I am trying to find the volume above $\ z = 0 $ under $\ z = x^2 + y^2 $ and inside the cylinder $\ x^2+y^2 = 2x $ using double integral.
the intersection between the cylnder and the paraboloid is the plane $\ z = 2x $ and the project on $\ xy $ plane is just circle at $\ 1,0 $ . so suppose $\ y $ values run from $\ -1 $ to $\ 1 $ and $\ x $ from $\ y=\sqrt{2x-x^2} $ to $\ y= - \sqrt{2x-x^2} $ I get
$$\ \int_{-1}^1 \int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}} 2x \ dy \ dx $$
so doing such integral is obviously not the purpsoe of the excercise. So I tried to convert to polar form which should be
$$\ \int\int_R 2r \cos \theta \ r \ dr \ d\theta $$
but here I get stuck, I mean i can imagine the circle at $\ 1,0 $ but how do I know exactly what are the angles of $\theta $ and I know the smallest $\ r $ is the $\ 0 $ and the largest $\ 2 $ but not sure if this is what I should use?
Note that the base of your cylinder is the inside of the circle $$\ x^2+y^2 = 2x$$ which in polar coordinates is described by $$r=2\cos(\theta)$$
Also your upper surface $$z=x^2+y^2$$ is simply $$z=r^2$$.
Thus in polar coordinates you find the volume by $$v=2\int _{0}^{\pi/2}\int_{0}^{2\cos(\theta)}(r^2)rdrd\theta$$