Calculus III:

Divergence theorem

Green’s Theorem relates line integral to double integral over a region delimited by an oriented curve.

Stokes' theorem relates a line integral to surface integral over a region delimited by an oriented curve.

Divergence theorem relates surface integral to triple integral over a region delimited by an oriented surface.

Divergence theorem

E is a simple solide reigion . S is the boundary surface of E with positive orientation. is a vector field with continuous first order partial derivatives components:

Example

Use the divergence theorem to evaluate ∫∫_S . , where: = xy + xz (3 - y)z , and the surface S is a wrapped dome, that has three parts:
. On the top: paraboloid z = 10 - 2 x² - 2 y² , 2 ≤ z ≤ 10
. On the side: Cylinder x² + y² = 4 , 0 ≤ z ≤ 2
. On the bottom: disk x² + y² = 4 , z = 0



∫∫_S . = ∫∫∫_E div dV . Let's work on ∫∫∫_E div dV.

The region E for the triple integral is then the solid enclosed by these three surfaces.

The cylindrical coordinates system is the appropriate system for this region:

x = r cos θ
y = r sin θ
z = 10 - 2 r²cos² θ - 2 r² sin² θ = 10 - 2 r²

The limits for the ranges are:

0 ≤ r ≤ 2
0 ≤ θ ≤ 2π
0 ≤ z ≤ 10 - 2 r²

We have:

dV = r dr dθ dz.

= 〈xy , xz , (3 - y)z 〉. Hence div = y + 0 + 3 - y = 3

div = 3. Therefore

∫∫∫_E div dV =
∫₀^2Ï€ ∫₀² ∫₀^{10 - 2 r2} 3 dz r dr dθ =
3 ∫₀^2Ï€ ∫₀² (10 - 2 r²) r dr dθ =
6 ∫₀^2Ï€ ((5/2) r² - (1/3)r³)|₀² dθ =
6 ∫₀^2Ï€ ((10 - 8/3) dθ =
44 ∫₀^2Ï€ dθ = 88 Ï€

∫∫_S . = ∫∫∫_E div dV = 88 π