volume between a plane, a cone and two cylinders Compute the volume which is between:
The plane $z=0$ the cone $z=\sqrt{x^2+y^2}$ and the cylinders $x^2+y^2=x$, $x^2+y^2=2x$.
I feel that it may be helpful to change the variables to cylindrical coordinates, but I find it hard to identify the domain of each $r,\theta,z$.
Edit: Can we say:
$2x\geq x^2+y^2\geq x$, so $2r\cos\theta\geq r^2\geq r\cos\theta$, then $2\cos\theta\geq r\geq \cos\theta$.
Then $0\leq z\leq r$, and $-\pi/2\leq\theta\leq\pi/2$.
and now do the integration?
 A: Hint:
I'd rather try cylindrical coordinates, as the setting is clearly anisotropic.
The plane is $z=0$, the cone $z=r$ and the cylinders $r^2=r\cos\theta$ and $r^2=2r\cos\theta$ or $r=\cos\theta,r=2\cos\theta$, hence
$$\int_{\theta=\theta_0}^{\theta_1}\int_{r=\cos\theta}^{2\cos\theta}\int_{z=0}^rdV$$
Remains to determine the range of $\theta$ (such that $r\ge0$).

In spherical coordinates, for the sake of the exercise, the surfaces are
$$\rho\cos\phi=0,\rho\cos\phi=\rho\sin\phi,\rho^2\sin^2\phi=\rho\cos\theta\sin\phi,\rho^2\sin^2\phi=2\rho\cos\theta\sin\phi$$ or
$$\rho=0\lor\sin\phi=0\lor(\cot\phi=1,\rho\sin\phi=\cos\theta,\rho\sin\phi=2\cos\theta).$$
Then
$$\int_{\phi=\phi_0}^{\phi_1}\int_{\theta=\theta_0}^{\theta_1}\int_{\rho=\cos\theta/\sin\phi}^{2\cos\theta/\sin\phi}dV$$ and a deeper discussion of the signs and ranges is required.
A: Hint. 
Why don't you use cylindrical coordinates $(\rho,\theta,z)$?
Then the $xy$ domain $D=\{(x,y):x\leq x^2+y^2\leq 2x\}$ (the region between the circles $C_1(1)$ and $C_{1/2}(1/2)$ where $C_r(P)$ is the circle of radius $r$ and center $P$) becomes 
$$\cos(\theta)\leq \rho\leq 2\cos(\theta)$$
with $\theta\in [-\pi/2,\pi/2]$. Hence, by symmetry,
$$V=\iint_{D}\left(\int_{z=0}^{\rho} dz\right) \rho d\rho d\theta=2\int_{\theta=0}^{\pi/2}\int_{\rho=\cos(\theta)}^{2\cos(\theta)} \rho^2 d\rho d\theta.$$
Can you take it form here?
