# 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 $z$. – dakisbac Mar 10 '14 at 14:27
• I have given graphical explanation of the intersection of the spheres projected on x-z axis. Goodluck. Vote if possible. – Satish Ramanathan Mar 10 '14 at 16:12
• Thanks. I'm a couple reputation shy of being able to vote up the answer, but I will when I can. – dakisbac Mar 10 '14 at 16:17
• @dakisbac, I am very proud of you, You kept up your word!! – Satish Ramanathan Mar 13 '14 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). – hash man Jun 19 '20 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. – dakisbac Mar 10 '14 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. – Learning Apr 29 '20 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 – Ninad Munshi Apr 30 '20 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. – William Chang Mar 9 '14 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. – dakisbac Mar 10 '14 at 6:03
• Don't you think the z should run from $$\int_{\sqrt{1-r^2}}^{1+ \sqrt{1-r^2}}$$ – Satish Ramanathan Mar 10 '14 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 $3\pi/2$ – dakisbac Mar 10 '14 at 6:32
• @dakisbac, see the solution using cylindrical coordinates. – Satish Ramanathan Mar 10 '14 at 8:10