Since my previous, introductory question Calculate the flux through a surface S from a field described by vectors about this example raised even more questions that I had initially - I was advised to post a new question and below I present my solution of the example, just to make sure if I did that correctly.
The given data: The $F$ and $S$ are as follows ($S$ is oriented outwards): $$\vec{F}=r^2 \cdot \vec{r}$$ $$S: x^2+y^2+z^2=R^2$$ $$\iint\limits_{S} \vec{F} \vec{ds} =\text{ ?}$$
I began with rejecting the use of vector normal to the surface: $$\vec{n}= \frac{\vec{r}}{R}$$ since I have not seen it applied in any other example exploiting Divergence theorem.
Am I right with this?
My solution: (applying advice from @ $$\vec{F}=r^2 \cdot \vec{r}= (r^2x, r^2y, r^2z)$$ $$div\vec{F}=5r^2$$ Then I determined my new set of coordinates and their range: $$V: \left\{ (r, \varphi, \theta) \quad 0 \le r \le R; 0 \le \varphi \le 2\pi; \frac{-\pi}{2} \le \theta \le \frac{\pi}{2}\right\}$$ Then, I calculated the divergence of $\vec{F}$ and substituted the result into the triple integral over the volume described by $S$:
$$\iint\limits_{S} \vec{F} \vec{ds} = \iiint\limits_{V} div\vec{F}\vec{ds}=\iiint\limits_{V} 5r^2 dxdydz = \int_{0}^{R} \left[ \int_{0}^{2\pi} \left[ \int_{ \frac{-\pi}{2} }^{ \frac{\pi}{2} } 5r^2 \cdot R^{2}cos \theta d \theta \right] d \varphi \right] dr =$$
$$=\int_{0}^{R} \left[ \int_{0}^{2\pi} 10r^2 \ R^2 d \varphi \right] dr =20 \pi R^2\int_{0}^{R} r^2 dr=\frac{20}{3} \pi R^{5}$$
Is it the right answer?
Is the normal vector not supposed to be used here?
Your intepretation of $\vec{F}$ is wrong. You need to multiply the scalar by each component of the vector.
$$ \vec{F} = (r^2x, r^2y, r^2z)$$
Remember that the divergence is defined for a vector field - you can't apply it to a scalar. You can think of it as taking:
$$ \nabla\vec{F} = (\partial/\partial x, \partial/\partial y, \partial/\partial z) \cdot (r^2x, r^2y, r^2z) $$
In this case, each term is similar:
$$\frac{\partial}{\partial x}(r^2x) = \frac{\partial (r^2)}{\partial x}x + r^2 $$
where
$$\frac{\partial (r^2)}{\partial x} = 2x $$
Putting this together, we get:
$$ \nabla\vec{F} = 3r^2 + 2(x^2 + y^2 + z^2) = 5r^2 $$