )
F = (xu , yu , zu ) · (xv , yv , zv )
p so we have d = EG F 2 dudv. The reader is referred to [2] for details on the conditions under which these arguments are valid but generally speaking we are safe when the functions are well-behaved (continuous with continuous partial derivatives). 4. Two-dimensional Line Integral 4.1. Line Integrals of Functions. This section borrows considerably from the presentation in [3]. Define the line integral with respect to arc length along the curve C as Z n X (35) f (x, y) ds = lim f (xi , yi ) i s i s!0
C
i=1
where s is measuring arc length along the curve C, i s is the i-th subdivision of the total arc length, and f (xi , yi ) is a value taken on at a point (xi , yi ) on the curve within the i-th subdivision. The curve C in parametric form is given by C = (x(t), y(t)) for a t b. We can write this in vector form: r¯(t) = x(t)ˆi + y(t)ˆj. We will assume r¯ is continuous (i.e., the component functions are continuous) and its derivative is non-zero for p a t b. Since ds = (dx/dt)2 + (dy/dt)2 dt we can rewrite (35) as Z Z b Z b q 2 2 (36) f (x, y) ds = f (xt , yt ) xt + yt dt = f (xt , yt )|¯ rt | dt C
a
a
For this i-th subdivision of C we can consider the related i x, i y, parallel to the x, y-axes, joining the endpoints of the subdivision. We define two integrals, the line integrals with respect to x and y respectively: Z Z b n X 0 (37) f (x, y) dx = lim f (xi , yi ) i x = f (xt , yt )x (t) dt C
(38)
Z
i x!0
f (x, y) dy = lim C
i y!0
i=1 n X i=1
a
f (xi , yi )
iy
=
Z
b
0
f (xt , yt )y (t) dt a
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 18
It is often convenient to work with the line integrals with respect to x and y at the same time and we write the sum in a shorthand notation as: Z Z Z (39) P (x, y) dx + Q(x, y) dy = P (x, y) dx + Q(x, y) dy C
C
C
Note: When a line integral is evaluated with respect to arc length, it has the same value whichever direction the curve is traced. However, when it is evaluated with respect to x or y, the value changes sign when the curve is traced in the opposite direction. 4.2. Line Integrals of Vector Fields. We define a two-dimensional vector field as a function F¯ (x, y) whose value is a 2-dimensional vector. That is, F¯ (x, y) = P (x, y)ˆi + Q(x, y)ˆj = (P, Q), where P, Q are scalar (real-valued) functions. For example, the gradient vector is a vector field; rF (x, y) = (Fx (x, y, z), Fy (x, y, z))). If we take a contour F (x, y) = a of F , then take the di↵erential at any point on the contour, dF = Fx dx + Fy dy = 0, we see that the gradient, rF (x, y), is normal to the tangent plane at that point. A vector field F¯ is called conservative if there exists a function f such that ¯ F = rf , in which case f is called a potential of F . For example F¯ = (y, x) is conservative with potential f (x, y) = xy but F¯ = ( y, x) is not conservative. So far we have looked at line integral of (scalar) functions. We can also define line integrals on vector fields. Let F¯ (¯ r(t)) = F¯ (x(t), y(t)) = F¯ (xt , yt ) be 2-D vector field, parameterized by r¯(t). That is, r¯(t) = (xt , yt ) defines a curve C as a t b. The line integral of F¯ along C is defined as: Z Z b 0 (40) F¯ · d¯ r= F¯ (¯ r(t)) · r¯ (t) dt C
a
0 0 As we saw in equation (18) we can write d¯ r/ds = Tˆ. Also, Tˆ = r¯ (t)/|¯ r (t)|. 0 Finally, we have |¯ r (t)| = |d¯ r/dt| = |d¯ r/ds · ds/dt| = |d¯ r/ds||ds/dt| = 1 · ds/dt as 0 long as s is increasing with increasing t. So ds = |¯ r (t)| dt. Thus we can write (40) as Z Z Z b r¯(t) 0 (41) F¯ · d¯ r= F¯ · Tˆ ds = F¯ · 0 |¯ r (t)|dt |¯ r (t)| C C a
Notice that if we calculate this integral in the reverse direction on C, that is, where s is increasing in the opposite direction from t, then the value of the integral changes sign because |ds/dt| = ds/dt. Suppose now that F¯ (x, y) = (P (x, y), Q(x, y)) and r¯ = (x, y) where x = x(t), y = y(t) and a t b. Then we have Z Z b Z b Z b Z 0 0 0 0 (42) F¯ ·d¯ r= (P, Q)·(x dt, y dt) = P x dt+ Qy dt = P dx+Q dy C
a
a
a
C
Notice that since Tˆ is a unit vector in (41), F¯ · Tˆ = FT is the component of F¯ in the direction of the tangent vector. Thus, we also write Z Z Z Z ¯ ¯ ˆ (43) F · d¯ r= F · T ds = FT ds = P dx + Q dy C
C
C
C
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 19
There is a Fundamental Theorem of Line Integrals which corresponds with the Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus states Rb 0 that a F (x) dx = F (b) F (a). It presumes that F has a continuous derivative 0 F on the interval from a to b. The corresponding theorem for line integrals over certain vector fields demand a continuous gradient on the curve. Theorem 1. If C is a continuous curve parameterized by r¯(t) for a t b where 0 r¯ (t) 6= 0 on the interval, and f has a continuous gradient on C, then Z (44) rf · d¯ r = f (¯ r(b)) f (¯ r(a)) C
Proof. The proof is simply an application of the Fundamental Theorem of Calculus applied to the line integral. Z Z b rf · d¯ r= (fx , fy , fz ) · (dx/dt, dy/dt, dz/dt) dt C
=
Z Z
a
b
(fx dx/dt + fy dy/dt + fz dz/dt) dt a b
d[f (¯ r(t)] dt dt = f (¯ r(b)) f (¯ r(a)) =
a
where the final step follows from the Fundamental Theorem of Calculus.
⇤
Notice that this theorem provides an evaluation of the integral that is independent of the path C. That is, any path joining a to b yields the same result. In fact, independence of path implies the existence of the gradient in the theorem. The theorem is useful for evaluating conservative vector fields F¯ , that is, those for which a potential exists. We still need to determine how we know when a vector field is conservative and then how to find the potential function that makes the integration easy by the theorem. A few more definitions concerning paths and regions will be useful. A path (curve) is closed if the initial point and end point coincide. A path is called simple if it does not cross itself. A region is open if it does not contain any of its boundary points. A region is connected if any two points in the region can be connected by a path that lies completely in the region. A region is called simply connected if it is connected and contains no holes. More precisely, simply connected means that the entire interior of any simple closed curve in the region is also in the region (so a hole can be a point). With these concepts in hand we can state the following facts. 1.
