I'm learning about the divergence theorem. If I have a vector function $f(x,y,z)=\sqrt {x^2+y^2} \cdot (x,y,z)$ and I want to get $\iint\limits_A f(x,y,z) \, d A $ (easy to evaluate, but I thought I'd practice converting from rectangular to another coordinate system), which - according to the theorem - is equal to $ \iiint\limits_V \nabla \cdot f(x,y,z) \, dV$, then if -in cylindrical coordinates- the volume is from $z=0 \ to \ 5$ and radius $r=2$, the result should be 80$\pi$. The function can be written as $f(r, \theta , z) = (r^2,\theta, 0)$, and so $\nabla \cdot f = (2r, 1/r, 0)$, as $\nabla = (\partial / \partial r , 1/r \cdot \partial / \partial \theta, \partial / \partial z)$.
Evaluating, the integral becomes ( as $dV = r\cdot drd\theta dz$) $\iiint \limits_V (2r^2 + 1) \, dr d\theta dz $, which is not $80 \pi$, it is $10\cdot \pi \cdot (16/3 + 2) $. What did I do wrong?
EDIT:
$\textbf{f}(x,y,z) = (\sqrt{x^2+y^2})\cdot (x \textbf{i} + y \textbf{j})$ is the function; its cylindrical form is
$\textbf{f}=r^2 \cdot \hat{r} + \theta \cdot \hat{\theta} + 0 \cdot \hat{z} $
I also know $\nabla$ for the new coordinates: $\nabla = \partial / \partial r \cdot \hat{r} + (1/r)\cdot \partial / \partial \theta \cdot \hat{\theta} + \partial / \partial z \cdot \hat{z}$
$\iiint \limits_V \nabla \cdot \textbf{f} \, dV = \iint\limits_A \textbf{f} \cdot \textbf{n}\, dA $. The second integral is easy, but with the volume integral I had the problem which is again: $\iiint\limits_V \nabla \cdot \textbf{f}\, dV $, in which I have already written the scalar product: it is 2r + 1/r, and as $dV = rdr d \theta dz$, I get $ 2 r^2 + 1 $ in the volume integral, which won't become $80 \pi$.
I don't know what you mean by $f(r,\theta,z)$. At any rate, we have the vector field $${\bf f}(x,y,z):=(\rho x,\rho y,\rho z),\qquad\rho:=\sqrt{x^2+y^2}\ .$$ One computes $$\rho_x={x\over\rho},\quad \rho_y={y\over\rho},\quad\rho_z=0\ ,$$ so that one obtains $${\rm div}\>{\bf f}(x,y,z)={x\over\rho} x+\rho+{y\over\rho} y+\rho+\rho=4\rho\ .$$ If $V$ denotes the given cylinder and $A$ its surface (mantle, top, and bottom) oriented outwards then Gauss' theorem says that $$\int_A {\bf f}\cdot {\bf n}\ {\rm d}\omega=\int_V {\rm div}\,{\bf f}\ {\rm d}(x,y,z)=2\pi\cdot 5\cdot\int_0^2 4\rho\>\rho d\rho={320\pi\over3}\ .$$