Calculation of total flux through an inverted hemisphere for a vector field in spherical unit vectors

Question

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!

Answer 1

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 $$ \begin{gather} \iint_{S_1}\mathbf{F}\cdot\mathrm{d}\mathbf{S}=\iint_{S_1}R\cos\theta\,R^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi=R^3\cdot\frac12\cdot2\pi=\pi R^3\\ \iint_{S_2}\mathbf{F}\cdot\mathrm{d}\mathbf{S}=\iint_{S_2}r^2\,\mathrm{d}r\,\mathrm{d}\phi=\frac{2\pi R^3}{3}. \end{gather} $$