I have 2 questions regarding surface integral using Gauss's Divergence theorem.
My first is:
Evaluate $\int_{\tau} z$ $d\tau$ where $S: x^2+y^2+z^2=a^2$ and $z\geq 0$.
I found it to be $0$:
$\int_{\tau} z$ $d\tau= \iiint_\tau z$ $dxdydz= \iint_{R} \int_{z=0}^{z=\sqrt(a^2-x^2-y^2)} z$ $dzdR$. I then evaluated $\int_{z=0}^{z=\sqrt(a^2-x^2-y^2)} z$ $dz$ alone and use polar coordinates instead of cartesian to evaluate the resulting integral which is: $\int_{r=0}^{a} \int_{\theta=0}^{2\pi} a^2r - a^2r(sin^2\theta+ cos^2\theta) drd\theta= 0$.
Is my work correct?
My second is:
When we are asked to evaluate a surface integral using Gauss's theorem for a certain vector where $S: z= 4-x^2$ bounded by the planes $y=0$, $z=0$ & $y+z=5$, what should be the boundaries of $z$ while evaluating the integral if we want to evaluate it the same way as I did in the first one? ( This is our required method); i.e. we have $\iiint_{\tau} \nabla.\vec{V}$ $d\tau= \iint_{R} \int_{z=m}^{z=n} \nabla.\vec{V}$ $dzdR$. My question is what should be these $m$ and $n$?
Please answer me about this method and don't suggest any other ways as I need this in my exam that I will do.
Thanks!
hint for first
with spherical
$=\int_0^a\int_0^{2\pi}\int_0^{\pi /2}(r\cos (\phi) )(r^2\sin (phi))dr d\theta d\phi$
$$=\frac {\pi a^3}{3}\int_0^{\pi /2}\sin (2\phi)d\phi $$
$$=\frac {\pi}{3}a^3$$