This is the first time I'm doing this kind of problem, and I'm (understandably I think :)) hesitant about my answer. The question is:
Compute $\iint_{\partial Q}\ F\bullet ndS$ using the easiest method available where $Q$ is bounded by $z=4-x^2-y^2, z=1$ and $z=0$. $F=<z^3,x^2y,y^2z>$.
This is the work I did:
Divergence of F: $0+x^2+y^2\\
\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{1}r^3dz\ dr\ d\theta\\
=\int_{0}^{2\pi}\int_{0}^{2}r^3dr\ d\theta\\
=\int_{0}^{2\pi}4d\theta\\
=8\pi$
Is this right? Like I said it's the first time I'm doing it and so I'm looking for feedback - on my bounds, my set up etc.
If you do $z$ first this will need to be two integrals. Doing $r$ first makes it one:
$$I = \int_0^{2\pi} \int_0^1 \int_0^{\sqrt{4-z}} r^3\:dr\:dz\:d\theta =\frac{37\pi}{6}$$