Question on Divergence Theorem Evaluate $\iint F.ds$  
where $F= \frac{59}{3} x^3 \hat i + \frac{59}{3} y^3 \hat j + \frac{59}{3} z^3 \hat k$, and $S$ is the surface: 
$S= \{(x,y,z) | x^2 + y^2 +z^2 =9\}$ 
Please explain in detail how to get the answer.
 A: Simple application of the Divergence theorem:
$$\iint F\cdot ds = \iiint \nabla \cdot  F dV$$
$$=\iiint \nabla \cdot (\frac{59}{3} x^3 \hat i + \frac{59}{3} y^3 \hat j + \frac{59}{3} z^3 \hat k) dV$$
$$=\iiint (59 x^2 \hat i + 59 y^2 \hat j + 59 z^2 \hat k) dV$$
$$=\iiint 59(x^2+y^2+z^2) dV$$
$$=59 \cdot \iiint r^2 \cdot r^2 \sin \theta \, dr \, d\theta \, d\phi$$
$$=59 \cdot \int_{0}^{3} r^4 \, dr \int_{0}^{\pi} \sin \theta \, d\theta \int_{0}^{2\pi} d\phi$$
$$=59 \cdot \frac{3^5}{5} \cdot 2 \cdot 2\pi $$
A: Let's go from the beginning.
Your surface $S= \{(x,y,z) \in \mathbb{R}^3: x^2+y^2+z^2=9\}$, as you may see is a sphere with radius $3$ and centred in the origin. So, we can write this surface as in a simpler way, to do so, we change the variables $(x, y, z) \mapsto (\rho, \theta, \varphi)$, in this way:
\begin{align}
\begin{cases}
x= \rho \sin (\varphi) \cos (\theta)\\
y= \rho \sin (\varphi) \sin (\theta)\\
z= \rho \cos (\varphi),
\end{cases}
\end{align}
such that $x^2 + y^2 + z^2 = \rho^2$
In the image below, from Pauls, you can see how it works.

This means we can rewrite $$S = \{(\rho, \theta, \varphi) \in [0,3] \times [0,2\pi] \times [0,\pi]\}$$ and yields the differential volume element $$dV = dx \, dy \, dz = \rho^2 \sin(\varphi) \, d\rho \, d\theta \, d\varphi.$$
Now we have everything set up to solve this problem. From Gauss-Ostrogradsky theorem
\begin{align}
\int_{\partial S} \overrightarrow{F} \cdot ds
&= \int_{S} \left( \nabla \cdot \overrightarrow{F} \right)dV\\
&= \int_{S} 59 \left( x^2 + y^2 + z^2 \right)dV\\
&=59 \int_{0}^{3} \int_{0}^{2 \pi} \int_{0}^{\pi} \left(\rho^2\right) \rho^2 \sin(\varphi) \, d\rho \, d\theta \, d\varphi.\\
&=59\int_{0}^{3} \int_{0}^{2 \pi} \int_{0}^{\pi} \rho^4 \sin(\varphi) \, d\rho \, d\theta \, d\varphi\\
\end{align}
By Fubini's
\begin{align}
\hphantom{\int_{\partial S} \overrightarrow{F} \cdot ds}
&=59\int_{0}^{3} \rho^4 \, d\rho \int_{0}^{2 \pi}  d\theta \int_{0}^{\pi}  \sin(\varphi)  \, d\varphi.\\
&=59 \left[\frac{\rho^5}{5} \right]_{0}^{3} \times \left[\theta \right]_{0}^{2\pi} \times \left[-\cos(\varphi) \right]_{0}^{\pi}\\
&=59 \times \frac{3^5}{5} \times 2\pi \times 2\\
&=\frac{57348 \pi}{5}
\end{align}
