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

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?

• Are you sure you didn't read $\vec{F}\cdot d\vec{s}$ (a dot-product of vectors) and mis-copy it as $\vec{F}\,\vec{ds}$? And should it not just be a single integral along the boundary, i.e. $\displaystyle\int\limits_{\partial S}$ rather than $\displaystyle\iint\limits_S$? ${}\qquad{}$ – Michael Hardy Sep 3 '15 at 11:55
• Yes, you are right with the first, that it should be $\vec{F} \cdot \vec{ds}$. But regarding the second point - it should have a symbol of $\unicode{x222F}$ but I couldn't make it work. – Peter Cerba Sep 3 '15 at 11:59
• 2. No. You need to look carefully at the definition of the divergence theorem - note that the integral is $\iiint\limits_{V}dV$, where $dV$ is a scalar. We are changing from an integral of the flux over a surface to integrating the divergence (which is a scalar) throughout the volume contained by the surface. – michaelrccurtis Sep 3 '15 at 23:37
• 1. Not quite. You haven't quite calculated the divergence correctly, it should be $5r^2$, I think, otherwise your method looks correct. You should be careful to note the difference between $r$ and $R$ in this question. – michaelrccurtis Sep 3 '15 at 23:41
• @michaelrccurtis Thank you for answering. I just want to clarify if "2. No" means "No, this normal vector is not supposed to be used here". And regarding the 1. - I still can't figure out how did you get that $div\vec{F}$ is $5r^{2}$ from $(x^2 + y^2 + z^2) (x,y,z)$ which for me is equal to $x^3+y^3+z^3$? – Peter Cerba Sep 4 '15 at 7:43

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.

• Thank you. Why does it take the form of $(x^2+y^2+z^2)$ and not $(x^2,y^2,z^2)$ - I mean why do you add the vector components together? Ad. 3 I would have calculated it, but I wanted to make sure the $\vec{F}$ is correct. – Peter Cerba Sep 3 '15 at 12:06
• Because $r^2$ here is clearly a scalar, and the usual convention is that $r \equiv |\vec{r}|$, which you have called $R$. – michaelrccurtis Sep 3 '15 at 12:07
• Here, the dot between $r^2$ and $\vec{r}$ just indicates normal multiplication. – michaelrccurtis Sep 3 '15 at 12:09
• Please check my EDIT, I have added a solution and I don't know if it is correct. – Peter Cerba Sep 3 '15 at 22:33
• @PeterKowalski I have commented above. – michaelrccurtis Sep 3 '15 at 23:38