Question

Find the flux of f = x3 i + y3 j + z3 k out of the closed surface z = x2 + y2 , 0 ≤ z ≤ 5 oriented outward. g

Answer 1

The surface is closed, so you can use the divergence theorem:

$\vec f=x^3\,\vec\imath+y^3\,\vec\jmath+z^3\,\vec k\implies\nabla\cdot\vec f=3x^2+3y^2+3z^2$

Then the flux is

$\displaystyle\iint_S\vec f\cdot\mathrm d\vec S=3\iiint_Rx^2+y^2+z^2\,\mathrm dV$

where $R$ is the region with boundary $S$.

Convert to cylindrical coordinates, taking

$x=u\cos v$

$y=u\sin v$

$z=z$

Then the integral is

$\displaystyle3\int_0^{2\pi}\int_0^1\int_{u^2}^5(u^2+z^2)u\,\mathrm dz\,\mathrm du\,\mathrm dv=\frac{525\pi}4$