Question

Et f(x, y, z) = z tan−1(y2)i + z3 ln(x2 + 5)j + zk. find the flux of f across s, the part of the paraboloid x2 + y2 + z = 11 that lies above the plane z = 2 and is oriented upward.

Answer 1

Since the surface is closed, and the vector field is rather complicated, you can use the divergence theorem. The flux of $\mathbf f(x,y,z)$ across $S$ is given by a surface integral, which the divergence theorem asserts is equivalent to a volume integral:

$\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\iiint_R(\nabla\cdot\mathbf f)\,\mathrm dV$

where $R$ denotes the space with boundary $S$. We have

$\nabla\cdot\mathbf f(x,y,z)=\dfrac{\partial z\tan^{-1}(y^2)}{\partial x}+\dfrac{\partial z^3\ln(x^2+5)}{\partial y}+\dfrac{\partial z}{\partial z}=0+0+1=1$

So in fact the flux across $S$ happens to be equal (in magnitude) to the volume encased by $S$.

$\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\int_{x=-3}^{x=3}\int_{y=-\sqrt{9-x^2}}^{y=\sqrt{9-x^2}}\int_{z=2}^{z=11-x^2-y^2}\mathrm dz\,\mathrm dy\,\mathrm dx$

Convert to cylindrical coordinates, setting

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

$\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\int_{\zeta=2}^{\zeta=11-r^2}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=3}r(11-r^2-2)\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=3}(9r-r^3)\,\mathrm dr=\dfrac{81\pi}2$