Calculate the flux through a surface S from a field described by vectors

Question

I have encountered yet another example which is not that typical.

I need to calculate: $$\iint\limits_{S} \vec{F} \vec{ds} =\text{ ?}$$ Where the $F$ and $S$ are as follows ($S$ is oriented outwards): $$\vec{F}=r^2 \cdot \vec{r}$$ $$S: x^2+y^2+z^2=R^2$$ My questions:

1. How should I interpret: $$r^2 \cdot \vec{r}$$ Should it be? $$(x^2,y^2,z^2) \cdot (x,y,z) = (x^3,y^3,z^3)$$
2. Do I need to calculate a vector normal to the surface and substitute it in the integral? I assume if needed it should look like this: $$\vec{n}= \frac{\vec{r}}{R}$$
3. Can I use divergence (Gauss) theorem in that example? Or maybe I should rather stick to following formula for flux calculation through a surface?:

   $$\iint R-PF_x-QF_y$$

EDIT, based on the answer from @michaelrccurtis, below I present my solution of the example. Could you check whether it's correct?

I began with rejecting the use of $$\vec{n}= \frac{\vec{r}}{R}$$ since I have not seen it applied in any other example exploiting Divergence theorem. Am I right with this?

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 $S$:

$$\iint\limits_{S} \vec{F} \vec{ds} = \iiint\limits_{V} div\vec{F}\vec{ds}=\iiint\limits_{V} 3x^{2}+3y^{2}+3z^{2} dxdydz = \int_{0}^{R} \left[ \int_{0}^{2\pi} \left[ \int_{ \frac{-\pi}{2} }^{ \frac{\pi}{2} } 3R^{2} \cdot R^{2}cos \theta d \theta \right] d \varphi \right] dr =$$

$$=\int_{0}^{R} \left[ \int_{0}^{2\pi} 6R^{4} d \varphi \right] dr =\int_{0}^{R} 12 \pi R^{4} dr=12 \pi R^{5}$$

1. Is it the right answer?

2. Is the normal vector not necessary here?

Answer 1

1. Given the notation, I would assume that $$r^{2} \cdot \vec{r}$$ means:

   $$(x^2 + y^2 + z^2) (x,y,z)$$

2. Yes, since $$\vec{dS} = \vec{\hat{n}}dS$$ where $dS$ is the surface element and $\vec{\hat{n}}$ is the unit normal of the surface, which assuming $R\equiv|\vec{r}|$ you have correct.

3. The divergence theorem is in principle useful in problems like this, provided that the divergence exists and assuming that it gives you a form that is easier to integrate. In this case, why don't you try calculating the divergence? It has a fairly simple form.