R
C
rf · d¯ r is independent of path. [This follows directly from Theorem (1)]
R 2. If F¯ is a conservative vector field, then C F¯ · d¯ r is independent of path. [F¯ ¯ conservative means F = rf , some f , then use 1]
R 3. If F¯ is a continuous vector field on an open connected region D and C F¯ · d¯ r is independent of path for any path in D, then F¯ is a conservative vector field on D.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 20
R R Proof: [From [2]] Let (x0 , y0 ) be any point in D where C F¯ · d¯ r = P dx + Q dy R (x,y) is independent of path (IOP) in D. Define f (x, y) = (x0 ,y0 ) P dx + Q dy where the integral is taken on an arbitrary path in D joining (x0 , y0 ) and (x, y). Since this defining integral is IOP, it depends only on (x, y) so it does define a function f . We will show that /nablaf = (fx , fy ) = (P, Q) in D. For a particular (x, y) in D, choose x1 6= x, such that the segment joining (x1 , y) and (x, y) is in D. (We know we can do this because D is open - so there is a ball around (x, y) that is in D. Also, since D is connected there is a path in D R (x ,y) from (x0 , y0 ) to (x1 , y).) Again, using IOP, f (x, y) = (x01,y0 ) (P dx + Q dy) + R (x,y) (P dx + Q dy) because it is just another path joining the endpoints. Here we (x1 ,y) think of x1 and y as fixed while (x, y) varies along the segment. That is, on this segment, y has been restricted to a constant value and f (x, y) is being considered as a function of x near a particular choice of x. The first of the two integrals on the right is independent of x, while the second can Rbe integrated along the line x segment. Hence, for this fixed y, f (x, y) = constant + x1 P (x, y) dx, (y = constant so dy = 0), so by the Fundamental Theorem of Calculus, we have fx = P (x, y). Similarly, fy = Q(x, y). QED R R 4. If C F¯ · d¯ r is independent of path then C F¯ · d¯ r = 0 for every closed path C. [Closed path made up of two segments with same endpoints, one traced in reverse so sum is zero] R R 5. If C F¯ · d¯ r = 0 for every closed path C then C F¯ · d¯ r is independent of path. [If C1 , C2 are two non-intersecting paths joining endpoints then closed path made R R of C1 and -C2 has zero sum, so C1 = C2 . If they intersect in a finite number of points, just apply above argument a number of times. If they intersect and infinite number of times, need limiting argument where paths are approximated by segments.] R 6. If P, Q have continuous partials in D and C F¯ ·d¯ r is independent of path then Py = Qx . [IOP means there is F with Fx = P, Fy = Q, so Py = Fxy = Fyx = Qx .] For our case of a 2-D vector field, the following theorem (almost the converse of 6 but with the added condition of simple connectedness) gives us an easy means of testing the vector field for being conservative. Theorem 2. Let F¯ = (P, Q) be a vector field on an open and simply connected (no holes) region D. If P and Q have continuous partial derivatives in D, and Py = Qx , then F¯ is conservative. Proof as in [2]. Let C be any piecewise smooth simple closed curve in D with an interior,R. Since D is simply connected, have continuous partials R R is in D, so P, QRR in R, and Green”s Theorem applies. C P dx + Q dy = R (Qx Py ) dxdy = 0. Since this is true for any such C, we have, by 5 above, that the line integral is independent of path, which by 3 tells us that the vector field is conservative. ⇤ These results can be somewhat confusing in the abstract so it is useful to tie them to something practical. Let F¯ be the gravitational force of the earth with mass M on an object m at height h1 above the surface of the earth. Then F¯ = (GM m/r2 )kˆ
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 21
where kˆ is the unit vector in the direction joining the center of the earth to the center of the object and r is the length of that vector. In particular, F is a onedimensional vector field, or just a function of r. Let f = GM m/r be a scalar function. We have df /dr = F , so f is a potential of F . (Actually, any f + constant is such a potential so we can assign the constant for convenience of the problem. For example, we could set the potential to be zero at the surface of the earth or at the center of the earth or the top of the lab table and then study potentials relative to this assignment.) For this discussion we will take constant = 0. We call the scalar potential, f , the potential energy, f = GM m/r. Taking g = GM/r2 , we have, F = gm and f = gmr. We have shown that the gravitational force is conservative (f exists with F = rf = df /dr) so we know that the line integral of F is independent of path. R This means that the work done in moving an object from point A to point B, C F¯ · d¯ r, does not depend on the path taken from A to B. Also, we know that if we move an object around but end up at the starting point, the net work done is zero. Another very nice application of this theory is given in [4] where electrostatic fields are considered. In general, if we know F¯ = (P, Q) is conservative, how do we find the potential function f such that F¯ = rf R ? We know that rf R = (fx , fy ) = (P, Q) so fx = P and fy = Q, so f (x, y) = P dx or f (x, y) = Q dy where these are a type of iterated integral where one variable is treated as a constant. 3 3 ¯ Example from [3]: F¯ = (2x3 y 4 +x, 2x4 y 3 +y). Py = 8xR3 y 3 , Qx = R 8x 3y 4, so F conservative. Now to find the potential function. f (x, y) = p dx = (2x y +x) dx = 0 (1/2)x4 y 4 +(1/2)x2 +g(y). So Q = fy = 2x4 y 3 +g (y). But we know Q = 2x4 y 3 +y, 0 so g (y) = y or g(y) = y 2 /2+k. That is, f (x, y) = (1/2)x4 y 4 +(1/2)x2 +(1/2)y 2 +k. R The line integral C P dx+Q dy was seen to be interpreted as the vector integral R F¯ · Tˆ ds where Tˆ is the unit tangent vector. That is C Z
C
F¯ · d¯ r=
(45)
=
Z
P dx + Q dy =
ZC C
Z
(P dx/ds + Q dy/ds) ds Z Z ¯ ˆ (P, Q) · (dx/ds, dy/ds) ds = F · T ds = FT ds C
C
C
¯ . That is, if F¯ = We can also interpret it in terms of the normal vector N ¯ (P, Q), consider the vector field G = (Q, P ). The unit tangent Tˆ = d¯ r/ds = ˆ = (dy/ds, dx/ds) 90 degrees behind Tˆ. That (dx/ds, dy, ds) has unit normal N ˆ where kˆ is the unit vector perpendicular to the x, y-plane. We have, ˆ = Tˆ ⇥ k, is, N Z
P dx + Q dy = C
(46)
=
Z
(P dx/ds + Qdy/ds) ds
ZC C
(Q, P ) · (dy/ds, dx/ds) ds =
Z
Finally, notice that Z Z Z (47) FN ds = (P, Q) · (dy/ds, dx/ds) ds = C
C
C
¯·N ˆ ds = G
Z
GN ds C
Q dx + P dy C
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 22
Now we are ready to establish one of the three main theorems about line integrals.
4.3. Green’s Theorem in the plane relating line integral on a closed curve with ordinary double integral over enclosed region. Theorem 3. Let D be a domain in the xy-plane and let C be a piecewise smooth simple closed curve in D whose interior is also in D. Let P (x, y), Q(x, y) be functions defined and continuous and having continuous first partial derivatives in D. Then Z ZZ (48) P dx + Q dy = (Qx Py ) dxdy C
R
where R is the closed region bounded by C.
Proof. From [2]. First we assume that R can be represented in both the following ways: (49) (50)
a x b, c y d,
f1 (x) y f2 (x)
g1 (y) x g2 (y)
We will prove the theorem for this special region, then extend it to regions which can be broken into subregions of this form. The most general regions, can be treated as limits of these regions, but will not be handled here. The double integral RR RbRf P dxdy, can be evaluated as the iterated integral a f12 Py dydx which gives R y Rb Ra Rb usR a [P (x, f2 (x)) P (x, f1 (x))] dxRR= P (x, f2R(x)) dx P (x, f1 (x)) dx = b a P (x, y) dx. In the same way, Q dxdy = Q(x, y) dy. Adding these we C R x C have the equation in the theorem for this type of region. Now suppose the region of interest, R, can be decomposed into subregions R1 , R2 , . . . , Rn of the type above and let C1 , C2 , . . . , Cn be the corresponding boundaries of these regions. We can use theR result already established for each R R of these regions and add them together to get C1 (P dx + Q dy) + C2 () + · · · + Cn = RR RR (Qx Py ) dxdy + · · · Rn () dxdy. The new parts of the boundaries on the R1 left hand side, established in dividing R into subregions, are each counted twice, in reverse directions, in the total sum. Thus, these parts of the total line integral on the left cancel and we are left with justR the line integral over the original boundary C of R. That is the left hand side is C P dx + QRRdy. The right hand side is just the sum of all the areas of the subregions so it is R (Qx Py ) dxdy. ⇤ Note: We assume here that we have a positive orientation for the closed curve C. Usually we think of this as the counterclockwise direction however we generalize this to the notion of positive orientation being the direction which puts the interior of the curve on the left. For example, if we have a region made by the area between two concentric circles of radii r = 1 and r = 2 , then the positive direction on the outer circumference is ccw but that of the inner is clockwise - so that the interior of the region is always on the left as we trace the boundary. In particular, we can cut over from the outside to the inside, circle about the inside, then cross back on the same path but in the opposite direction to continue tracing R R the remainder of the boundary. Since the line integral of C P dx + Q dy = P dx + Q dy, the C
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 23
crossover paths cancel each other out. 2 2 RR Example 1: Let C be the RR ellipse: x +4y = 4. Then (1 ( 1)) dxdy = 2 R dxdy = 2⇡(2)(1) = 4⇡. R
R
C
(2x y) dx+(x+3y) dy =
R y x Example 2: Let C be the ellipse: 4x2 +y 2 = 4. Consider C x2 +y 2 dx x2 +y 2 dy. Since P, Q are not continuous at (0, 0), Green’s Theorem does not apply. We can also consider the vector interpretation of Green’s Theorem. First for vector field F (P, Q) we define: 1. The gradient of F¯ , grad F¯ = rF¯ = (Px , Qy ), another vector field. 2. The divergence of F¯ , div F¯ = Px + Qy , a scalar. ˆ 3. The curl of of F¯ , curl F¯ = (Qx Py )k. If F¯ = (P, Q) then Qx Py is the z-component of curl F¯ . That is curlz F¯ = Qx Py and we have the vector form of the theorem Z
(51)
FT ds = C
ZZ
curlz F¯ dxdy R
R R We also saw above that C P dx + Q dy = C GN ds where G = (Q, P ) and ˆ = Tˆ ⇥ kˆ is the outward normal. Now (Qx Py ) = div G ¯ and we have N Z
(52)
GN ds = C
ZZ
¯ dxdy div G R
or, from equation (47).
