Question

Use the divergence theorem to calculate the surface integral s f · ds; that is, calculate the flux of f across s. f(x, y, z) = x4i − x3z2j + 4xy2zk, s is the surface of the solid bounded by the cylinder x2 + y2 = 1 and the planes z = x + 7 and z = 0.

Answer 1

$\mathbf f(x,y,z)=x^4\,\mathbf i-x^3z^2\,\mathbf j+4xy^2z\,\mathbf k$
$\mathrm{div}(\mathbf f)=\dfrac{\partial(x^4)}{\partial x}+\dfrac{\partial(-x^3z^2)}{\partial y}+\dfrac{\partial(4xy^2z)}{\partial z}=4x^3+0+4xy^2=4x(x^2+y^2)$


Let $\mathcal D$ be the region whose boundary is $\mathcal S$. Then by the divergence theorem,

$\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_{\mathcal D}4x(x^2+y^2)\,\mathrm dV$

Convert to cylindrical coordinates, setting

$x=r\cos\theta$
$y=r\sin\theta$

and keeping $z$ as is. Then the volume element becomes


$\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$

and the integral is

$\displaystyle\iiint_{\mathcal D}4x(x^2+y^2)\,\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=0}^{z=r\cos\theta+7}4r\cos\theta\cdot r^2\cdot r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle4\iiint_{\mathcal D}r^4\cos\theta\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\dfrac{2\pi}3$