Compute $$\iiint\frac{xz}{1+x^2+y^2}\,dz\,dy\,dx,$$ where $1≤x^2+y^2≤3, 0≤z≤3$.
I've tried it. But I'm only confused with $\theta$. I think it should be $0$ to $2\pi$, but that'll make the whole value zero. We have to solve this by the cylindrical method. Without solving for theta I get $\frac{9}{2{(π/6)-4}}$.
Passing to cylindrical coordinates $$\begin{matrix} x&=& r \cos \theta \\ y&=& r \sin \theta \\ z&=& z \\ \end{matrix}$$ you get $$I= \int_0^3 \left( \int_0^{2 \pi}\left( \int_1^3 r \cdot \frac{zr \cos \theta}{1+r^2} dr\right)d \theta \right)dz$$ You can separate everything and get three integrals $$I= \int_0^3 zdz \int_0^{2 \pi} \cos \theta d \theta \int_1^3 \frac{r^2}{1+r^2} dr$$ Since $\int_0^{2 \pi} \cos \theta d \theta =0$ the whole integral is zero.