Using the divergence theorem to calculate the surface area of a sphere I want to calculate:
$$\iiint_V div (\overrightarrow F \cdot \space  dV) $$
with $\overrightarrow F=x^3\hat i+ y^3 \hat j+z^3\hat k$   and Surface of sphere given as $x^2+y^2+z^2=r^2$
So, first I calculate $div.\overrightarrow F=3(x^2+y^2+z^2)$
Then using spherical co-ordinates, we know:
$x= rsin\theta \cdot cos\phi$
$y= rsin\theta \cdot sin\phi$
$z= rcos\phi$
Now what I don't understand is the following given according the textbook:
$dxdydz=r^2sin\theta \cdot dr \cdot d\theta \cdot d\phi$
Where does $dxdydz$ come from?
and How are these limits are calculated?


*

*For $r$, the limit is $0$ to a  

*For $\theta$ the limit is $0$ to $\pi$  

*For $\phi$ the limit is $0$ to $2\pi$


Using these limits, the result can be calculated using integration.

Not exactly an answer to your question, but a bit long for a comment:
Since $r$ denotes the radius of your sphere, it's best to use a different symbol, such as $\rho$, to denote the "radius function" (i.e., the distance to the origin).
Be that as it may, you can avoid the nuisance of integrating the angular variables in your question because $\nabla \cdot F = 3\rho^{2}$ is constant on each spherical shell $S_{\rho}$. Consequently,
$$
\iiint_{V} (\nabla \cdot F)\, dV
  = \int_{0}^{r} \left(\iint_{S_{\rho}} 3\rho^{2}\, dA\right) d\rho
  = \int_{0}^{r} 3\rho^{2} \left(\iint_{S_{\rho}} dA\right) d\rho
  = 3(4\pi)\int_{0}^{r} \rho^{4}\, d\rho.
$$
