Let $F(x,y,z)=(x^2\sin(y^2-z^3),xy^4z+y,e^{-x^2-y^2}+yz)$. Find the flux integral $\int_D F\cdot n dS$, where $n$ is the outward normal and $D$ is the part of the sphere $x^2+y^2+z^2=r^2$ in $z>0$.
I'm applying the divergence theorem. $$div F=2x\sin(y^2-z^3)+4xy^3z+1+y;$$ Let $x=r\cos t,y=r\sin t, z=+\sqrt{r^2-(x^2+y^2)}=\sqrt{r^2-r^2}=0$. So the integral is $$\int_0^{2\pi}\int_0^r\int_0^{z=\sqrt{r^2-x^2+y^2}=0}(2r\cos t\sin (r^2\cos^2t)+1+r\sin t)rdzdrdt,$$ and the inner integral (and so the whole integral) is zero. This makes me doubt I'm doing everything correctly, am I?
The fixed integral is $$\int_0^{2\pi}\int_0^R\int_0^{\sqrt{R^2-r^2}}(2r\cos t\sin (r^2\cos^2t+(R^2-r^2)^{3/2})+1+r\sin t)rdzdrdt,$$ where the sphere has radius $R$. But now another problem appears: how to compute $$\int_0^R 2r^2\cos t\sin(r^2\cos^2 t-(R^2-r^2)^{3/2})\sqrt{R^2-r^2}+r\sqrt{R^2-r^2}+r^2\sqrt{R^2-r^2}\sin t) dt?$$
The second summand can be integrated by substitution $\xi=r^2$, but how about the first and the third summands?
$z = \sqrt {r^2 - x^2 - y^2}\\ dS = (\frac {x}{z}, \frac {y}{z}, 1)\ dx\ dy$
$F\cdot dS = \frac {x^3}{z}\sin(y^2+ z^3) + xy^5 + \frac {y^2}{z} + e^{-x^2-y^2} + yz$
Before converting to polar, note that some of these terms are odd. When we integrate over a symmetric region they will equal 0. i.e.
$\int_{-r}^r \frac {x^3}{z}\sin(y^2+ z^3)\ dx = 0$
Dropping the odd terms we are left with
$\iint \frac {y^2}{\sqrt {r^2 - x^2 - y^2}} + e^{-x^2 - y^2} \ dA$
Now convert to polar and proceed.
If you want to use the divergence theorem, you will need to seal it off. $\iint F\cdot dS + \iint F\cdot dD = \iiint \nabla \cdot F \ dV$
D is the disk of radius r and the normal points downward $(0,0,-1)$
$\iint F\cdot dS = \iiint \nabla \cdot F \ dV + \iint e^{-x^2-y^2}\ dD$
using the same argument above on the odd terms of $\nabla\cdot F$
$\iiint 1 \ dV + \iint e^{-x^2-y^2} \ dD$