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.
Best Answer
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
$\begin{align*} I &= \displaystyle \int_{0}^{2\pi} \int_0^{\pi/2} \int_0^1 2 \rho^5 \cos \phi \sin^3 \phi \sin^2 \theta d\rho d\phi d\theta \\ &= \frac{1}{3} \displaystyle \int_{0}^{2\pi} \int_0^{\pi/2} \cos \phi \sin^3 \phi \sin^2 \theta d\phi d\theta\\ &=\frac{1}{12} \displaystyle \int_{0}^{2\pi} \int_0^{\pi/2} \sin^2 \theta d\theta\\ &=\frac{\pi}{12} \end{align*}$