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.



hint for first

with spherical

$=\int_0^a\int_0^{2\pi}\int_0^{\pi /2}(r\cos (\phi) )(r^2\sin (phi))dr d\theta d\phi$

$$=\frac {\pi a^3}{3}\int_0^{\pi /2}\sin (2\phi)d\phi $$

$$=\frac {\pi}{3}a^3$$

  • $\begingroup$ Yess you are right. Is my answer correct? $\endgroup$ – Nour May 26 '17 at 11:30
  • $\begingroup$ @Nour is it okay now for the first. $\endgroup$ – hamam_Abdallah May 26 '17 at 11:32
  • $\begingroup$ Yes it is okay, but what is my mistake? This is the way our doctor solves problems. $\endgroup$ – Nour May 26 '17 at 11:42
  • $\begingroup$ @Nour what is your Jaccobian. $\endgroup$ – hamam_Abdallah May 26 '17 at 11:44

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