Gauss’ Divergence Theorem. The Gauss Divergence theorem relates flux integrals and triple integrals. Recall that when we evaluate a double integral over a surface we compute the integral over the *skin* of the object enclosed by the surface (the boundary). In triple integrals we integrate inside of the object: Theorem 1 Gauss’ divergence theorem: Let T be a closed bounded region in the space whose boundary ∂T is a piecewise smooth orientable surface. Let F~ be a continuous vector field whose partial derivatives are continuous in some domain containing T . Then: ZZZ
div F~ dV =
T
ZZ
F~ · ~n dA
∂T
Example 1 Verify the divergence theorem when F~ (x, y, z) = (7x, 0, −z) and T is the region enclosed by the sphere of radius 2. 1
In this case: T = {(x, y, z) : x2 + y 2 + z 2 ≤ 4} and ∂T = {(x, y, z) : x2 + y 2 + z 2 = 4} We first compute the triple integral over T using a change of variable to spherical coordinates: (x, y, z) = (r cos θ cos φ, r sin θ cos φ, r sin φ), r ∈ [0, 2], θ ∈ [0, 2π], φ ∈ [ and the Jacobian of change ∂x ∂r J = ∂x ∂θ ∂x ∂φ
−π π , ] 2 2
of variables is: ∂y ∂z ∂r ∂r ∂y ∂z = ∂(x, y, z) = r2 cos φ. ∂θ ∂θ ∂(r, θ, φ) ∂y ∂z ∂φ ∂φ
Then: RRR T
div F~ dV
= = = = =
RRR ∂(7x) ∂0 ∂(−z) + + dx dy dz T ∂x ∂y ∂z RRR R 2π R π2 R 2 2 r cos φdr dφ dθ 6 dxdydz = 6 0 −π 0 T 2 h i 3 r=2 R 2π R π2 6 0 −π cos φ r3 dφdθ 2 r=0 R 2π R π2 16 0 −π cos φdφ dθ 2 π R 2π 2 16 0 [sin φ] −π dθ = 64π (4) 2
On the other hand, a parametric representation for ∂T , the sphere of radius 2, could be: ~r(u, v) = (2 cos u cos v, 2 sin u cos v, 2 sin v), u ∈ [0, 2π], v ∈ [ Then: ~ru = (−2 sin u cos v, 2 cos u cos v, 0) ~rv = (−2 cos u sin v, −2 sin u sin v, 2 cos v) 2
−π π , ] 2 2
therefore ~ru × ~rv = (4 cos u cos2 v, 4 sin u cos2 v, 4 cos v sin v) and F~ (~r(u, v)) = F~ (2 cos u cos v, 2 sin u cos v, 2 sin v) = (14 cos u cos v, 0, −2 sin v) RR
~ du dv = F~ (~r) · N R
R 2π R π2
=
R 2π R π2
0
0
4 cos u cos2 v
2 −π (14 cos u cos v, 0, −2 sin v) 4 sin u cos v 2 4 cos v sin v −π 2
dv du
(56 cos3 v cos2 u − 8 cos v sin2 v)dvdu
Notice that: R 2π R π2 0
−π 2
= = = =
R 2π R π2
cos2 u cos v(1 − sin2 v)dvdu R 2π R R 2π R π2 56 0 cos2 u cos vdvdu − 56 0 −π cos2 u cos v sin2 vdvdu 2 π π 3 R 2π R 2π 2 2 56 0 cos2 u[sin v] −π du − 56 0 cos2 u[ sin3 v ] −π du 2 2 R 2π 56(2 − 23 ) 0 cos2 udu R 2π 2u du 56(2 − 23 ) 0 1 + cos 2
56 cos3 v cos2 udvdu = 56
−π 2 π 2 −π 2
0
= 56π(2 − 32 ) because
R 2π 0
(♣)
cos 2u = 12 [sin 2u]2π 0 = 0.
