For an assignment I am asked to find the volume of a given volume R, namely $R=\left\{(x,y,z):0\leq z\leq\sqrt{4-x^2-y^2},(x-1)^2+y^2\leq1\right\}$.
I have attempted solving this using cylindrical coördinates, which I believe is the way to go, but I don't quite understand the volume transformation.
For solving the volume I have tried solving $\int_{-\pi/2}^{\pi/2}\int_0^{2\cos{\varphi}}\int_0^{\sqrt{4-r^2}}r\,dz\,dr\,d\varphi$, but this gives the wrong answer (it results in $\frac{8}{3}\pi$ whereas it should be $\frac{8}{3}\pi-\frac{32}{9}$).
Integrating with respect to $\varphi$ from $0$ to $\pi$ however yields the correct result. This seems odd to me as the circle in the $xy$ plane over which we integrate the function is in the domain $\varphi=[-\pi/2,\pi/2]$.
How come the $\varphi$ integration bounds are $0$ and $\pi$?
Your approach is correct: $$\int_{-\pi/2}^{\pi/2}\int_{r=0}^{2\cos{\varphi}}r\int_{z=0}^{\sqrt{4-r^2}} dz dr d\varphi=\int_{-\pi/2}^{\pi/2}\int_{r=0}^{2\cos{\varphi}}r\sqrt{4-r^2} dr d\varphi\\= \frac{8}{3}\int_{-\pi/2}^{\pi/2}(1-|\sin(\varphi)|^3) d\varphi =\frac{8}{3}\pi-\frac{32}{9}.$$ Have you considered the modulus at the last step?