Z
FN ds = C
Z
Q dx + P dy = C
ZZ
(Px + Qy ) dxdy = R
ZZ
div F¯ dxdy R
Green’s Theorem can be extended to multiply-connected domains (domains with holes). Let C1 be the boundary of a domain with holes with boundaries C2 , ..., Cn . Then if we take the line integral about C1 counterclockwise and the line integral about the hole boundaries in a clockwise direction (thus R R R always RRin the positive orientation as discussed above), then C1 + C2 + · · · + Cn = R (Qx Py ) dxdy. (Again, the boundaries introduced by extending C1 to the holes and back cancel each other and we get a decomposition similar to the 2nd case of the proof above.) R Example: Evaluate C y 3 dx x3 dy where C is comprised of the two circle of radius r = 2 and r = 1 centered at the origin with positive orientation.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 24
The region of interest, R is the ring between the two circles. Z ZZ y 3 dx x3 dy = 3x2 3y 2 dxdy C R ZZ = 3 (x2 + y 2 ) dxdy = =
3 3
Z Z
R 2⇡
0 2⇡
Z
2
r2 r dr d✓ 1
(16/4
1/4) d✓ =
45/4(2⇡) = ( 45⇡/2)
0
A useful application of Green’s Theorem is to find areas of plane figures. That RR is, the area of a region is given by R dA = A. The left hand side looks like the term in Green’s Theorem when Qx Py = 1. There are many functions P and Q that satisfy this such as (P, Q) = (0, x), ( y, 0), ( y/2, x/2), etc. So, by Green’s Theorem the area can be computed as Z Z Z (53) A= x dy = y dx = (1/2) x dy y dx C
C
C
where C is the boundary of the region R.
Example: Find the area of the disk of radius a. In this case, C is given by (x(t), y(t)) = a cos t, a sin t) for 0 t 2⇡. UsR R 2⇡ ing A = (1/2) C x dy y dx = (1/2) 0 a cos t(a cos t) dt a sin t( a sin t) dt = R 2⇡ (1/2) 0 a2 dt = (a2 /2)(2⇡) = ⇡a2 .
We can also compute this area using any other (P, Q) with the defining characteristic R such as:R 2⇡ R 2⇡ R 2⇡ A = C x dy = 0 a cos t(a cos t) dt = a2 0 cos2 t dt = a2 /2 0 (1 + cos(t/2)) dt =a2 /2(2⇡ + 0) = ⇡a2 . 5. Three-dimensional Line Integral We need to extend the definitions, concepts of the last section to space. Again, we have included material in this section from [3] but in general this is all very standard material found in many texts. The presentation cited is very nicely done and recommended to the reader. We define a three-dimensional vector field as a function F¯ (x, y, z) whose value is a 3-dimensional vector. That is, F¯ (x, y, z) = P (x, y, z)ˆi+Q(x, y, z)ˆj +R(x, y, z)kˆ = (P, Q, R), where P, Q, R are scalar (real-valued) functions. First for vector field F (P, Q, R) we define treating r = ( / x, / y, / z) : 1. The gradient of F¯ , grad F¯ = rF¯ = (Px , Qy , Rz ), another vector field. 2. The divergence of F¯ , div F¯ = r · F¯ = Px + Qy + Rz , a scalar. 3. The curl of of F¯ , curl F¯ = r ⇥ F¯ = ((Ry Qz ), (Pz Rx ), (Qx Py )). Regarding the curl, it is easy to see that if f has continuous 2nd order partial derivatives, we have rf = (fx , fy , fz ), so curl rf = (fzy fyz , fxz fzx , fyx fxy ) = (0, 0, 0) = ¯ 0. Thus, if F¯ is conservative, meaning there exists an f such that ¯ F = rf , we must have curl F¯ = curl rf = ¯0. The converse is harder to prove but we
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 25
state it as a fact: If F¯ is defined on a subset of R3 on which the component functions have continuous first order partials, then if curl F¯ = ¯0 then F¯ is conservative. Physical interpretation of curl: Let F¯ be the velocity field of a flowing fluid. Then curl F¯ represents the tendency of particles at (x, y, z) to rotate about the axis pointing in the direction of curl F¯ . if curl F¯ = ¯0 then F¯ is called irrotational. Regarding divergence, we notice that div (curl F¯ ) = Ryx Qzx + Pzy Rxy + Qxz Pyz = 0 when the 2nd order partials of F¯ are continuous. Physical interpretation of divergence: In the scenario discussed for the interpretation of the curl, div F¯ represents the rate of change of the mass of the fluid flowing from the point (x, y, z) per unit volume. That is, the tendency of the fluid to diverge from a point. If div F¯ = 0 at a point then F¯ is called incompressible at that point. Define the line integral with respect to arc length along the curve C as Z n X (54) f (x, y, z) ds = lim f (xi , yi , zi ) i s i s!0
C
i=1
where s is measuring arc length along the curve C, i s is the i-th subdivision of the total arc length, and f (xi , yi , zi ) is a value taken on at a point (xi , yi , zi ) on the curve within the i-th subdivision. The curve C in parametric form is given by C = (x(t), y(t), z(t)) for a t b. ˆ We will assume We can write this in vector form: r¯(t) = x(t)ˆi + y(t)ˆj + z(t)k. r¯ is continuous (i.e., the component functions are continuous) and its derivative p is non-zero for a t b. Since ds = (dx/dt)2 + (dy/dt)2 + (dz/dt)2 dt we can rewrite (54) as Z Z b Z b q (55) f (x, y, z) ds = f (xt , yt , zt ) x2t + yt2 + zt2 dt = f (xt , yt , zt )|¯ rt | dt C
a
a
For this i-th subdivision of C we can consider the related i x, i y, i z, parallel to the x, y, z-axes, joining the endpoints of the subdivision. We define three integrals, the line integrals with respect to x,y and z respectively: Z Z b n X 0 (56) f (x, y, z) dx = lim f (xi , yi , zi ) i x = f (xt , yt , zt )x (t) dt i x!0
C
(57) (58)
Z
f (x, y, z) dy = lim
i y!0
C
Z
f (x, y, z) dz = lim
i z!0
C
i=1 n X
i=1 n X
a
f (xi , yi , zi ) f (xi , yi , zi )
i=1
iy
iz
= =
Z
b
0
f (xt , yt , zt )y (t) dt a
Z
b
0
f (xt , yt , zt )z (t) dt a
It is often convenient to work with the line integrals with respect to x,y and z at the same time and we write the sum in a shorthand notation as: Z P (x, y, z) dx + Q(x, y, z) + R(x, y, z) dy C Z Z Z (59) = P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz C
C
C
Note: When a line integral is evaluated with respect to arc length, it has the same value whichever direction the curve is traced. However, when it is evaluated
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 26
with respect to x or y or z, the value changes sign when the curve is traced in the opposite direction. As before, we can write the integral in vector form: Z Z P dx + Q dy + R dz = (P dx/ds + Q dy/ds + R dz/ds) ds C Z Z ˆ (60) = (P, Q, R) · T ds = FT ds Z X = F¯ · d¯ r = lim F¯ (xi , yi , zi ) i¯i
However, the concept of the normal is more involved in three dimensions so we don’t yet have a corresponding form for equation (47). The Fundamental Theorem for Line Integrals carries over to three variable without change as do the five facts mentioned there about independence of path. The test for being conservative with three variables is a bit more involved than that for two variables (need to show curl F¯ = ¯0 as discussed above) but actually finding the potential function follows the same basic idea. Note in particular that Green’s Theorem applies to two dimensional situations only but, as we will see in the next section, it has generalizations on surfaces. Example 1: (from [3]): Find a potential function f for F¯ = (P, Q, R) = (2xy 3 z 4 , 3x2 y 2 z 4 , 4x2 y 3 z 3 ). First, we see that curl F¯ = (12x2 y 2 z 3 12x2 y 2 z 3 , 8xy 3 z 3 8xy 3 z 3 , 6xy 2 z 4 6xy 2 z 4 ) = (0, 0, 0), so F¯ is conservative. Must have fx = P = 2xy 3 z 4 ; fy = Q = 3x2 y 2 z 4 , fz = R = 4x2 y 3 z 3 . Start by integrating the first equation to get f (x, y, z) = x2 y 3 z 4 + g(y, z). Di↵erentiating this with respect to y gives us 3x2 y 2 z 4 + gy (y, z) which we know is equal to Q, so gy (y, z) = 0 which means g(y, z) is constant with respect to y, say, g(y, z) = h(z). Now we have f = x2 y 3 z 4 + h(z). 0 Di↵erentiating this with respect to z gives us 4x2 y 3 z 3 + h (z) and we know this is 0 equal to R, so h (z) = 0 and h(z) = c, a constant. Thus f (x, y, z) = x2 y 3 z 4 + c is a potential function for F¯ for any constant c. Example 2: R (from [3]): Evaluate C F¯ · d¯ r where F¯ = (2x3 y 4 + x, 2x4 y 3 + y) and C is the curve r¯(t) = (t cos(⇡t) 1, sin(⇡t/2)) for 0 t 1. We note that Qx = 8x3 y 3 = Py . Thus a potential function exists which we can R 4 4 2 2 ¯ · d¯ find, as above, is f (x, y) = (1/2)x y + (1/2)x + (1/2)y + c. Thus F r = C R rf · d¯ r , which by the Fundamental Theorem is equal to f (¯ r (1)) f (¯ r (0)) = C f ( 2, 1) f ( 1, 0) = 10. We have been looking at integrals of functions or vector fields over curves in two or three dimensions. Next we will look at such integrals over surfaces.
