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Week 10 Discussion Notes

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Divergence Theorem

Recall Stokes' Theorem:

∬S(∇×F⃗)⋅dS⃗=∮∂SF⃗⋅dr⃗\iint_{\mathscr{S}} \p{\nabla \times \vec{F}} \cdot \diff\vec{S} = \oint_{\partial\mathscr{S}} \vec{F} \cdot \diff\vec{r}

This relates an integral on the surface to an integral on the boundary. The divergence theorem has a similar form:

∭W∇⋅F⃗ dV=∯∂WF⃗⋅dS⃗\iiint_{\mathscr{W}} \nabla \cdot \vec{F} \,\diff{V} = \oiint_{\partial\mathscr{W}} \vec{F} \cdot \diff\vec{S}

It relates an integral on a solid region to an integral on the boundary.

These theorems are helpful since many times, a region or surface is complicated, but its boundary is very simple or vice versa. For example, the boundary of a (solid) cube is complicated since it's made up of 66 different surfaces, but the cube itself is a simple region to integrate over.

Example 1.

Let W\mathscr{W} be the region between the sphere of radius 44 and the cube of side length 11, both centered at the origin. Calculate the flux through the boundary S=∂W\mathscr{S} = \partial\mathscr{W} of a vector field F⃗\vec{F} whose divergence has constant value ∇⋅F⃗=−4\nabla \cdot \vec{F} = -4.
Solution.

At first, the problem looks like it shouldn't be possible since we don't have a formula for F⃗\vec{F}. However, because of the divergence theorem, we actually have enough information to do this problem:

∬∂WF⃗⋅dS⃗=∭W∇⋅F⃗ dV=∭W−4 dV=−4⋅vol⁡(W)=−4⋅(43π⋅43−1).\begin{aligned} \iint_{\partial\mathscr{W}} \vec{F} \cdot \diff\vec{S} &= \iiint_{\mathscr{W}} \nabla \cdot \vec{F} \,\diff{V} \\ &= \iiint_{\mathscr{W}} -4 \,\diff{V} \\ &= -4 \cdot \operatorname{vol}\p{\mathscr{W}} \\ &= \boxed{-4 \cdot \p{\frac{4}{3} \pi \cdot 4^3 - 1}}. \end{aligned}