And:
R 2π sin3 v π2 [ 3 ] −π du 0 0 2 32 = − 3 π (F) Adding (♣) and (F) we obtain the same result that we got doing the triple −8
R 2π R π2
−π 2
cos v sin2 vdvdu = −8
integral of the divergence of F~ over T (4). Example 2 Use the Divergence Theorem to calculate
RR ∂T
F~ · ~n dA, when
∂T = {(x, y, z) : x2 + y 2 = z 2 , 0 ≤ z ≤ 2} and F~ (x, y, z) = (4x, 3z, 5y) 3
By the Divergence Theorem: RR ∂T
F~ · ~ndA =
RRR T
divF~ dV,
where T = {(x, y, z) : x2 + y 2 ≤ z 2 , 0 ≤ z ≤ 2} RRR T
divF~ dV
= 4 = 4
RRR
1dxdydz T R 2 R z R √z2 −x2 0
−z
√ − z 2 −x2
1dydxdz,
Using polar coordinates, x = r cos θ and y = r sin θ and r ∈ [0, z] and θ ∈ [0, 2π]: 4
R 2 R z R √z2 −x2 0
0
√ − z 2 −x2
1dydxdz = 4
R 2 R z R 2π
= 8π
0
0
0
R2Rz 0
−z
rdθdrdz
rdrdz
z 2 dz = 32π 3 RR Example 3 For F~ (x, y, z) = (x3 , x2 y, x2 z), evaluate ∂T F~ · ~n dA when: = 4π
R2 0
∂T = {x2 + y 2 = a2 , 0 ≤ z ≤ b} ∪ {x2 + y 2 ≤ a2 , z = 0} ∪ {x2 + y 2 ≤ a2 , z = b} 4
in two different ways. By Gauss divergence theorem: ZZ
F~ · ~n dA =
ZZZ
divF~ dV
T
∂T
where T = {x2 + y 2 ≤ a2 , 0 ≤ z ≤ b} and divF~ = 5x2 . Let us compute the triple integral first noticing that x and y are in the disc of radius a, x2 + y 2 ≤ a2 and then we can write: ZZZ T
divF~ dV =
Z b Z Z
2
5x dxdy dz x2 +y 2 ≤a2
0
We can use polar coordinates in the double integral as follows: x = r cos θ, y = r sin θ, r ∈ [0, a], θ ∈ [0, 2π]
5
Recall that the Jacobian of change of variable in this case is r. R b RR 0
R b R 2π R a 3 2 2 5x dx dy dz = 5 r cos θdr dθ dz x2 +y 2 ≤a2 0 0 0 h i r=a R b R 2π r4 2 = 5 0 cos θ 0 4 r=0 dθ dz 4 R b R 2π 5a 1 + cos 2θ = 4 0 0 dθ dz 2 4 4 Rb 5πa 5bπa4 θ=2π [sin 2θ] dz = = 5bπa + θ=0 4 8 0 4
()
On the other hand: RR ∂T
F~ · ~n dA =
RR {x2 +y 2 =a2 ,0≤z≤b}
+
RR {x2 +y 2 ≤a2 ,z=b}
F~ · ~n dA +
RR
F~ · ~n dA
We compute these three integrals independently. To compute: ZZ
F~ · ~n dA
{x2 +y 2 =a2 ,0≤z≤b}
6
{x2 +y 2 ≤a2 ,z=0}
F~ · ~n dA
we need a parametric representation of the circular cylinder of radius a, for example: ~r(u, v) = (a cos u, a sin u, v), u ∈ [0, 2π], v ∈ [0, b] Then ~ru = (−a sin u, a cos u, 0) and ~rv = (0, 0, 1) obtaining: ~ = ~ru × ~rv = (a cos u, a sin u, 0) N Furthermore F~ (x, y, z) = (x3 , x2 y, x2 z) and therefore: F~ (~r(u, v)) = (a3 cos3 u, a3 cos2 u sin u, a2 v cos2 u) Then:
a cos u ~ = (a3 cos3 u, a3 cos2 u sin u, a2 v cos2 u) F~ (~r(u, v))·N a sin u 0
= a4 cos4 u+a4 cos2 u sin2 u
But a4 cos4 u + a4 cos2 u sin2 u = a4 cos2 u(cos2 u + sin2 u) = a4 cos2 u Hence: RR {x2 +y 2 =a2 ,0≤z≤b}
because 2
R 2π 0
F~ · ~n dA =
R b R 2π
a2 cos2 u du dv R b R 2π = a2 0 0 (1 + cos 2u) du dv = a2 bπ, 0 2
0
cos 2udu = [sin 2u]2π 0 = 0.
To compute: ZZ
F~ · ~n dA
{x2 +y 2 ≤a2 ,≤z=0}
we need a parametric representation of the {x2 + y 2 ≤ a2 , z = 0}, for example: ~r(u, v) = (u cos v, u sin v, 0), u ∈ [0, a], v ∈ [0, 2π]. 7
Then: ~ru = (cos v, sin v, 0) and ~rv = (−u sin v, u cos v, 0) obtaining: ~ = ~ru × ~rv = (0, 0, u) N and F~ (~r(u, v)) = (u3 cos3 v, u3 cos2 v sin v, 0) Therefore:
0
3 3 3 2 ~ ~ F (~r(u.v)) · N = (u cos v, u cos v sin v, 0) 0 = 0 u Hence: ZZ
F~ · ~n dA = 0
{x2 +y 2 ≤a2 ,≤z=0}
Finally, to compute: ZZ
F~ · ~n dA
{x2 +y 2 ≤a2 ,≤z=b}
we need a parametric representation of the {x2 + y 2 ≤ a2 , z = 0}, for example: ~r(u, v) = (u cos v, u sin v, b), u ∈ [0, a], v ∈ [0, 2π]. Then: ~ru = (cos v, sin v, 0) and ~rv = (−u sin v, u cos v, 0) obtaining: ~ = ~ru × ~rv = (0, 0, u) N F~ (~r(u, v)) = (u3 cos3 v, u3 cos2 v sin v, bu2 cos2 v)
8
and
0
~ = (u3 cos3 v, u3 cos2 v sin v, bu2 cos2 v) F~ (~r(u.v)) · N 0 = bu3 cos2 v u Therefore: RR {x2 +y 2 ≤a2 ,≤z=b}
R 2π R a
F~ · ~n dA =
0
0
4 bu3 cos2 vdu dv = ba4 π
(Check this last integral, we have done similar ones before!) Hence: RR ∂T
F~ · ~n dA =
RR {x2 +y 2 =a2 ,0≤z≤b}
+
RR {x2 +y 2 ≤a2 ,z=b} 4
F~ · ~n dA + F~ · ~n dA
4 = ba4 π + 0 + ba4 π = 5ba4 π ,
as we expected from ()
9
RR {x2 +y 2 ≤a2 ,z=0}
F~ · ~n dA