Let $H=\{(x,y,z):z>0, x^2+y^2+z^2=R^2\}$ and $$F(x,y,z)=(x^2(y^2-z^3),xzy^4+e^{-x^2}y^4+y,x^2y(y^2x^3+3)z+e^{-x^2-y^2})$$
Find $\int_H F\cdot n\,dS$ where $n$ is the outward unit normal and $dS$ is the area element.
I believe $$\int_H F\cdot n\,dS=\iint_DF\cdot(r_\phi\times r_\theta) \, dA$$ where $$r(\phi,\theta)=R(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi).$$
(Or do I need to add the integral over the circle in the $z=0$ plane?)
We have $$r_\phi\times r_\theta = R^2(\sin^2\phi\cos\theta, \sin^2\phi\sin\theta, \sin\phi\cos\phi)$$
Even computing $F(r(\phi,\theta))$ yields cumbersome expressions, not to mention the dot product of $F(r(\phi,\theta))$ with $r_\phi\times r_\theta$. Is there a way to simplify this without computing? There must be some symmetry which I've always have hard time seeing and understanding. I also tried the divergence theorem, but the expressions there are cumbersome too.
If you apply the divergence theorem you obtain a volume integral over the upper half $B$ of the ball of radius $R$ and another surface integral over the disc $D$ of radius $R$ in the $(x,y)$-plane. Write out the integrands in terms of $x$, $y$, $z$, and cross out all terms that vanish trivially, or by symmetry. Note that the half ball $B$ is symmetric with respect to the $x$- and $y$-axes. At the end not much will remain that actually has to be computed.
One obtains $${\rm div}(F)=2x(y^2-z^3)+4(xz+e^{-x^2}) y^3+1+x^2 y(y^2x^3+3)\ .$$ Only the constant $1$ term gives a contribution, by symmetry. But this is not all. Gauss's theorem applies to the half ball $B$ and its complete boundary surface $\partial B$. One part of $\partial B$ is the surface $H$ defined in the question, and the other part is the disc $D$. We therefore have to calculate and take into account $$\int_D F(x,y,0)\cdot(0,0,1)\>{\rm d}(x,y)=\int_D e^{-x^2-y^2}\>{\rm d}(x,y)\ .$$ Here we can introduce polar coordinates and compute $$\int_0^R r e^{-r^2}\>dr\ .$$ Now put it all together, taking care of the signs.