I want to find the volume of the region inside the ellipsoid $$\frac{x^2}{4}+\frac{y^2}{4}+z^2=1$$ and the cylinder $$x^2+(y-1)^2=1$$
I tried shifting the axes so that the cylinder was centered at the origin, then evaluating an integral in cylindrical/polar coordinates. $$\int^{2\pi}_{0}\int^{1}_{0}\int^{\frac{1}{2}\sqrt{4-r^2-2rsin(\theta)-1}}_{-\frac{1}{2}\sqrt{4-r^2-2rsin(\theta)-1}}r dzdrd\theta$$
$$\int^{2\pi}_{0}\int^{1}_{0}r\sqrt{3-r^2-2rsin(\theta)} drd\theta$$
However, this integral gets messy. Is there an easier method of finding the volume?
Best Answer
You can do it using ordinary cylindrical coordinates. Since the projection of that cylinder onto the plane $z=0$ is the circle centered at $(0,1)$ with radius $1$, which is the union if the first and the second quadrants, $0\leqslant\theta\leqslant\pi$. The condition$$\frac{x^2}4+\frac{y^2}4+z^2\leqslant1\quad\text{becomes}\quad\frac{r^2}4+z^2\leqslant1$$and the condition$$x^2+(y-1)^2\leqslant1\quad\text{becomes}\quad r^2-2r\sin\theta\leqslant0.$$So, compute the integral$$\int_0^\pi\int_0^{2\sin\theta}\int_{-\sqrt{1-r^2/4}}^{\sqrt{1-r^2/4}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$You should get $\dfrac83\pi-\dfrac{32}9$.