TITLE: Find volume between two spheres using cylindrical & spherical coordinates
QUESTION [8 upvotes]: I've got two spheres, one of which is the other sphere just shifted, and I'm trying to find the volume of the shared region. The spheres are $x^2 + y^2 +z^2 = 1$ and $x^2 + y^2 +(z-1)^2 = 1$ 
I know how to transform the variables into cylindrical and spherical coordinates but I'm having trouble figuring out the bounds.
How do I do this?
EDIT: Based on Kaladin's answer, which helped me realize the bounds for $r$, would it be correct to express the volume of the region as follows? (as cylindrical coordinates)
$$V = 2\int_0^{2\pi} \int_{1/2}^1 \int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} rdzdrd\theta$$
EDIT 2: Assuming I integrated the above integral properly, that equals $\frac{2\pi\sqrt{2}}{3}$, which is obviously not Kaladin's answer. What's the problem?

REPLY [2 votes]: There is a way to do this problem with only one integral in spherical coordinates, and it is easier than the cylindrical coordinates version because there are no square roots to contend with. It's
$$\int_0^{2\pi} \int_0^1 \int_0^{\cos^{-1}\left(\frac{\rho}{2}\right)} \rho^2 \sin\varphi d\varphi d\rho d\theta$$
which integrates to
$$ 2\pi \int_0^1 -\rho^2\cos\varphi\Bigr|_0^{\cos^{-1}\left(\frac{\rho}{2}\right)} d\rho = 2\pi \int_0^1 \rho^2 - \frac{1}{2}\rho^3 d\rho = \frac{5\pi}{12}$$