Divergence theorem examples

Example 1

Compute ∫∫SF ⋅ dS where

F = (3x + z77,y2 - sin x2z,xz + yex5 )
and S is surface of box
0 ≤ x ≤ 1, 0 ≤ y ≤ 3, 0 ≤ z ≤ 2.
Use outward normal n.

Solution: Given the ugly nature of the vector field, it would be hard to compute this integral directly. However, the divergence of F is nice:

div F = 3 + 2y + x.
We use the divergence theorem to convert the surface integral into a triple integral
∫ ∫SF ⋅ dS = ∫ ∫∫B div FdV
where B is the box
0 ≤ x ≤ 1, 0 ≤ y ≤ 3, 0 ≤ z ≤ 2.
We compute the triple integral of div F = 3 + 2y + x over the box B:
∫∫SF ⋅ dS = ∫ 01 ∫ 03 ∫ 02(3 + 2y + x)dzdydx
= ∫ 01 ∫ 03(6 + 4y + 2x)dydx
= ∫ 01(18 + 18 + 6x)dx
= 36 + 3 = 39.

Example 2

For F = (xy2,yz2,x2z), use the divergence theorem to evaluate

∫ ∫SF ⋅ dS
where S is the sphere of radius 3 centered at origin. Orient the surface with the outward pointing normal vector.

Solution: Since I am given a surface integral (over a closed surface) and told to use the divergence theorem, I must convert the surface integral into a triple integral over the region inside the surface.

Since div F = y2 + z2 + x2, the surface integral is equal to the triple integral

∫ ∫∫B(y2 + z2 + x2)dV
where B is ball of radius 3.

To evaluate the triple integral, we can change variables to spherical coordinates. In spherical coordinates, the ball is

0 ≤ ρ ≤ 3, 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π.
The integral is simply x2 + y2 + z2 = ρ2. For spherical coordinates, we know that the Jacobian determinant is dV = ρ2 sin ϕdϕdθdρ. Therefore, the integral is
∫ 03 ∫ 02π ∫ 0πρ4 sin ϕdϕdθdρ = ∫ 03 ∫ 02π[ ] - ρ4 cosϕ ϕ=π ϕ=0 dθdρ
= ∫ 03 ∫ 02π2ρ4dθdρ
= ∫ 034πρ4dρ = |3 4π-ρ5|| 5 | 0 = 972π- 5.