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