i'm trying to find the volume of the solid which is the intersection of the solid sphere $x^2+y^2+z^2\leqq1 \; , 2x^2+y^2-2x=0$
My attempt:
I tried with cylindrical coordinates $$x=rcos(\phi)\;,y=rsin(\phi)\;,z=z$$
$$(rcos(\phi))^2+(rsin(\phi))^2+z^2=r^2+z^2 \leqq 1 \Rightarrow -\sqrt{(1-z^2)} \leqq r \leqq \sqrt{(1-z^2)} $$ I'm stuck here, because i don't know how to use $2x^2+y^2-2x=0$ to find the other limits of integration.
Well, first of all, the second equation describes a shifted elliptical cylinder. From $2x^2+y^2-2x=0$, divide by $2$ and complete the square, yielding $$ (x-1/2)^2+y^2/2=1/4, $$ which writing in standard form yields $$ \frac{(x-1/2)^2}{1/4}+\frac{y^2}{1/2}=1. $$ Hence, the natural $(x,y)$ parameterization is \begin{align} x(r,\theta)&=\frac12+\frac{r}{4}\cos\theta \\ y(r,\theta)&=\frac{r}{2}\sin\theta \end{align} with $0\leq r \leq 1$ and $0\leq \theta \leq 2\pi$. Then calculate the Jacobian and produce a triple integral. Using the two equations, you should be able to solve for the $z$ bounds in terms of $x$ and $y$ (and therefore, in terms of $r$ and $\theta$).