This is from the UChicago GRE problem sets.
Let $C$ be the cylinder bounded by $x^2 + y^2 = 9$ and $z = 0, z = 5$.
If $F(x, y, z) = (3x, y^3, -2z^2)$, then calculate the flux of $F$
through $C$, i.e., integrate the normal vector dotted with the field
over the cylinder.(A) $-\frac{40}{2}\pi$
(B) $-\frac{45}{2}\pi$
(C) $0$
(D) $\frac{-36}{2}\pi$
(E) $\frac{45}{2}\pi$
I tried following the method outlined in this answer and I get:
\begin{align}\Phi&=\int\vec F\cdot d^2\vec A=\int_0^{5}\int_0^{2\pi}\langle 3\cos\theta,\sin^3\theta,-2z^2\rangle\cdot\langle3\cos\theta,3\sin\theta,0\rangle\,d\theta \ dz\\
&=\int_0^{5}\int_0^{2\pi} (9\cos^2\theta + 3\sin^4\theta) \ d\theta \ dz = \int_0^5\frac{45}{4}\pi \ dz = \frac{225\pi}{4}.\end{align}
This result does not match any of the answers given in the options. Is something wrong?
I used Wolfram Alpha to evaluate the inner integral.
Best Answer
using the divergence theorem
$div(F)=3+3y^2-4z$
$\alpha(h,r,\theta)=(rcos(\theta),rsin(\theta),h)$
$ J(\alpha(h,r,\theta))=r$
$\int_0^5\int_0^3\int_0^{2\pi}(3+3r^2sin^2(\theta)-4h)r d\theta dr dh=-\frac{45\pi}{4}$