6. Surface Integrals 6.1. Parameterized Surfaces. Note: We have discussed the topic of surface integrals of functions previously under geometric applications.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 27
We parameterized a curve in two or three dimensions by letting a variable t take values in an interval [a, b] and assigning points r¯ = (x(t), y(t)) or r¯ = (x(t), y(t), z(t)) to trace the curve. We parameterize a surface in three dimensions by letting point (u, v) take values in a two-dimensional region, Ruv , and assigning points r¯ = (x(u, v), y(u, v), z(u, v)) to trace the surface. The set of equations:x = x(u, v); y = y(u, v); z = z(u, v) are called the parametric equations of the surface. We can also describe a surface as a function z = f (x, y) for (x, y) in a region Rxy or implicitly by the equation F (x, y, z) = 0. Notice that z = f (x, y) is just an explicit form of F (x, y, z) = z f (x, y) = 0. Examples from [3]. Example 1: Find parametric equations for the elliptic paraboloid: x = 5y 2 + 2z 10. Here we have the surface defined as x = f (y, z). We can take y(u, v) = u; z(u, v) = v; x(u, v) = 5u2 + 2v 2 10. So r¯(u, v) = (5u2 + 2v 2 10, u, v). Note, this surface is defined for all (u, v) on the uv-plane. For a portion of the surface we would restrict the domain to a portion of the plane. 2
Example 2: Find parametric equations for the sphere x2 + y 2 + z 2 = r2 . In spherical coordinates, the variables x, y, z are given by x = r sin cos ✓; y = r sin sin ✓; z = r cos , where is the angle from the z-axis to the position vector (x, y, z) and ✓ is the angle in the xy-plane from the x-axis to the projection of the position vector onto the xy-plane. Thus, parametric equations for the sphere are: x( , ✓) = r sin cos ✓; y( , ✓) = r sin sin ✓; z( , ✓) = r cos where 0 ⇡ and 0 ✓ 2⇡. Example 3: Find the tangent plane to the surface r¯(u, v) = (u, 2v 2 , u2 + v) at (2, 2, 3). First we want to know what point (u, v) is mapped to (2, 2, 3). Solving for U and v we get u = 2 from the first coordinates, 2v 2 = 2 or v = ±1 from the second coordinates, and u2 + v = 4 ± 1 = 3 from the third equation. From this we conclude that u = 2, v = 1. Now we determine the tangent plane: In section (3.7.3), equation (32) we saw that the normal to the tangent plane is given by n ¯ = T¯u0 (v0 )⇥T¯v0 (u0 ) = (xu , yu , zu )⇥ (xv , yv , zv ) where the partials are evaluated at (u0 , v0 ) = (2, 1). That is, (1, 0, 2u)⇥ (0, 4v, 1) at (2, 1). So n ¯ = (16, 1, 4) and the tangent plane is 16(x 2) (y 2) 4(z 3) = 0 or 16x y 4z = 18. Example 4: Find the surface area of the portion of the sphere of radios 4 that lies inside the cylinder x2 + y 2 = 12 and above the xy-plane. In example 2 we wrote the parametric equations for the sphere: x( , ✓) = r sin cos ✓; y( , ✓) = r sin sin ✓; z( , ✓) = r cos where 0 ⇡ and 0 ✓ 2⇡. In this case we p still have 0 ✓ 2⇡ but we can see that the cylinder intersects the sphere at (0, 12, 2) so that is restricted to the interval 0 ⇡/3. In section (3.7.3), equation (33) we saw that the surface area for a surface given RR in parametric form is R ✓ |¯ r ⇥ r¯✓ | d d✓.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 28
r¯ = (4 cos cos ✓, 4 cos sin ✓, 4 sin ) and r¯✓ = ( 4 sin sin ✓, 4 sin cos ✓, 0), R 2⇡ R ⇡/3 R 2⇡ so |¯ r ⇥ r¯✓ | = 16 sin . Then 0 0 16 sin d d✓ = 16 0 [cos(0) cos(⇡/3)] d✓ = 8(2⇡) = 16⇡. 6.2. Orientable Surfaces. Recall that a line integral has an orientation. We defined a positive orientation on a closed curve to be the direction that leaves the interior on the left. This allowed us to generalize from the notion of a simple closed curve with positive orientation being a counterclockwise direction around the interior to more complicated curves with holes and still keep a consistent meaning of positive orientation. We also want to define a positive orientation on surfaces. However, it is not so easy. Some types of surfaces just don’t fit what we otherwise think of as a standard notion of orientation. For this reason we separate surfaces into those that are and those that are not orientable under a useful but not all-inclusive definition of orientability. For a smooth (continuous partials) surface S in space, if we can choose a unit normal vector, n ¯ (x, y, z), at each point of S so that n ¯ varies continuously on S then we say S is orientable. Once oriented (and the choice of orientation is not necessarily unique) we can choose a positive orientation/direction for angles at each point on the surface (say, counterclockwise about n ˆ ). Closed surfaces are those that form the boundary of some solid region in space. For example a sphere is closed. For closed surfaces the convention is that the positive orientation is obtained by taking the normals pointing away from the enclosed region and the negative orientation is obtained by taking the normals pointing toward the enclosed region. More generally, with a positive direction defined for angles, we can assign a positive orientation/direction to simple closed curves forming the boundary of S. Then, at each point P on the closed boundary curve C of the surface S, we can ¯ in the choose the tangent T¯ in the positive direction chosen and inner normal N ˆ ˆ tangent plane at P . Then T , N , n ˆ form a positive triple at P of C. If S is piecewise smooth (the union of smooth pieces) then one cannot choose a continuously varying normal vector for the entire surface (there are discontinuities where the pieces come together, for example the top and bottom of a cylinder). In this case we say that S is orientable if each smooth piece can be oriented in such a way that for each common boundary curve between two pieces, the positive direction along that curve is the opposite for the two pieces sharing the common boundary curve (i.e., one of the surfaces induces the opposite direction as positive than the other). Note: The Moebius Strip is a nonorientable surface. If you define n ˆ at any one point and move continuously around the surface back to the starting point, the direction of n ˆ will have switched to n ˆ. The following are examples of surfaces with continuously varying normal, n ˆ. 1. z = f (x, y), choose either the upper normal (z-coordinate q positive) or the lower normal (z-component negative) in n ˆ = ±( zx , zy , 1)/ 1 + zx2 + zy2 . 2. x = x(u, v), y = y(u, v), z = z(u, v), choose plus or minus for n ˆ = ±(P1 ⇥ P2 )/|P1 ⇥ P2 |, where P1 = (xu , yu , zu ), P2 = (xv , yv , zv ), as long as EG F 2 6= 0.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 29
