Question

Evaluate the surface integral s f · ds for the given vector field f and the oriented surface s. in other words, find the flux of f across s. for closed surfaces, use the positive (outward) orientation. f(x, y, z) = x i + y j + 10 k s is the boundary of the region enclosed by the cylinder x2 + z2 = 1 and the planes y = 0 and x + y = 2

Answer 1

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+10\,\mathbf k$
$\implies\nabla\cdot\mathbf f(x,y,z)=1+1+0=2$

By the divergence theorem, the surface integral along $S$ is equivalent to the triple integral over the region $R$ bounded by $S$:

$\displaystyle\iint_S\mathbf f(x,y,z)\,\mathrm dS=\iiint_R\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=2\iiint_R\mathrm dV$

Convert to cylindrical coordinates, setting

$\begin{cases}x=r\cos\theta\\y=Y\\z=r\sin\theta\end{cases}\implies\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dY$

The triple integral is then equivalent to

$=\displaystyle2\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{Y=0}^{Y=2-r\cos\theta}r\,\mathrm dY\,\mathrm dr\,\mathrm\theta$
$=\displaystyle2\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r(2-r\cos\theta)\,\mathrm dr\,\mathrm\theta$
$=\displaystyle\frac23\int_{\theta=0}^{\theta=2\pi}(3-\cos\theta)\,\mathrm dr\,\mathrm\theta$
$=4\pi$