# 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?

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

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

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

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

• Looks good, thanks. I had already corrected the bounds for $r$, but I'm still not quite sure how to get the bounds for $z$. Commented Mar 10, 2014 at 14:27
• I have given graphical explanation of the intersection of the spheres projected on x-z axis. Goodluck. Vote if possible. Commented Mar 10, 2014 at 16:12
• Thanks. I'm a couple reputation shy of being able to vote up the answer, but I will when I can. Commented Mar 10, 2014 at 16:17
• @dakisbac, I am very proud of you, You kept up your word!! Commented Mar 13, 2014 at 19:42
• I believe this to be not exact, the limit of the outer integral should be $\pi$ and then from symmetry, you get the final result (doubling by 2). Commented Jun 19, 2020 at 12:58

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

• Yep, this is what I have. Commented Mar 10, 2014 at 14:25

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

$$2\pi \int_0^1 -\rho^2\cos\varphi\Bigr|_0^{\cos^{-1}\left(\frac{\rho}{2}\right)} d\rho = 2\pi \int_0^1 \rho^2 - \frac{1}{2}\rho^3 d\rho = \frac{5\pi}{12}$$

• +1)Cleverly used change of order. Commented Apr 29, 2020 at 13:57
• @Learning thank you! I didn't know if anyone would see this answer but I'm glad someone benefitted from this perspective Commented Apr 30, 2020 at 2:18

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

• Darn! Beat me to it. Commented Mar 9, 2014 at 23:06
• This answer is correct, and I have verified it using spherical coordinates. I still can't figure out how to do it using cylindrical coordinates though. Commented Mar 10, 2014 at 6:03
• Don't you think the z should run from $$\int_{\sqrt{1-r^2}}^{1+ \sqrt{1-r^2}}$$ Commented Mar 10, 2014 at 6:16
• @satishramanathan: That doesn't work either, and does not make sense to me. Using your bounds for $z$ and the same bounds above for $r$ and $\theta$ I got $3\pi/2$ Commented Mar 10, 2014 at 6:32
• @dakisbac, see the solution using cylindrical coordinates. Commented Mar 10, 2014 at 8:10