3. F (x, y, z) = 0, choose plus or minus for n ˆ = ±rF/|rF , provided rF 6= 0.
6.3. Surface Integrals of Vector Fields. We have seen surface integrals of functions in computing surface areas above. Now we will look at surface integrals for vector fields where the points come from a surface in space and define them in terms of the standard surface integral of a function with which we are already familiar. Recall then that a standard surface integral of a function f (x, y, z) is given by ZZ n X (61) f d = lim f (xi , yi , zi ) i n!1
S
i=1
For z = g(x, y), or f (x, y, z) = z g(x, y), we have seen, ZZ ZZ (62) f d = f (x, y, g(x, y)) sec dxdy S
S
ˆ We get similar results when where is the angle between the upper normal and k. ˆ we start with x = g(y, z), with respect i or y = g(x, z) with respect to ˆj. In parametric form, we have seen ZZ ZZ p (63) f d = f (x(u, v), y(u, v), z(u, v))) EG F 2 dudv S
S
In the case ofR line integral, R we defined R the line integral for a vector field in terms of the tangent C F¯ · d¯ r = FT ds = C P dx + Q dy. For the surface integral, it is the normal that takes the central role. Let F¯ = ˆ = (L, M, N ) be a vector field where L, M, N are functions of x, y, z and let N (cos ↵, cos , cos ) be the continuous unit normal to the surface. Define the following surface integrals on functions L, M, N : ZZ ZZ (64) L dydz = L cos ↵ d ZZ S Z ZS (65) M dxdz = M cos d S S ZZ ZZ (66) N dxdy = N cos d S
S
which follow directly from the results above as we have just moved sec to the other side as cos. We add these to get (67) ZZ ZZ ZZ ZZ ˆ ¯ ˆ L dydz+M dxdz+N dxdy = (L, M, N )· N d = F ·N d = FN d S
S
s
6.3.1. Definition of surface integral of a vector field. Let F¯ be a vector field with ˆ , then the surface integral of F¯ over the surface S is continuous unit normal N ZZ ZZ ZZ ˆ d = (68) F¯ · d¯ = F¯ · N FN d S
S
S
ˆd . where the double integral on the right is a standard surface integral and d¯ = N ¯ This integral is sometimes called the flux of F across S.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 30
Physical application: If v¯ is the velocity field of a fluid, then the surface integral represents the volume of fluid flowing through S per unit time, again, the flux of v¯ across S. 6.3.2. Evaluation. The evaluation of the surface integral reduces to getting an expression for the continuous normal. As we have seen, the choice of orientation is key to the evaluation. Example 1: Let F¯ = (L, M, N ), the surface be z = g(x, y), and the orientation upwards (z > 0). ZZ
S
F¯ · d¯ =
(69)
=
where we note that
ZZ ZZ
( zx , zy , 1) q 2 (L, M, N ) · q zx + zy2 + 1 dA 2 2 Rxy zx + zy + 1 Lzx
M zy + N dxdy
Rxy
q zx2 + zy2 + 1 = |rf | where f (x, y, z) = z
( gx , gy , 1) = ( zx , zy , 1) .
g(x, y); rf =
Example 2: Let F¯ = (L, M, N ), the surface be r¯(u, v), and the orientation upwards (z > 0). ZZ
S
F¯ · d¯ = =
ZZ
ZZ
r¯u ⇥ r¯v |¯ ru ⇥ r¯v |dA |¯ ru ⇥ r¯v |
Ruv
(L, M, N ) ·
Rxy
F¯ · r¯u ⇥ r¯v dudv
where we have chosen the positive orientation. RR Example 3: Evaluate F¯ · d¯ where F¯ = (0, y, z) and the surface S is the combination of the paraboloid y = x2 + z 2 for 0 y 1 and the disk x2 + z 2 1 at y = 1 (the cap on the paraboloid) and has a positive orientation. Here we have a closed surface so the positive orientation is outward. We need to treat the surface as the union of two smooth surfaces, the paraboloid S1 and the cap, S2 . A sketch shows that outward on S1 means the normals will have a negative y-component, and on the cap the normals will have a positive y-component, in fact, they will be parallel to ˆj. The surface integral will be the sum of the surface integrals on S1 and S2 . ZZ ZZ ˆ d = F¯ · N (0, y, z) · (2x, 1, 2z) dxdz S1
Rxz
where we have taken the normal in the direction of y < 0. Then ZZ ZZ 2 ( y 2z ) dxdz = ( (x2 + z 2 ) 2z 2 ) dxdz Rxz Rxz ZZ (70) = x2 + 3z 2 dxdz Rxz
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 31
Now we evaluate this integral using polar coordinates x = r cos ✓; z = r sin ✓; dxdz = r drd✓, to get ZZ Z 2⇡ Z 1 2 2 x + 3z dxdz = (r2 cos2 ✓ + 3r2 sin2 ✓)r drd✓ Rxz
=
Z
0
0
2⇡
(cos2 ✓ + 3 sin2 ✓)[
0
= ( 1/4) = ( 1/4) =
⇡
Z Z
2⇡
Z
1
r3 dr] d✓ 0
[(1 + cos(2✓)/2 + (3/2)(1
cos(2✓)] d✓
0 2⇡
[2
cos(2✓)] d✓ = (1/4)[2(2⇡)]
0
Now we must compute the surface integral on S2 . The unit outer normal is now (0, 1, 0) on the disk so we have ZZ ZZ F¯ · ¯ = F¯ · (0, 1, 0) d S2 S2 ZZ ZZ = y dxdz = dxdz Rxz
(71)
Rxz
=⇡
Thus the surface integral over the entire closed surface S is
⇡ + ⇡ = 0.
6.4. Stokes’ Theorem - Higher dimensional version of Green’s Theorem. Green’s Theorem related a line integral around an oriented boundary curve to a double integral over the enclosed region. In the case of a surface, the orientation of the surface induces the positive orientation of the boundary curve. (If you are walking along the curve and the surface is to your left when your head is pointing in the direction of the continuous unit normals then you are walking in the positive direction on C. Theorem 4. If F¯ is a vector field defined on a smooth surface S with a simple, closed, smooth boundary curve C with the orientation induced by S, then Z ZZ ¯ (72) F · d¯ r= curl F¯ · d¯ C
S
Examples from [3]. RR Example 1: Evaluate S curl F¯ · d¯ where F¯ = (z 2 , 3xy, x3 y 3 ) and S is the part of z = 5 x2 y 2 above the plane z = 1. Assume S is positively oriented. As a closed surface, positively oriented means normals directed to the exterior of the surface. We take the boundary curve to be the circle of intersection of plane z = 1 and surface z = 5 x2 y 2 , that is, 1 = 5 x2 y 2 or x2 + y 2 = 4. This boundary curve is parameterized as r¯(t) = (2 cos t, 2 sin t, 1) for 0 t 2⇡. Thus 0 r¯ (t) = ( 2 sin t, 2 cos t, 0). RR R R 2⇡ curl F¯ · d = C F¯ · d¯ r = 0 (1, 12 cos t sin t, 64 cos3 t sin3 t)·( 2 sin t, 2 cos t, 0) dt = S R 2⇡ R 2⇡ R 2⇡ [ 2 sin t 24 cos2 t sin t] dt = 2 0 sin t + 24 0 cos2 t( sin t) dt = 0. 0
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 32
R Example 2: Evaluate C F¯ · d¯ r where F¯ = (z 2 , y 2 , x) and C is the triangle with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1) with a counterclockwise orientation. RR We know, by Stokes’ Theorem that this integral is equal to S curl F¯ · d¯ . The ˆ curl of F¯ is (0, 2z 1, 0) and the unit normal can p be computed as N = ±( 1, 1, 0) ⇥ ( 1, 0, 1)/|( 1, 1, 0) ⇥ ( 1, 0, 1)| = ±(1, 1, 1)/ 3. The counterclockwise orientation ˆ . So, with the surface S being the plane is obtained by taking the + value for N through the three points, that is, x + y + z = 1, we have z = 1 x y and: ZZ ZZ ˆ d curl F¯ · d¯ = curl F¯ · N S
= = = (73)
=
Z Z Z
0
0
0
Rxy 1Z 1 x 1Z 1Z
(2z
1) dydx
0
1 x
(2(1
x
y)
1) dydx
0 1 x
(1
2x
2y) dydx
0
1/6
6.5. Divergence Theorem or Gauss’ Theorem - Another Higher Dimensional version of Green’s Theorem. Recall from the section on Green’s Theorem, we had the expressions: Z Z ZZ ¯ F · d¯ r= FT ds = curlz F¯ dxdy C C R Z ZZ FN ds = div F¯ dxdy C
R
ˆ = Tˆ ⇥ kˆ is the outer normal. where N The first of these looks like a 2-dimensional version of Stokes’ Theorem, that is, where the (plane) surface is parallel to the xy-plane. The Divergence Theorem, also reminiscent of Green’s Theorem, relates the surface integral to a triple integral over a solid region enclosed by the surface.
