1
$\begingroup$

Here is a question which I came across recently

Calculate the flux of the vector field $\vec{v} = r\cos\theta (\hat{r}) + r\sin\theta (\hat{\theta}) + r\sin\theta \cos\phi (\hat{\phi})$ over an inverted hemisphere of radius $R$ with its base on the $x\text{-}y$ plane and centred at the origin.

Using Gauss's Divergence Theorem, I can easily calculate the flux through the entire surface. But I wish to calculate it using the normal definition- $\iint \vec{v}\cdot d\vec{A}$ over the closed surface $S$. Now, $S$ can be written as an union of two boundaries $\{S_{1} \cup S_{2}\}$, where $S_{1}$ would be the outer bounding surface, and $S_{2}$ would be the bounding circle in the $x\text{-}y$ plane. Calculating the flux through through $S_{1}$, I get the value as $\frac{2\pi r^{3}}{3}$. I had taken the unit normal vector $\hat{n}$ to be $\hat{r}$ for $S_{1}$, but this won't work for $S_{2}$. The unit normal vector for $S_{2}$ would be $-\hat{k}$ in Cartesian coordinates, so I changed it to unit vectors in spherical coordinates, as the vector field $\vec{v}$ is also in the spherical unit vectors. But after that, I can't work out the surface integral through $S_{2}$. The answer should be $\frac{5\pi r^{3}}{3}$, as obtained from the Gauss's Divergence Theorem, but I'm not being able to reach that. Can someone please guide me how to proceed from here? I'm still just learning these things!

PS: Please don't discard this question as a repeated one. It's true that there have been questions before on this forum related to flux through a hemisphere, but in most of them, the vector fields were in Cartesian coordinates and it was easy to work with. I found a few answers on the spherical unit vectors, but they didn't include the flux due to the bounding circle in $x\text{-}y$ plane, for any reason whatsoever!

$\endgroup$
9
  • $\begingroup$ You mean the bounding disc, not the bounding circle. And the question is asking for the flux across the hemisphere $S_1$ so why bother with closing it with $S_2$?. $\endgroup$ Commented Apr 2, 2021 at 8:50
  • $\begingroup$ The answer is $\frac{5\pi r^{3}}{3}$. This indicates that I need to include the flux through $S_{2}$ as well, I guess. $\endgroup$ Commented Apr 2, 2021 at 8:52
  • $\begingroup$ Divergence theorem gives me the answer, but I want to verify this using the conventional method as well. Integrating over $S_{1}$ gives me $\frac{2\pi r^{3}}{3}$, if I've done it correctly. $\endgroup$ Commented Apr 2, 2021 at 8:54
  • 1
    $\begingroup$ On $S_2$, you have $\theta=\pi/2$ and $\hat\theta=-\hat{k}$ (assuming "physicist" convention $(x,y,z)=(r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta)$), so you are just integrating $r\cdot r\,\mathrm{d}r\,\mathrm{d}\phi$ over the disc (I'm not sure whether "inverted" hemisphere means the positive-$z$ or negative-$z$ half of the sphere, so I'll leave it at that). $\endgroup$ Commented Apr 2, 2021 at 9:21
  • 1
    $\begingroup$ Unit normal vector $\hat n = (0, 0, -1)$ can be written in spherical coordinates as $\hat {n} = (\sin \theta \ \hat {\theta} - \cos \theta \ \hat {r})$. As $\theta = \pi/2$, the dot product of $\vec v \cdot \hat{n} = r$. $\endgroup$
    – Math Lover
    Commented Apr 2, 2021 at 10:35

1 Answer 1

1
$\begingroup$

We compute \begin{align*} \operatorname{div}\mathbf{F} &=\frac1{r^2}\frac{\partial}{\partial r}(r^2F_r)+\frac1{r\sin\theta}\frac{\partial}{\partial\theta}(F_\theta\sin\theta)+\frac1{r\sin\theta}\frac{\partial}{\partial\phi}F_\phi\\ &=\frac1{r^2}\frac{\partial}{\partial r}(r^3\cos\theta)+\frac1{r\sin\theta}\frac{\partial}{\partial\theta}(r\sin^2\theta)+\frac1{r\sin\theta}\frac{\partial}{\partial\phi}(r\sin\theta\cos\phi)\\ &=3\cos\theta+2\cos\theta-\sin\phi \end{align*} So for $\mathcal{V}=\{(r,\theta,\phi)\in(0,R)\times(0,\pi/2)\times(0,2\pi)\}$, $$\iiint_{\mathcal{V}}\operatorname{div}\mathbf{F}\,\mathrm{d}\mathrm{vol}=\frac{5\pi R^3}3$$ and $$ \begin{gather} \iint_{S_1}\mathbf{F}\cdot\mathrm{d}\mathbf{S}=\iint_{S_1}R\cos\theta\,R^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi=R^3\cdot\frac12\cdot2\pi=\pi R^3\\ \iint_{S_2}\mathbf{F}\cdot\mathrm{d}\mathbf{S}=\iint_{S_2}r^2\,\mathrm{d}r\,\mathrm{d}\phi=\frac{2\pi R^3}{3}. \end{gather} $$

$\endgroup$
1
  • $\begingroup$ I got it now, thanks a lot. I've committed a dumb error in calculating for $S_{1}$, I realized just now :( $\endgroup$ Commented Apr 2, 2021 at 10:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .