Find volume between two spheres using cylindrical & spherical coordinates

Question

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?

Answer 1

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 $$u = 1-r^2 ; r = 0 => u = \frac{1}{4} ; r = \sqrt{\frac{3}{4}} => u = 1$$

$$V = 2\pi [-\int_1^{\frac{1}{4}} u^{\frac{1}{2}} du - \int_{0}^{\sqrt{\frac{3}{4}}} rdr]$$

$$V= 2\pi (\frac{2}{3}u^{\frac{3}{2}}) - (\frac{r^2}{2})$$ $$V =2\pi*( \frac{2}{3}(1-\frac{1}{8}) - \frac{3}{8})$$

$$V = 2\pi*(\frac{14}{24} - \frac{3}{8}) = 2\pi*\frac{5}{24} = \frac{5}{12} \pi$$

Answer 2

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 $$ \text{area of the intersection}=2\int_{1/2}^{1}\pi \sqrt{1-z^2}^2dz=2\pi\int _{1/2}^{1}1- z^2dz=2\pi(1-\frac{1}{2}-\frac{1}{3}+\frac{1}{24})=\frac{5}{12}\pi $$

Answer 3

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 $\rho=2\cos\varphi$: $$ V= \int_0^{2\pi}\int_0^{\pi/3}\int_0^1\rho^2\sin\varphi d\rho d\varphi d\theta+ \int_0^{2\pi}\int_{\pi/3}^{\pi/2}\int_0^{2\cos\varphi}\rho^2\sin\varphi d\rho d\varphi d\theta=2\pi\left({1\over 6}+{1\over 24}\right) $$