Theorem 5. If R is a simple solid region, S the boundary surface of R with a positive orientation, and F¯ a vector field with continuous first order partial derivatives, then ZZ ZZ ZZZ (74) F¯ · d¯ = FN d = div F¯ dV S
Example 1. Evaluate consists of three surfaces:
RR
S
R
F¯ · d¯ where F¯ = (xy, y 2 /2, z) and the surface S z=4
3x2
3y 2 , 1 z 4
x2 + y 2 = 1, 0 z 1 z=0
on the top, sides and bottom, respectively. Again, to evaluate the surface integral directly, we would have to consider each of the three surfaces (with outer normals) separately and then add these parts together to get the desired value. Instead, we compute the volume of the solid
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 33
enclosed by these surfaces. For this we simply need to express the height at any point over the base, the circle of radius one centered at the origin on the xy-plane. We will need div F¯ = y y + 1 = 1.Thus we have, using cylindrical coordinates where dV = rdrd✓dz: ZZ ZZZ F¯ · d¯ = 1 rdrd✓dz S
= (75)
Z
R
2⇡
0
= 5⇡/2
Z
1
r[ 0
Z
4 3r 2
dz] drd✓ 0
7. Review and Discussion 7.1. Three theorems and proofs. The three theorems (Green, Stokes, Divergence) we have studied all equate two well-defined integrals which might be evaluated on their own terms. The power of the theorems is that they present a choice where in many cases one way is much less complicated than the other. We will see this in some examples below. Green’s Theorem: If D is a domain in the xy-plane, C a piecewise smooth, simple, closed, positive oriented curve in D, with interior R in D and P, Q defined, continuous, with continuous partial derivatives in D, then Z ZZ (76) P dx + Q dy = (Qx Py ) dxdy C
Rxy
where Rxy is the closed region bounded by C. In vector form, with F¯ = (P, Q), Tˆ ˆ the unit outer normal, the unit tangent and N Z ZZ (77) FT ds = curlz F¯ dA C R Z ZZ (78) FN ds = div F¯ dA C
R
Stokes’ Theorem: If S is a piecewise smooth oriented surface in space with a piecewise smooth simple closed boundary curve C directed in accordance with the given orientation in S and F¯ = (L, M, N ) is a vector field with continuous, di↵erentiable components in domain D of a space including S then Z ZZ ZZ ˆ d (79) FT ds = curl F¯ · d¯ = curl F¯ · N C
or (80) Z
L dx + M dy + N dz =
C
S
ZZ
(Ny
S
Mz ) dydz + (Lz
Nx ) dzdx + (Mx
Ly ) dxdy
S
Divergence Theorem (Gauss): If F¯ = (L, M, N ) is a vector field in domain D of space, L, M, N are continuous with continuous derivatives in D and S a piecewise smooth surface in D which forms the complete boundary of a bounded, closed region R in D, then ZZ ZZZ (81) FN d = div F¯ dV S
R
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 34
or (82)
ZZ
L dydz + M dzdx + N dxdy = S
ZZZ
Lx + My + Nz dxdydz R
The proofs of these theorems have a familiar aspect in that they establish the result for an easy to handle case and then extend the result to the more general case. In the case of Green’s Theorem, we first assume the region R can be represented both as a lex b, f1 (x) y f2 (x) and c ley d, g1 (y) RR x g2 (y). This R allows us toRRiterate the double integrals. That is, we show that P dxdy = P dx R y C R and R Qx dxdy = C Q dy so their di↵erence gives us the desired result. We extend this type of region to regions made up of finite combinations of such regions and then make a limiting argument for general regions. In the case of Stokes’ Theorem, we first assume S can be represented as z = f (x, y), (x, y) 2 Rxy . We define Cxy to be the projection of the boundary curve C onto the xy-plane, thus forming the boundary of Rxy and getting its orientation from C. If upper normal chosen on S,R then direction on Cxy is positive. R R Then we have C L(x, y, z) dx = Cxy L(x, y, f (x, y)) dx = Cxy L1 (x, y) dx + RR RR 0 dy which by Green’s Theorem is equal to (L1 )y dxdy = (Ly + Rxy Rxy Lz fy ) dxdy. RR RR Also, we have S Lz dzdx L dxdy = 0 dzdy + Lz dzdx + ( RRLy ) dxdy. y S RR From equation (69), this is Rxy 0( zx ) + Lz ( zy ) + ( Ly ) dxdy = (Ly + Rxy Lz fy ) dxdy. R RR From the last two paragraphs we have CRRL(x, y, z) dx = S Lz dzdx Ly dxdy. R Similarly we show that M (x, y, z) dy = S Mx dxdy Mz dydz and C R RR N (x, y, z) dx = N dydz Nx dzdx. Then we add these together to get the y C S desired result. We extend this type of region to regions made up of finite combinations of such regions and then making a limiting argument for general regions. NOTE: can extend to arbitrary orientable surfaceS whose boundary is formed of distinct simple closed curves C1 , ..., Cn In the case of the Divergence Theorem, we first assume that R is representable in the form f1 (x, y) z f2 (x, y), (x, y) 2 Rxy where Rxy is a bounded, closed region in the xy-plane bounded by the simple, closed curve C. We show that RR RRR N dxdy = N z dxdydz and the other two cases follow similarly. S Rxy The surface is composed of 3 parts: The bottom, S1 : z = f1 (x, y), (x, y) 2 Rxy ; the top, S2 : z = f2 (x, y), (x, y) 2 Rxy ; and the sides, S3 : f1 (x, y) z f2 (x, y), (x, y) 2 C. Notice that S3 may degenerate to simply the curve C, for example the equator of a sphere. ˆ The crux of RR the proof is handling the direction of the normal with respect to k. RR 0 ˆ We know that S N dxdy = S N cos d , where = \(ˆ n, k). Define to be ˆ the angle between the upper normal and k. On S3 , = ⇡/2, so cos = 0. On S2 , 0 the top, = since the outer normal is the upper normal. On S1 , the bottom, 0 =⇡ since the outer normal is the lower normal. 0 Now d = sec dxdy on S1 , S2 because for areas we always consider upper 0 normal (sec > 0).
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 35
Thus we have ZZ ZZ N dxdy = N cos d S Z ZS ZZ = N cos d + N cos d S1 S2 ZZ ZZ 0 0 = N cos(⇡ ) sec dxdy + (83)
=
ZZ
Rxy
[N (x, y, f2 (x, y))
0
sec
0
dxdy
Rxy
N (x, y, f1 (x, y))] dxdy
Rxy
Also we have ZZZ ZZ Nz dzdxdy = Rxy
(84)
N cos
=
ZZ
[ Rxy
Z
z=f2 (x,y)
Nz dz] dxdy z=f1 (x,y)
[N (x, y, f2 (x, y))
N (x, y, f1 (x, y))] dxdy
Rxy
RR RRR RR Hence S N dxdy = Nz dxdydz. Similarly we show that S M dzdx = RRR RR Rxy RRR My dxdydz and S L dydz = L dxdydz. Then we add these toRxy Rxy x gether to get the desired result. We extend this type of region to regions made up of finite combinations of such regions and then making a limiting argument for general regions. Thus all three proofs follow a similar pattern, establishing the theorem in a simplified case and then extending to more complicated cases - a useful lesson in itself! We see then that the generalization of Green’s Theorem to space has two forms, corresponding to the two forms of Green’s Theorem in vector form: For vector fields u ¯ = (P, Q), v¯ = (Q, P ), Green’s Theorem gives us Z ZZ Z ZZ (85) uT ds = curlz u ¯ dxdy = vN ds = div v¯ dxdy C
Rxy
C
Rxy
which is generalized to Stokes’ Theorem involving the curl of a 3-dimensional vector field and the Divergence Theorem involving the divergence of a 3-dimensional vector field.
