I am studying integration in polar coordinates using this textbook material:
Evaluate region between two cylinders $x^2+y^2=1$ and $x^2+y^2=4$ and between $z=x+2$ and $xy$-plane.
And they say that the answer is $0$. The formula used is this:
$$\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}} = \int_{{\,\alpha }}^{{\,\beta }}{{\int_{{{h_1}\left( \theta \right)}}^{{{h_2}\left( \theta \right)}}{{\int_{{{u_1}\left( {r\cos \theta ,r\sin \theta } \right)}}^{{{u_2}\left( {r\cos \theta ,r\sin \theta } \right)}}{{r\,f\left( {r\cos \theta ,r\sin \theta ,z} \right)\,dz}}\,dr}}\,d\theta }}$$
Although what is $f\left(r \sin\theta ,r \cos \theta ,z\right)$? I believe for cylinders it's just $rdzdrd\theta$.
Anyhow, here's my solution that shows
$$\int _0^{2pi}\:\int _1^2\:\int _0^{r\cos\theta +2}\:rdzdrd\theta = 6\pi $$
I've no idea if this is right, but surely it's not $0$?
Yes, it is $0$. What is being evaluated here is$$\iiint_E\color{red}y\,\mathrm dx\,\mathrm dy\,\mathrm dz,\tag1$$not the volume of $E$ (which is indeed greater than $0$). Note that $(x,y,z)\in E\iff(x,-y,z)\in E$; it follows from this that $(1)$ is indeed equal to $0$.