Triple integral of $|z|$

Question

I have to calculate the $$\int_A |z| \,dx dy dz $$ with $A=\{(x,y,z): x^2+y^2+z^2\le4, x^2+y^2-2y\le0\}$. Do I use cylindrical or spherical coordinates?

Answer 1

You could use cylindrical coordinates to get

$\displaystyle\int_0^{\pi}\int_0^{2\sin\theta}\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}|z|\cdot rdz drd\theta=2\int_0^{\pi}\int_0^{2\sin\theta}\int_0^{\sqrt{4-r^2}}z\cdot r\;dz\;dr\;d\theta$

$=\displaystyle2\int_0^{\pi}\int_0^{2\sin\theta}r\cdot\frac{1}{2}(4-r^2)dr\;d\theta=\int_0^{\pi}\int_0^{2\sin\theta}(4r-r^3)dr\;d\theta$

$\displaystyle=\int_0^{\pi}(8\sin^2\theta-4\sin^4\theta)d\theta=\int_0^{\pi}\left(\frac{5}{2}-2\cos2\theta-\frac{1}{2}\cos4\theta\right)d\theta=\frac{5\pi}{2}$