7.2. A special vector field, F⇤ . One of the complicating factors in reading mathematics is the number of conditions in the statement of a result. In basic geometry we had simple statements like ”the area of a circle is ⇡r2 . But when we look at a result like the Divergence Theorem above we see lots of qualifiers, all with very specific meaning, which set limitations to the scope of the result. That such qualifications are necessary is due to the fact that the world of mathematical objects is so varied and unless we define exactly what we are proving we can be led down false paths by assuming things hold in general that only hold under specific conditions. The vector field considered in this section shows us the importance of qualifications placed on some of the results we have studied.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 36
7.2.1. A family of velocity vector fields. We will consider the velocity vector field V0 (x, y) = ( y, x). This velocity (dx/dt, dy/dt) at the point P = (x, y) is perpendicular to the direction of the line segment from O = (0, 0) to P and in fact is directed 90 degrees counterclockwise of this segment. Since x0 = y and y 0 = x, we have x = y 0 = x00 and y 00 = x0 = y so a solution for the trajectory is x(t) p = r cos t, y(t) = r sin t. That is, V0 induces trajectories of ccw-circles of radius r = x2 + y 2 on each point (x, y) 2 R2 . We note that curl(V0 ) = 2kˆ and div(V0 ) = 0. Define Va (x, y) = ( y/ra , x/ra ) where a is a constant. Since Va (x, y) = V0 (x, y)/ra , we are just rescaling the vector field V0 so that each point of the vector field Va also points 90 degrees ccw from the direction (x, y) but now with a norm r/ra = r1 a . That is, the trajectories are the same but the period is scaled. 7.2.2. The F⇤ vector field. The vector field V0 is the special case of Va where a = 0. For all a 6= 0, the vector field Va di↵ers from V0 by a scaling factor 1/ra . In particular, all the trajectories are still ccw-circles. In general, with P = y/ra = y/(x2 + y 2 )a/2 and Q = x/ra = x/(x2 + y 2 )a/2 , we have (x(x2 + y 2 ) a/2 ) x = (x2 + y 2 ) a/2 · 1 + x( a/2)(x2 + y 2 )
Qx =
= Py =
(x2 y
1 + y 2 )a/2
( y(x2 + y 2 )
= (x2 + y 2 )
a/2
ax2 + y 2 )(a+2)/2
a/2
· 2x
)
· ( 1) + ( y)( a/2)(x2 + y 2 )
1 ay 2 + 2 2 a/2 +y ) (x + y 2 )(a+2)/2 2 a(x2 + y 2 ) Py = 2 2 a/2 2 (x + y ) (x + y 2 )(a+2)/2 2 a = 2 (x + y 2 )a/2 =
Qx
(x2
(a+2)/2
(a+2)/2
· 2y
(x2
that is, (86)
Qx
Py = (2
a)/ra
Thus, the scalar part of the curl is positive for a < 2 and negative for a > 2. At a = 2, the curl is zero, that is, we have an irrotational velocity field which we will call F⇤ . y x , ) (87) V2 (x, y) = F⇤ (x, y) = ( 2 x + y 2 x2 + y 2 7.2.3. F⇤ is not (globally) conservative. We have seen that every gradient vector field F = rf is irrotational because curl(F ) = curl(rf ) = ¯0 whenever the second order partials are continuous. However, not every irrotational field is a gradient field as we will see below. Consider the vector field F⇤ (x, y). We know from section (7.2.2) that curl(F⇤ ) = 0 (since Qx = Py ).
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 37
The circle x2 + y 2 = R2 of radius R about the origin is a simple, closed R curve C. This curve is parameterized by r¯(t) = (r cos t, r sin t), 0 t 2⇡, so C F⇤ · d¯ r= R 2⇡ R 2⇡ 0 F⇤ (x(t), y(t)) · r¯ (t) dt = 0 dt = 2⇡. That is, for any circle about the origin, 0 the line integral of the vector function F⇤ is 2⇡. Thus, F⇤ is not a conservative field, and so, F⇤ is not a gradient field. Let’s see what happens when we try to find a potential for F⇤ in p the standard way. That is, we try to find a vector field q(x, y) such that q = y/ x 2 + y 2 , qy = x p 2 2 x/ x + y .R a. Using dt/(a2 + t2 ) = (1/a) arctan(t/a), we have qy = Q implies q(x, y) = arctan(y/x) + p(x). b. Using a.), we must have qx = p(x) =constant.
y/(x2 + y 2 ) + p0 (x) = P , so p0 (x) = 0 or
So, is q(x, y) = arctan(y/x) + k a potential for F⇤ on C? There is a problem. q(0, y) is not defined! But the circle about the origin crosses the y-axis so we do not have a continuous gradient on the curve. Recall in theorem (1) we needed a continuous gradient on the curve to get that result. On the other hand, if we took a closed path that didn’t cross the y-axis then the gradient would be continuous on C and the line integral would be zero. Our procedure for finding q such that F⇤ = r(q) seemed to work but it turns out to only work ”locally”. That is, near any point we can find a potential function, but in order for the vector field to be conservative, we need a single function q, defined and continuous on the entire region of definition of the vector field, and we saw that for F⇤ , we could not find a potential defined for all nonzero (x, y). Note: In applied terms, when we integrate the vector field around a ccw-circle there must be positive work done because the vector field is in the direction of movement, that is, the force and the velocity are parallel at every point so there will be positive work done, a gain in energy. 7.2.4. The angle function. Of course, q(x, y) = arctan(y/x) is a very familiar function, it is just ✓(x, y) = angular coordinate of (x, y). But on a closed path that winds around the origin, ✓ is not a continuous function because it is not a welldefined function, that is, it is not single-valued. We can make it single-valued by restricting the range of ✓ to, say, 0 ✓ < 2⇡. In such case, for any (x, y) 2 R2 , we have a single-value, ✓(x, y) assigned to this element. In so defining ✓ : R2 ! [0, 2⇡), we have a discontinuity at 2⇡. That is, if we start at ✓ = 0, then after one turn, as we approach ✓ = 2⇡, we see that as we cross the x-axis, ✓ jumps from a values increasingly near to 2⇡ back to values near zero. That is, by making the restriction which makes ✓ well-defined, we introduce a discontinuity along the x axis. In di↵erential form, the line integral takes the form: d✓ = ✓x dx + ✓y dy = R R 2⇡ P dx+Q dy, so C P dx+Q dy = 0 d✓ = 2⇡, where qx = P = y/(x2 +y 2 ), qy = Q = x/(x2 + y 2 ). In fact, line integrals of F⇤ are almost conservative in the sense that the line integral of F⇤ along any closed curve is 2n⇡ where n is the winding number, the number of times the curve winds around the origin. That is, let C to be any simple closed curve circling ccw about the origin, and let Ce be a very small circle about the origin lying inside C. From above, we know
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 38
R that Ce F⇤ · d¯ r = 2⇡. Consider the region R between C and Ce . The boundary, R, of R is given by C Ce (that is, a combination of ccw-C and cw-Ce ). So Green’s Theorem says Z Z Z F⇤ · d¯ r F⇤ · d¯ r= F⇤ · d¯ r C Ce Z C Ce = F⇤ · d¯ r R ZZ = curl(F⇤ ) dA = 0 R
since curl(F R ⇤ ) = 0. Thus C F⇤ · d¯ r = 2⇡ for this more general C. Similarly, the integral of any simple closed curve circling clockwise about the origin will have the line integral 2⇡. If you can show that any closed curve can be broken up into finitely many simple closed curves which either wind ccw around the origin once or cw around the origin once, or do not circle around the origin at all (in which case the line integral is zero), it follows that the integral of F⇤ along any closed curve is 2⇡n for some integer n (positive or negative) called the winding number of C about the origin. 7.2.5. Extensions to complex variables. The discussions in this section borrow freely and appreciatively from [5].
