This question has plagued me since the start of the semester more than 2 months ago. I find myself without any solution to this problem.
$$\mathbf{F}=(-x,-y,z)$$ across the surface S that is part of the cone $$z=\sqrt{x^{2}+y^{2}}$$ that lies between the planes $z=1$ and $z=2$ and has inward orientation. Find the flux.
I've tried using spherical coordinates but found myself stuck. Cylindrical might work but suppose I'd like the see the solution developed as a consequence of solving the problem in Spherical coordinates, is that possibile?
In spherical coordinates, the vector field is $$ F = (-r\sin{\theta}\cos{\phi},-r\sin{\theta}\sin{\phi},r\cos{\theta}). $$ Simple enough. The cone is given by $r\cos{\theta}=\sqrt{r^2 \sin^2{\theta}(\cos^2{\phi}+\sin^2{\phi})} = r\sin{\theta},$ i.e. $\tan{\theta}=1$, or $\theta=\pi/4$. The surface we integrate over has $0<\phi<2\pi$, and $1<r\cos{\theta}= \frac{r}{\sqrt{2}}<2$.
Now for the surface element. The normal to the surface is $-e_{\theta}$, by staring at it (it's a $\theta=$const surface, after all...), and the area element is $r \sin{\theta} dr d\phi$ (from the all-time classic Wikipedia article, Del in cylindrical and spherical coordinates). So after all that, we can set up the integral, after remembering $e_{\theta} = (\cos{\theta}\cos{\phi},\cos{\theta}\sin{\phi},-\sin{\theta})$:
$$ -F \cdot e_{\theta} = r(\cos^2{\phi}\cos{\theta}\sin{\theta}+\sin^2{\phi}\cos{\theta}\sin{\theta}+\sin{\theta}\cos{\theta} ) = 2r\cos{\theta}\sin{\theta} = r\sin{2\theta} = -r. $$ Therefore the $\phi$ integral is trivial and gives the usual $2\pi$. The rest of the integral is $$ \frac{1}{\sqrt{2}} \int_{r=\sqrt{2}}^{2\sqrt{2}} r \, dr = \frac{1}{2\sqrt{2}}(8-2), $$ I think, so I get $ \frac{6\pi}{\sqrt{2}}, $ if I haven't forgotten anything.