I need to find the volume of $x^2+y^2=y$ bounded by $x^2+y^2+z^2=2$. I am capable of doing the integral, but I'm not sure how do I find the bounds for the integral. I know it's a triple integral. Should I just convert into cylindrical coordinates and calculate?
Any help?
Note that $x^2+y^2\leqslant y\iff x^2+\left(y-\frac12\right)^2\leqslant\frac14$. So, $0\leqslant y\leqslant1$ and therefore if you express the volume that you want to compute in cylindrical coordinates, you will have $0\leqslant\theta\leqslant\pi$. On the other hand,$$x^2+y^2\leqslant y\iff r^2\leqslant r\sin(\theta)\iff r<\sin(\theta)$$and$$x^2+y^2+z^2\leqslant2\iff z^2\leqslant2-r^2.$$So, compute the integral$$\int_0^\pi\int_0^{\sin(\theta)}\int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$