Question

Let

F(x, y, z) = z tan?1(y2)i + z3 ln(x2 + 7)j + zk.

Find the flux of F across S, the part of the paraboloid

x2 + y2 + z = 27

that lies above the plane

z = 2

and is oriented upward.

\int \int_{s}^{} F*ds

Answer 1

$\vec F(x,y,z)=z\tan^{-1}(y^2)\,\vec\imath+z^3\ln(x^2+7)\,\vec\jmath+z\,\vec k$

has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z\tan^{-1}(y^2))}{\partial x}+\dfrac{\partial(z^3\ln(x^2+7))}{\partial y}+\dfrac{\partial z}{\partial z}=1$

If $S$ includes the disk $x^2+y^2=25$ which is contained in the paraboloid, then by the divergence theorem the flux is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^5\int_2^{27-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{625\pi}2}$

If the disk is not part of $S$, then subtract the flux of $\vec F$ across the disk (with downward orientation; call it $D$). Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+2\,\vec k$

where $0\le u\le5$ and $0\le v\le2\pi$, with normal vector

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k$

Then the flux of $\vec F$ across $D$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=-2\int_0^{2\pi}\int_0^5u\,\mathrm du\,\mathrm dv=-50\pi$

which makes the flux across $S$, $\dfrac{625\pi}2-50\pi=\boxed{\frac{615\pi}2}$.