Facing issue in rectifying the solution. Gauss Divergence Theorem problem Evaluate
$$
\iint_S (y^2z^2 \textbf{i} \, +z^2x^2\textbf{j}+z^2y^2\textbf{k}).\textbf{n} ~\mathrm{d}S 
$$
where S is the part of sphere $x^2+y^2+z^2=1$ above the $xy$ plane and bounded by this plane.
I've tried solving this using Gauss Divergence Theorem, as follows,
$$
\iiint_V \text{div}  (y^2z^2\textbf{i} \, +z^2x^2\textbf{j}+z^2y^2\textbf{k})~\mathrm{d}V  = \iiint_V (2zy^2)~\mathrm{d}V 
$$
For the limits of the volume V,
$-1\le x\le 1$
$-\sqrt{1-x^2}\le y \le \sqrt{1-x^2}$
$0\le z\le \sqrt{1-x^2-y^2}$
Upon integrating with these limits, I'm getting the asnwer as $\frac{\pi}{10}$, which is not the correct answer. And I'm unable to make out where I'm going wrong.
Kindly guide me to understand my error and rectify the solution. Thank you.
 A: If we use spherical coordinates, then
$x = \rho \sin \phi \cos \theta$
$y = \rho \sin \phi \sin \theta$
$z = \rho \cos \phi$
And the volume integral is
$\begin{align*}
I &= \displaystyle \int_{0}^{2\pi} \int_0^{\pi/2} \int_0^1 2 \rho^5 \cos \phi \sin^3 \phi \sin^2 \theta   d\rho d\phi d\theta \\
&= \frac{1}{3} \displaystyle \int_{0}^{2\pi} \int_0^{\pi/2}    \cos \phi \sin^3 \phi \sin^2 \theta    d\phi d\theta\\
&=\frac{1}{12} \displaystyle \int_{0}^{2\pi} \int_0^{\pi/2}   \sin^2 \theta   d\theta\\
&=\frac{\pi}{12}
\end{align*}$
A: $ \displaystyle \iiint_V \nabla \cdot  (y^2z^2, z^2x^2, z^2y^2) \ dV  = \iiint_V 2zy^2 \ dV$
Your bounds are correct so you may have made some mistake in evaluating it.
$ \displaystyle \iiint_V 2 z y^2 \ dV = 2 \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{0}^{\sqrt{1-x^2-y^2}} z  y^2 \ dz \ dy \ dx$
$ \displaystyle  = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} y^2 (1 - x^2 - y^2) \ dy \ dx$
Converting to polar coordinates,
$x = r \cos\theta, y = r \sin\theta$
$ \displaystyle  = \int_0^{2\pi} \int_0^1 r^2 \sin^2\theta (1 - r^2) \ r \ dr \ d\theta$
$ \displaystyle  = \int_0^{2\pi} \int_0^1  \sin^2\theta (r^3 - r^5) \ dr \ d\theta$
$ \displaystyle  = \int_0^{2\pi}  \cfrac{\sin^2\theta}{12} \ d\theta = \cfrac{\pi}{12}$
Alternatively use spherical coordinates, as the other answer shows.
