Surface integral calculation doubts

Question

I have vector field $\vec{F}=\frac{\vec{r}}{\lvert \vec{r} \rvert^2}=\frac{\vec{r}}{r^2}$ and I want to calculate net flux from a sphere of radius $R$ using spherical coordinates. This is what I tried

$$\int \int_S \vec{F} \cdot d\vec{S}$$

$$\int_0^{2\pi} \int_0^\pi \frac{1}{r^2} r^2 \sin \theta d\theta d\phi$$

$$\int_0^{2\pi} \int_0^\pi \sin \theta d\theta d\phi = 4\pi$$

I am not sure if this is correct because I got $4\pi R$ from divergence theorem. The divergence of $\vec{F}$ is $\frac{1}{r^2}$ and the integral calculation went like this

$$\int \int \int_V \nabla \cdot \vec{F} dV$$

$$\int_0^{2\pi} \int_0^\pi \int_0^R \frac{1}{r^2} r^2 \sin \theta d\theta d\phi = 4\pi R$$

Where did I go wrong?

Answer 1

Too long for a comment.

...

You need to understand the conversion tables, definitions of divergence, curl etc. in cartesian, cylindrical and spherical coordinates to be able to solve these kind of questions. They are listed here: https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

In my analysis courses I always used $\rho$ in spherical coordinates not $r$. I think it is because of physicists that in the above link they used $\rho$ in cyclindrical coordinates. It confuses me.

I am confused with your question too. In what coordinates did you give your vector field $\vec{F}(x,y,z)=\frac{\vec{r}}{r^2}$? I think in cartesian coordinates. Because, $\vec{r}=(x,y,z)$ and $r=\sqrt{x^2+y^2+z^2}$. And your vector field is $\vec{F}(x,y,z)=\frac{(x,y,z)}{x^2+y^2+z^2}$. It is cartesian coordinates!

Now, we have to write this vector field in spherical coordinates. This is the difficult part of the question. As J.G. noted in the comments, the conversion is $\vec{F}(r,\phi,\theta)=\frac{\hat{r}}{r}$. You may use Conversion between unit vectors in Cartesian, cylindrical, and spherical coordinate systems in terms of source coordinates" table in the above link to do that. I am still working on this conversion. But, I think the trick is to notice that $(x,y,z)=r\hat{r}$.

Then, we can take the divergence of $\vec{F}$ by using "Table with the del operator in cartesian, cylindrical and spherical coordinates" of the link above in spherical coordinates. We find $\nabla\cdot\vec{F}=\frac{1}{r^2}$. Then, we can do your second calculation as you did. (For the singularity at the origin, we delete an $\epsilon$-ball centered at the origin. The divergence integral on the remaining solid is $4\pi (R-\epsilon)$ and we take the limit $\epsilon\rightarrow 0.$)

For your first calculation, we need to understant the surface differential element $d\vec{S}$ in spherical coordinates, arguably the most complicated differential element in analysis. Anyway, I think, in our problem it is $d\vec{S}=(R^2sin\theta d\theta d\phi)\hat{r}$ and also on $S$, we have $\vec{F}=\frac{1}{R}\hat{r}$. So, on $S$, $\vec{F}\cdot d\vec{S}=Rsin\theta d\theta d\phi$. And $$\int\int_S \vec{F}\cdot d\vec{S}=\int_0^{2\pi}\int_0^{\pi}Rsin\theta d\theta d\phi=\int_0^{2\pi}2Rd\phi=4\pi R.$$

Answer 2

You cannot use the divergence theorem to solve this problem! The divergence theorem requires that the vector field is continuous in the region of space enclosed by $S$, which isn't happening here.

You can solve this problem without appealing to parametric surfaces. If we realize that $$\vec{F}(x,y,z)=\frac{(x,y,z)}{R^2} \text{ and }\vec{n}=\frac{(x,y,z)}{R}$$ at any $(x,y,z)\in S$ then $$\begin{eqnarray*}\iint_{S}(F\cdot \vec{n})\mathrm{d}S &=& \iint_{S}\left[\frac{(x,y,z)}{R^2}\cdot \frac{(x,y,z)}{R}\right]\mathrm{d}S \\ &=& \iint_{x^2+y^2+z^2=R^2}\frac{x^2+y^2+z^2}{R^3}\mathrm{d}S \\ &=& \frac{1}{R}\iint_{x^2+y^2+z^2=R^2}\mathrm{d}S \\ &=& \frac{1}{R}\times 4\pi R^2\end{eqnarray*}$$