For the vector field $\mathbf{F}=(xy^2,yz^2,x^2z)$ use the divergent theorem to evaluate $$\iint_S\mathbf{F}\cdot d\mathbf{S}$$ where $S$ is the sphere of radius $1$ centered at the origin.
We use the divergence theorem and get that the divergence is, $div\mathbf{F}=y^2+z^2+x^2$ and hence the surface integral is equal to the triple integral $$\iiint_B(y^2+z^2+x^2)dV$$ where $B$ is the ball of radius $1$. Now to evaluate the tripple integral, we can change to variables of spherical coordinates. In our case $$0 \leq \rho \leq 1, 0 \leq \theta \leq 2 \pi, 0 \leq \phi \leq \pi$$
Now why we get that the integral is simply $x^2+y^2+z^2=\rho^2$? Here I got stucked.
EDIT: We than know that the Jacobian determinant is $dV=\rho^2 \cdot sin(\phi)\cdot d\phi \cdot d \theta\cdot d \rho$
In spherical coordinates, $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho\cos\phi$
So $x^2+y^2+z^2 = \rho^2 \ $ where $0 \leq \rho \leq 1$
The region is a unit ball and the equation is written as $x^2 + y^2 + z^2 \leq 1$. Note the inequality sign. When you do surface integral, your integral is over the surface and hence $\rho = 1$. But in this case, you are applying divergence theorem which is volume integral and hence $0 \leq \rho \leq 1$
So your integral is $\displaystyle \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^4 \sin \phi \ d\rho \ d\phi \ d\theta = \frac{4\pi}{5} \ $ ($\rho^2 \sin\phi$ is the Jacobian)