Question

Find the outward flux of the vector field f=(x3,y3,z2) across the surface of the region that is enclosed by the circular cylinder x2+y2=64 and the planes z=0 and z=6.

Answer 1

Use the divergence theorem.

$\mathbf f(x,y,z)=(x^3,y^3,z^3)\implies \nabla\cdot\mathbf f=3(x^2+y^2+z^2)$

Calling the closed region $\mathcal R$ and its boundary the surface $\partial\mathcal R$, we have by the divergence theorem,

$\displaystyle\underbrace{\iint_{\partial\mathcal R}\mathbf f\cdot\mathrm d\mathbf S}_{\text{flux}}=\iiint_{\mathcal R}\nabla\cdot\mathbf f\,\mathrm dV$

We set up the volume integral in cylindrical coordinates, using

$\begin{cases}x=u\cos v\\y=u\sin v\\z=w\end{cases}\implies\mathrm dV=u\,\mathrm du\,\mathrm dv\,\mathrm dw$

$\displaystyle\iiint_{\mathcal R}\nabla\cdot\mathbf f\,\mathrm dV=3\int_{w=0}^{w=6}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=8}(u^2\cos^2v+u^2\sin^2v+w^2)u\,\mathrm du\,\mathrm dv\,\mathrm dw$
$=6\pi\displaystyle\int_{w=0}^{w=6}\int_{u=0}^{u=8}(u^3+uw^2)\,\mathrm du\,\mathrm dw=50,688\pi$