Divergence theorem and a hemisphere

I'm confused about applying the Divergence theorem to hemispheres.

Here is the statement:

As far as I understand, this question asks to compute $\int\int_{S_1}\mathbf F\cdot d\mathbf S$ over $$S_1=\{(x,y,z):z>0, x^2+y^2+z^2=R^2\}.$$ Here $E=\{(x,y,z): z>0,\ x^2+y^2+z^2\le R^2\}$. The boundary of $E$ is $S_1\cup S_2$, where $$S_2=\{(x,y,0):x^2+y^2=R^2\}.$$ So the divergence theorem reads $$\int\int_{S_1}\mathbf F\cdot d\mathbf S+\int\int_{S_2}\mathbf F\cdot d\mathbf S=\int\int\int_E\operatorname{div}\mathbf F\ dV.$$It is shown that for $\mathbf F$ in the cited question, $\operatorname {div}\mathbf F=1$ so $$\int\int\int_E \operatorname {div}\mathbf F\ dV=\text{volume of the hemisphere $E$}$$ and $$\int\int_{S_2}\mathbf F\cdot d\mathbf S=\int\int_{S_2}\mathbf F\cdot \mathbf n\ d\ S=\\ \int\int_{S_2} \mathbf F(x,y,0)\cdot (0,0,-1)\ dS=\int_0^{2\pi}\int_0^R-e^{-r^2}rdrd\phi=\pi (e^{-R^2}-1),$$ $\mathbf n$ being the outward unit normal.

And thus the integral we need equals $$\int\int_{S_1}=-\int\int_{S_2}+\int\int\int_V=\pi(1-e^{-R^2})+\frac{2}{3}\pi R^3.$$

My questions are:

• Is my interpretation correct?
• Did I use somehow that $z>0$ (as opposed to $z\ge 0$) in the definition of $S_1$? What if I had $z\ge 0$? Would the integral be the same?
• In this question, should I also subtract the integral over the circle in the $xy$-plane from the divergence integral to obtain the final answer?

Solution 1:

$S_1$ really should be the closed hemisphere (i.e., $z\ge 0$). (And $E$ should likewise include $z=0$.) However, $S_2$ also needs to be a surface, not a curve. You did the computation for the disk $x^2+y^2\le R^2$, $z=0$, so your computation seems to be in the right direction. Since you didn't include $\mathbf F$ in your post (I can't chase it down while I'm typing an answer), I won't check the details. One way to think about this physically/intuitively, is that whatever material flows across the bottom disk plus all the interior sources will tell you what flows out across the top hemisphere.