Find volume between two spheres using cylindrical & spherical coordinates 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?
 A: Answer using Cylindrical Coordinates:
Volume of the Shared region = 
Equating both the equations for z, you get z = 1/2.  Now substitute z = 1/2 in in one of the equations and you get r = $\sqrt{\frac{3}{4}}$.
Now the sphere is shifted by 1 in the z-direction, Hence
Volume of the Shared region = $$\int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{3}{4}}} \int_{1-\sqrt{1-r^2}}^{\sqrt{1-r^2}} rdzdrd\theta$$
$$V=2\pi \int_{0}^{\sqrt{\frac{3}{4}}} [2{\sqrt{1-r^2}}-1] rdr$$
substitute $$u = 1-r^2  ; r = 0 => u = \frac{1}{4} ; r = \sqrt{\frac{3}{4}} => u = 1$$
$$V = 2\pi [-\int_1^{\frac{1}{4}} u^{\frac{1}{2}} du - \int_{0}^{\sqrt{\frac{3}{4}}} rdr]$$
$$V= 2\pi (\frac{2}{3}u^{\frac{3}{2}}) - (\frac{r^2}{2})$$
$$V  =2\pi*( \frac{2}{3}(1-\frac{1}{8}) - \frac{3}{8})$$
$$V = 2\pi*(\frac{14}{24} - \frac{3}{8}) = 2\pi*\frac{5}{24} = \frac{5}{12} \pi$$

A: In spherical coordinates the intersection points $r=\sqrt 3/2$, $z=1/2$ have colatitude $\varphi_0=\arctan\sqrt 3=\pi/3$ and the second sphere is $\rho=2\cos\varphi$:
$$
V=
\int_0^{2\pi}\int_0^{\pi/3}\int_0^1\rho^2\sin\varphi d\rho d\varphi d\theta+
\int_0^{2\pi}\int_{\pi/3}^{\pi/2}\int_0^{2\cos\varphi}\rho^2\sin\varphi d\rho d\varphi d\theta=2\pi\left({1\over 6}+{1\over 24}\right)
$$
A: 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
$$ 2\pi \int_0^1 -\rho^2\cos\varphi\Bigr|_0^{\cos^{-1}\left(\frac{\rho}{2}\right)} d\rho = 2\pi \int_0^1 \rho^2 - \frac{1}{2}\rho^3 d\rho = \frac{5\pi}{12}$$
A: When the circle if you take $z$ fixed has radius $\sqrt{1-z^2}$ (pythagoras) so that circle has as area $\pi \sqrt{1-z^2}^2$. Now note that the area of the intersection is twice the area of the part of the sphere where $1/2\leq z\leq 1$ Therefor
$$
\text{area of the intersection}=2\int_{1/2}^{1}\pi \sqrt{1-z^2}^2dz=2\pi\int _{1/2}^{1}1- z^2dz=2\pi(1-\frac{1}{2}-\frac{1}{3}+\frac{1}{24})=\frac{5}{12}\pi
$$
