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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?

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

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

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

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

enter image description here

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  • $\begingroup$ Looks good, thanks. I had already corrected the bounds for $r$, but I'm still not quite sure how to get the bounds for $z$. $\endgroup$ – dakisbac Mar 10 '14 at 14:27
  • $\begingroup$ I have given graphical explanation of the intersection of the spheres projected on x-z axis. Goodluck. Vote if possible. $\endgroup$ – Satish Ramanathan Mar 10 '14 at 16:12
  • $\begingroup$ Thanks. I'm a couple reputation shy of being able to vote up the answer, but I will when I can. $\endgroup$ – dakisbac Mar 10 '14 at 16:17
  • $\begingroup$ @dakisbac, I am very proud of you, You kept up your word!! $\endgroup$ – Satish Ramanathan Mar 13 '14 at 19:42
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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 $\rho=2\cos\varphi$: $$ V= \int_0^{2\pi}\int_0^{\pi/3}\int_0^1\rho^2\sin\varphi d\rho d\varphi d\theta+ \int_0^{2\pi}\int_{\pi/3}^{\pi/2}\int_0^{2\cos\varphi}\rho^2\sin\varphi d\rho d\varphi d\theta=2\pi\left({1\over 6}+{1\over 24}\right) $$

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  • $\begingroup$ Yep, this is what I have. $\endgroup$ – dakisbac Mar 10 '14 at 14:25
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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 $$ \text{area of the intersection}=2\int_{1/2}^{1}\pi \sqrt{1-z^2}^2dz=2\pi\int _{1/2}^{1}1- z^2dz=2\pi(1-\frac{1}{2}-\frac{1}{3}+\frac{1}{24})=\frac{5}{12}\pi $$

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    $\begingroup$ Darn! Beat me to it. $\endgroup$ – William Chang Mar 9 '14 at 23:06
  • $\begingroup$ 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. $\endgroup$ – dakisbac Mar 10 '14 at 6:03
  • $\begingroup$ Don't you think the z should run from $$\int_{\sqrt{1-r^2}}^{1+ \sqrt{1-r^2}}$$ $\endgroup$ – Satish Ramanathan Mar 10 '14 at 6:16
  • $\begingroup$ @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 $3\pi/2$ $\endgroup$ – dakisbac Mar 10 '14 at 6:32
  • $\begingroup$ @dakisbac, see the solution using cylindrical coordinates. $\endgroup$ – Satish Ramanathan Mar 10 '14 at 8:10

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