I'm trying to find the volume bounded by a sphere and an elyptical cylinder. The sphere is given by $x^2+y^2+z^2=1$ and the elyptical cylinder by $2x^2+y^2-2x=0$.
My first attempt with spherical coordinates didn't go well, now I did some work with cylindrical coordinates but I'm not sure if I have the right answer.
Playing with the equation for the cylinder I found the canonical form of the ellipse: $\displaystyle\frac{\left(x-\frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2}+ \displaystyle\frac{y^2}{\left(\frac{1}{\sqrt{2}}\right)^2}=1$.
And this can be se clearer setting $z=0$ and looking the ellipse drawn on the $xy$-plane
Which seems to involve some kind of ugly looking function if I were to describe the ellipse in polar coordinates, which in some way I have to.
Now to find the volume I changed to cylindrical coordinates $$x=\rho\cos\theta, y=\rho\sin\theta,z=z$$
The equation for the ellipse states $x^2+y^2-2x=0$ then $2\rho^2\cos^2\theta+\rho^2\sin^2\theta-\rho\cos\theta$ from which I have $\rho^2(2\cos^2\theta+\sin^2\theta)-\rho\cos\theta\implies \rho(\theta)=\displaystyle\frac{\cos\theta}{2\cos^2\theta+\sin^2\theta} = \displaystyle\frac{\cos\theta}{1+\cos^2\theta}$ with $0\leq\theta\leq\pi$.
Now, for $z$ considering the sphere $x^2+y^2+z^2=1$ follows $\rho^2+z^2=1\implies -\sqrt{1-\rho^2}\leq z \leq \sqrt{1-\rho^2}$.
The volume seems to be given by $\displaystyle\int_0^{\pi}\displaystyle\int_0^{\frac{\cos\theta}{1+\cos^2\theta}}\rho\displaystyle\int_{-\sqrt{1-\rho^2}}^{\sqrt{1-\rho^2}}\;dzd\rho d\theta$.
For me here is when this become tricky, I'm going to calculate this integrals one at a time.
(1)$\displaystyle\int_{-\sqrt{1-\rho^2}}^{\sqrt{1-\rho^2}}\;dz = 2\sqrt{1-\rho^2}$
(2)$\displaystyle\int_0^{\frac{\cos\theta}{1+\cos^2\theta}}\rho (2\sqrt{1-\rho^2})d\rho = \displaystyle\int_0^{\frac{\cos\theta}{1+\cos^2\theta}}2\rho\sqrt{1-\rho^2} d\rho = -\displaystyle\frac{2}{3}(1-\rho^2)^{3/2}|_{\rho=0}^{\rho=\frac{\cos\theta}{1+\cos^2\theta}} = \left[1-\left(\displaystyle\frac{\cos\theta}{1+\cos^2\theta}\right)^2\right]^{3/2} + \displaystyle\frac{2}{3}$ $=\left[\displaystyle\frac{\cos^2\theta}{1+2\cos^2\theta+\cos^2\theta}\right]^{3/2} + \displaystyle\frac{2}{3}$
(3) Stuck. I don't know what could I do to integrate $\left[1-\left(\displaystyle\frac{\cos\theta}{1+\cos^2\theta}\right)^2\right]^{3/2}$
For the term $\left[1-\left(\displaystyle\frac{\cos\theta}{1+\cos^2\theta}\right)^2\right]$ I considered using partial fractions as follows: $\left(\displaystyle\frac{\cos\theta}{1+\cos^2\theta}\right)^2 =\displaystyle\frac{1+\cos^2\theta+\cos^4\theta}{(1+\cos^2\theta)^2} = 1 + \displaystyle\frac{1}{(1+\cos^2\theta)^2} + \displaystyle\frac{1}{1+\cos^2\theta}$
But this would make the last integral harder because I'd have to integrate $\displaystyle\int_0^{\pi} \left[1 + \displaystyle\frac{1}{(1+\cos^2\theta)^2} + \displaystyle\frac{1}{1+\cos^2\theta}\right]^{3/2}\;d\theta$
Any ideas?.