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