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bounded by plane $x+y+z=1$ and $x^2 +y^2+z^2 = 1$.

I need this part to solve https://math.stackexchange.com/questions/325143/change-of-variables-multiple-integrals-volume-question

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sarah
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2 Answers2

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It is a spherical cap. You know distance of the plane from centre. $V = \frac{\pi h}{6} (3a^2 + h^2)$(Picture from Wikipedia) How do we get this?

1From Wiki

Change your z axis to this. Now use Cylindrical co-ordinates V=$\int\int\int r1d\theta d(r1)dz. $ r1 is the radius variable.

What will be the limits? Look at this like making the volume through discs. It is clear that $\theta$ will change from 0 to 2$\pi$. r1 will change with z. The limits of r1 will involve z somewhere and z will change from R(radius of sphere)-h to h. Check your answer with the one above mentioned. http://en.wikipedia.org/wiki/Spherical_cap

  • I still dont understand. Could you explain more in detail? I guess we use the distance from orgin to plane and radius of sphere equal to 1 to find a by pythagorean thm? – sarah Mar 09 '13 at 15:29
  • I still dont understand how to find the limits for r and z for the integral. – sarah Mar 09 '13 at 15:31
  • I think limit for z is from (1-x-y) to (sqrt(1-v^2-w^2)) – sarah Mar 09 '13 at 15:33
  • and r is from 0 to a? – sarah Mar 09 '13 at 15:34
  • How do we change z axis to the h axis? – sarah Mar 09 '13 at 15:44
  • SO i get a= root(2/3). and distance from plane to origin=1/root(3) – sarah Mar 09 '13 at 15:50
  • Sorry for the late reply.1)Volume is independent of the choice of axis.(Essentially, rotating the sphere,.2)Regarding the limits- Here is a link regarding Cylindrical co-ods.http://en.wikipedia.org/wiki/Cylindrical_coordinate_system. Now r1 which is the radial component, changes from 0 to sqrt( R^2-z^2) and z will change from R-h to h. PS Don't confuse the variables of the figure with the other variables. –  Mar 18 '13 at 16:49
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In general, $V=\pi \int_a^b[f(x)]^2dx$. $$a^2=R^2-(R-x)^2=2Rx-x^2$$ $$V=\pi \int_0^h[a(x)]^2dx=\frac{\pi h^2}{3}(3R-h)$$.

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