Suppose I want to find the flux of $F$ through the portion of the cylinder $z^2 + y^2 = 16$ from $x = 1$ and $x = 3$ and normal points outwards of the cylinder.
$F= (z^2 + y)i+(x^2 + z)j+5z^2k$
I solved for the divergence
$div \ F$ = $10z$
How do I go about solving this triple integral in cylindrical coordinates $dxdrdθ$? I'm not quite sure how to convert the $z$ and make it in terms of $x$ or $r$. Do I just treat the $z$ as $r$?
I am assuming you are calculating flux including flux through the part of the planes $x = 0, x = 1$ in the cylinder.
Given $z$ is an odd function and the cylinder is symmetric with respect to $XY$ plane, your integral will be zero.
You can use cylindrical coordinates as below -
$z = r \cos \theta, y = r \sin \theta, x$
$\displaystyle \iiint_S div F \, dV = 10 \int_0^{2\pi} \int_0^4 \int_1^3 \, r^2 \cos\theta \, dx \, dr \, d\theta = 0$