Divergence Theorem

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Let E be a simple solid region and let S be the boundary surface of E, given with positive (outward) orientation. Let $\bar{F}$ be a vector field whose component functions have continuous partial derivatives on an open region that contains E. Then,

$\iint_{\mathrm{S}}\overline{F}\cdot d\overline{S}=\iiint_{E}\mathrm{div}\overline{F}dV.$

Thus, the Divergence Theorem states that, under the given conditions, the flux of  $\bar{F}$ across the boundary surface of E is equal to the triple integral of the divergence of $\bar{F}$ over E.

Example 1. Let S be the circle given by $x^{2}+y^{2}\leq4$ and $z=5$ and let $\overline{F}=y^{2}\overline{i}+x^{2}\overline{j}+z^{2}\overline{k} .$ Compute $\iint_{S}\bar{F}\cdot d\overline{s} .$

Solution. We use parametric coordinates as follows:

$\begin{aligned} &\overline{r}(u,v)=\langle u,v,5\rangle, \\ &\bar{T}_{u}=\langle1,0,0\rangle, \\ &\overline{T}_{\nu}=\langle0,1,0\rangle, \\ &\overline{N}=\overline{T}_{u}\times\overline{T}_{v}=& \begin{vmatrix}\overline{i}&\overline{j}&\overline{k}\\1&0&0\\0&1&0\end{vmatrix}=\overline{k}.  \end{aligned}$

We then calculate and obtain

$\iint_{S}\overline{F}\cdot d\overline{s}=\iint_{x^{2}+y^{2}\leq4}\langle\nu^{2},u^{2},25\rangle\cdot\langle0,0,1\rangle dA=\iint_{x^{2}+y^{2}\leq4}25dA=25(16\pi)=400\pi.$

Example 2. Let W be the cylinder given by $x^{2}+y^{2}\leq4$ and $0\leq z\leq5$ and let $\overline{F}=xy^{2}\overline{i}+y^{3}\overline{j}+4x^{2}z\overline{k} .$ Compute the flux of $\bar{F}$ across the boundary surface of W.

Solution. We begin by calculating the divergence of $\bar{F}$

$\mathrm{div}\overline{F}=\frac{\partial xy^{2}}{\partial x}+\frac{\partial y^{3}}{\partial y}+\frac{\partial4x^{2}z}{\partial z}=y^{2}+3y^{2}+4x^{2}=4y^{2}+4x^{2}$.

Then, applying the Divergence Theorem, we obtain

$\begin{aligned} \iint_{\mathrm{S}}\bar{F}\cdot d\overline{s}& =\iint_{\mathrm{S}}\langle xy^{2},y^{3},4x^{2}z\rangle\cdot d\overline{s} \\ &=\iiint_{W}\mathrm{div}\overline{F}dV \\ &=\iiint_{W}(4y^{2}+4x^{2})dV \\ &=4\iint_{x^{2}+y^{2}\leq4}\biggl(\int_{0}^{5}(x^{2}+y^{2})dz\biggr)dA \\ &=20\iint_{x^{2}+y^{2}\leq4}(x^{2}+y^{2})dA \\ &=20\int_{0}^{2\pi}\int_{0}^{2}r^{2}rdrd\theta \\ &=20\int_{0}^{2\pi}\int_{0}^{2}r^{3}drd\theta \\ &=20\int_{0}^{2\pi}\biggl[\frac{1}{4}r^{4}\biggr]_{0}^{2}d\theta \\ &=20\times2\pi\times4 \\ &=160\pi, \end{aligned}$

where we’ve used the polar coordinates

$x=r\cos\theta,\quad y=r\sin\theta,\quad r^{2}=x^{2}+y^{2},\quad dA=rdrd\theta.$