Let R be a region in the plane. How many independent irrotational nonconservative vector fields can we define on R? Call the number N if we can find irrotational vector fieldsP F1 , ..., Fn such that, 1) For any constants c1 , ..., cn (not all zero), the vector field ci Fi is not a gradient field; and 2) For any P irrotational vector field G defined on R, there are constants k1 , ...kn such that G ki Fi = rf , for some f , is a gradient field. That is, any irrotational vector field on R depends on the Fi ’s up to a gradient vector field. Fact: N , the number of independent nonconservative irrotational vector fields is always equal to the number of holes of the region. Therefore, if we consider the plane minus the set of N points Pi = (xi , yi ), i = 1, ..., N as R, then there is a translated version of our F⇤ at each point: F⇤,Pi = [ (y yi ), x xi ]/[(x xi )2 + (y yi )2 ]. Since we have just translated everything to the point Pi , the line integral of F⇤,Pi around a small ccw-circle about Pi (and none of the other Pi ’s) is 2⇡. No linear combination of the F⇤,Pi ’s is conservative (since none are themselves and the gradient operator is linear), but given any irrotational vector field G with singularities at these points, by keeping track of the values of the line integrals Ii around the little circles about the Pi ’s, then Ii /2⇡ give the coefficients ki above. That is, by subtracting this linear combination of the F⇤,Pi ’s from G we get a gradient field. [Same result holds if the points are replaced with simple closed curve boundary elements of R. The idea that the number of irrotational vector fields up to gradient fields on a region R counts the number of holes in R is due to the early 20th century mathematician Gustave de Rahm and is called Rahm Theory or Rahm Cohomology. It is true for vector fields on surfaces and higher dimensions.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 39
Green’s Theorem in Complex Variable: For a complex line integral or contour integral of a function of a complex variable along a curve in the complex plane there is a complex analogue of the Fundamental Theorem of Calculus known as the Cauchy Integral Theorem. Theorem 6. Let R be a closed region of the plane with smooth boundary, R (any number of simple closed curves). Then R if f (z) is any complex function which is defined and holomorphic on all of R, R f (z) dz = 0i.e., IOP in D. .
Proof. Let f (z) be a function of a complex variable, z = x + iy. So f (z) = P (z)+iQ(z). In fact, if we identify R2 with the complex plane via (x, y) () x+iy, such functions correspond to vector fields F (x, y) = (P (x, y), Q(x, y)). In this sense, a complex line integral is however a pair of line integrals of vector fields. That is, letting C be an oriented curve in the plane, we can write its parameterization as z(t) = x(t) + iy(t). Then we define the complex integral: Z Z tmax (88) f (z) dz = (P (z) + iQ(z))(x0 (t) + iy 0 (t)) dt C
tmin
which can be written as two ordinary integrals Z tmax Z tmax (89) [P (z(t))x0 (t) Q(z(t))y 0 (t)] dt + i [P (z(t))y 0 (t) tmin
Q(z(t))y 0 (t)] dt
tmin
and we can still identify each integral as being the dot product of a certain vector field with the velocity vector v¯(t) = r¯0 (t) to get Z Z (90) (P, Q) · d¯ r + i (Q, P ) · d¯ r C
C
where the vector fields are F1 = (P, Q), F2 = (Q, P ). Suppose C is the boundary of a region R in the plane such that f (z) (and hence P (z), Q(z) as components) is defined on all of R, including its boundary. Then Green’s Theorem applies: curl(F1 ) = Qx Py and curl(F2 ) = Px Qy so we get Z ZZ ZZ (91) f (z) dz = (Qx + Py ) dA + (Px Qy ) dA C
R
R
NOTE: Neither integrand is the curl of the original F = (P, Q) so knowing F is irrotational or incompressible does not help us evaluate this integral. But, the Cauchy-Riemann equations are: (92)
Px = Q y , Q x =
Py
and a complex function f (z) = P (x+iy)+Q(x+iy) is holomorphic (or analytic) on region R (that is, the complex derivative limh!0 [f (z + h) f (z)]/h exists in a small enough neighborhood of each point of R) if and only if f (z) satisfies equations (92) in R. Thus, for holomorphic (analytic) functions, both integrands are identically zero, so the integrals are. ⇤ Thus, we treat the Cauchy-Riemann equations, (92), as the complex analogue of the vector equation curl F¯ = 0. That is, it is precisely the holomorphic (i.e. analytic) functions which play the analogous role with the theory of complex line integrals that the irrotational vector fields play in the usual theory.
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 40
R If R is a simple closed curve, then we are just getting RR f (z) dz = 0. So if f (z) is holomorphic on the complex plane (i.e. entire) then C f (z) dz = 0 for all such curves. If R has multiple boundary components C1 , ..., R Cn , then we R are just getting a relation between the integrals over these curves: C1 f + · · · + Cn f = 0. Example: Let f (z) = 1/z. This function has one singularity, at z = 0. If we integrate around any simple closed curve not enclosing zero, then the Cauchy Theorem guarantees the integral is zero. (Recall discussion about F⇤ .) But the R R 2⇡ integral around the unit circle z(t) = cos t + i sin t is C f (z) dz = 0 ( sin t + R 2⇡ i cos t)/(cos t+i sin t) dt = i 0 dt = 2⇡i. Now, as above, Green’s Theorem tells us that this is all we need to compute. That is, if C is any curve which winds n-times R ccw about the origin, then C dz/z = 2⇡in. Finally, as the analogue of the result we derived above, the function f (z) = 1/z is essentially the only holomorphic function on the complement of the origin for which line integrals around closed curves need not be zero, in the sense that every function g(z) holomorphic except at the origin can be written g(z) = R/z + h0 (z) [here the analogue of ”conservative vector fields are gradient fields” is the simpler ”conservative holomorphic functions are derivatives of other holomorphic functions”. In fact, even the reason we cannot integrate 1/z is the same as the reason we could not integrate our special field F⇤ - its antiderivative is defined locally, but involves the angular coordinate ✓ so cannot be continuously defined for an entire path about the origin.] Moreover, the number R has R the property that for any closed curve C winding n-times around the origin, C g(z) dz = 2⇡inR. This number R is called the residue of g(z) at the origin, and one learns techniques (Laurent expansion) for calculating it explicitly in the study of complex variables. We can generalize again to functions, g(z), having a finite number of singularities at z1 , ...zn in the complex plane: integrating g around a small ccw-circle centered at zi gives a number of the form 2⇡iRi where Ri is the residue of g at zi and get the Residue Theorem: Theorem 7. Let g(z) be a function holomorphic on the region R except for finitely many singularities at z1 , ..., zn . Let C be a curve in the region R, not passing through any of the singularities, which has winding numbers ni with respect to the zi . Let Ri be the residues of g(z) at zi . Then Z n X (93) g(z) dz = 2⇡iRj C
j=1
This theorem turns out to be a useful tool for evaluating (real-valued!) integrals and series and it is interesting to know that it is the complex version of the weak path-independence property of irrotational vector fields. See [2] for examples.
References [1] Lester R. Ford, Sr., Lester R. Ford Jr., Calculus, McGraw-Hill Book Company, Inc., 1963 Library Of Congress Catalog Card Number: 62-12480 [2] Wilfred Kaplan, Advanced Calculus, Addison-Wesley Publishing Company, Inc. 1952 Library Of Congress Catalog Card Number: 52-7667 [3] Paul’s Online Math Notes http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx
THE THEOREMS OF GREEN, GAUSS, AND STOKES: A STUDY GUIDE FROM FIRST PRINCIPLES 41
[4] K.E. Oughstun Electrostatic Fields: Coulomb’s Law & the Electric Field Intensity, 2012 http://www.cems.uvm.edu/~keoughst/LectureNotes141/Topic_01_(electrostatic_fund).pdf [5] Handout Eight: Green’s Theorem http://math.uga.edu/~pete/handouteight.pdf William G. Atwell, Silver Spring, MD
