Evaluate the inward flux of the vector field $F=$ over the surface $S$ of the solid bounded by $z=\sqrt{x^2+y^2}$ and $z=3$. Evaluate the inward flux of the vector field $F=<y,-x,z>$ over the surface $S$ of the solid bounded by $z=\sqrt{x^2+y^2}$ and $z=3$.
this is basically an inverted cone (right?)
So by changing to polar coordinates: $0\leq r\leq3$ and $0\leq \theta \leq 2\pi$
$$-\int_0^{2\pi} \int_0^3-y\frac{x}{\sqrt{x^2+y^2}} + x\frac{y}{\sqrt{x^2+y^2}}+\sqrt{x^2+y^2}\space \space \times rdrd\theta$$
Simplified to
$$-\int_0^{2\pi} \int_0^3r^2drd\theta $$
Did i make any mistakes here? because i solved this to be $-18\pi$ but the answer given is $-9\pi$
I'm thinking maybe the range for $\theta$ is wrong but then it can't be because the conditions require full circles.
 A: The solid body can be described in cylindrical coordinates as
$$\Omega  = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 3,0 \le \theta  \le 2\pi ,r \le z \le 3} \right\}$$
In fact, the surface surrounding your body, $\partial \Omega $, consist of two other surfaces 
$$\begin{array}{l}
\partial \Omega  = {S_1} \cup {S_2}\\
{S_1} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 3,0 \le \theta  \le 2\pi ,z = 3} \right\}\\
{S_2} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 3,0 \le \theta  \le 2\pi ,z = r} \right\}
\end{array}$$
You have just computed the flux toward $S_1$, i.e., the plane $z=3$! :)
A: If you want to practice on definition of surface integrals, I would suggest you not to use any "ready formulas" as Stokes' theorem, for example, but try first to gain basic understanding for the underlying geometry. A solution could look like the following:


*

*Split the vector field into two
$$
F=\underbrace{(y,-x,0)}_{F_1}+\underbrace{(0,0,z)}_{F_2}.
$$
Notice that the first one $F_1$ is a planar vector field that belongs to the tangent space for both surfaces, hence, its flux is zero, and we have to calculate only the flux of the second one (easier!).

*On the top surface $S_1$ (the disc in the plane $z=3$) we have $F_2=(0,0,3)$, which is (anti)parallel to the surface inward normal $n_1=(0,0,-1)$, and the scalar product of those vectors is $-3$. So the surface integral becomes
$$
\iint_{S_1} F_2\cdot n_1\,dA=-3\iint_{S_1}\,dA=-3\cdot\text{the disc area}=-3\cdot \pi 3^2=-27\pi.
$$

*On the side surface $S_2$ it is a bit trickier. The inward normal $n_2$ has the same angle $\pi/4$ with the vector field, so the scalar product is $F_2\cdot n_2=|F_2||n_2|\cos\pi/4=z/\sqrt{2}$, and the integral becomes
$$
\iint_{S_2}F_2\cdot n_2\,dA=\frac{1}{\sqrt2}\iint_{S_2}z\,dA.
$$
The surface area element $dA$ corresponds to the area of the narrow band around the cone between levels $z$ and $z+dz$ and can be approximated as $2\pi z\cdot \sqrt2 dz$ that gives
$$
\frac{1}{\sqrt2}\iint_{S_2}z\,dA=2\pi\int_0^3z^2\,dz=18\pi.
$$

*Adding together we get $-27\pi+18\pi=-9\pi$.


P.S. Using the Gauss-Ostrogradsky theorem, the problem is actually trivial, since the divergence of $F$ is $1$, so the space integral is simply the volume of the cone $\frac13\cdot\pi3^2\cdot 3=9\pi$. (Note the sign - the theorem is for the outward normal.)
