I want to find the volume remaining (the smaller volume) after the plane $P: x+y+z = 1$ cuts the sphere $S: x^{2} + y^{2} + z^{2} = 1$. I found that the problem could be solved by finding a plane (or line) perpendicular to the plane P, then treating the remaining volume as if it were the cap of a sphere (say when a plane $z = 1/2$ cuts the sphere giving a spherical cap), using as the "height" the distance between the sphere and the plane P (along that perpendicular plane).

However, I tried to solve the problem without using this fact by trying to use triple integrals (both in cylindrical and spherical coordinates). In cylindrical coordinates, I had that (I split up the integral into 4 different regions, the "bigger part" in the first octant and the three other smaller parts in the other octants).

For the first integral, we would have that the projection in the xy plane is the line $x+y=1$ and the circle $x^{2} + y^{2} + 1$:

$V = \displaystyle\int_0^{(\pi/2)} \displaystyle\int_{\frac {1}{\cos\theta + \sin\theta}}^1 \displaystyle\int_{1-r\cos\theta - r\sin\theta}^{\sqrt{1-r^{2}}} \! r \, \mathrm{d}z\mathrm{d}r\mathrm{d}\theta$

And then, for the "smaller volume" integral, the projection would be the same but in this case it is between the plane $ z = 0$ and the plane $x+y+z=1$. I assumed that due to symmetry, the volume of the smaller pieces would be equal (thus multiplying by 3):

$V = \displaystyle\int_0^{(\pi/2)} \displaystyle\int_{\frac {1}{\cos\theta + \sin\theta}}^1 \displaystyle\int_{1-r\cos\theta - r\sin\theta}^{0} \! r \, \mathrm{d}z\mathrm{d}r\mathrm{d}\theta$

However, I am not getting the correct answer, which should be:

$V = \pi(\displaystyle\frac{2}{3} - \displaystyle\frac{8}{9\sqrt{3}}) \approx 0.482$

Could someone give me a hint on how to set up the correct integral or as well explaining why the method I used is incorrect? I am sorry for the long post, but this problem has been troubling my mind for around a week and I cannot get it off my head.

Edit: I finally figured out how to set up the integral for the "larger volume" part. It seems that cross-sections must be constant when dealing with the triple integral. The volume will be:

$V = V = \displaystyle\int_0^1 \displaystyle\int_0^{1-x} \displaystyle\int_{1-x-y}^{{(1-x^2-y^2)}^{1/2}} \! 1 \, \mathrm{d}z\mathrm{d}y\mathrm{d}x + \displaystyle\int_0^1 \displaystyle\int_{1-x}^{{(1-x)}^{1/2}} \displaystyle\int_{0}^{{(1-x^2-y^2)}^{1/2}} \! 1 \, \mathrm{d}z\mathrm{d}y\mathrm{d}x = \displaystyle\frac{\pi}{6} - \displaystyle\frac{1}{6}$

I'm still wondering how to set up the integral for the "smaller parts" though. Thanks for everything thus far.


I recommend that you rotate your sphere so that the plane is perpendicular to the $z$ axis. The domain of integration then becomes quite natural in cylindrical coordinates and the resulting integral is quite manageable.

Since you asked for a hint, I'll leave it at that for the time being but could edit this answer to include a solution later.

  • $\begingroup$ Yes that can be done, but is it possible to set up an integral without rotating the axis/sphere? Isn't it that the reflection in the $xy$ plane is a line and part of a circle? Or why wouldn't this be the case? As well, do we have to change any terms or do any sort of transformation by rotating the sphere or we leave it as it is? Thank you for your answer. $\endgroup$ – arcbloom Mar 3 '14 at 1:15
  • $\begingroup$ I suppose that you could do a change of variables, which amounts to the same thing. I don't think you could over-emphasize the importance of choosing a correct description (i.e. coordinate system) for the this problem! $\endgroup$ – Mark McClure Mar 3 '14 at 1:39
  • $\begingroup$ Thanks for looking into the problem. I found it using your method. Now, out of pure curiosity, is it possible to evaluate this integral without using a transformation of variables? (except for cylindricals or sphericals). I got that the "higher volume" is $\approx 0.357$. I'm not sure on how to set up the other integrals of the "smaller volumes" though. $\endgroup$ – arcbloom Mar 3 '14 at 16:26

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