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Multiple integrals - volume of part of a unit sphere

  1. Sep 27, 2007 #1
    Find the volume of that portion of a unit sphere for which 0<theta<alpha, where theta is one of the spherical coordinates

    So i know the equation is z^2 = x^2 + y^2 , but what is the meaning of 0<theta<alpha?
    where do i start? I know i must convert to polar coordinates.
    z^2 = r^2
  2. jcsd
  3. Sep 27, 2007 #2


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    Are you sure that's the equation of a unit sphere?

    You are going to need to check what conventions you are using in your book or course for the angles in spherical coordinates. (Unhappily, there are a number of versions.) Is theta the "polar angle" or co-latitude? Is alpha the "azimuthal" angle? Or is it vice versa? Conventionally, both angles generally are started from zero. They want you to look at the region where alpha must be larger than theta, but which angle is which is going to make a difference!
  4. Sep 27, 2007 #3
    oh sorry z^2 = x^2 + y^2 - 1
    so z^2 = r^2 - 1
    will i need to isolate for z?

    it doesn't say, it just says it is from polar rectangle
    but it states that: if f is continuous on a polar rectangle R given by 0<=a<=r<=b, alpha <=theta<=beta, where 0<=beta - alpha <= 2pi, then
    [tex]\int\int_{R} f(x,y) dA = \int^{\beta}_{\alpha}\int^{b}_{a} f( rcos \vartheta , r sin\vartheta ) r dr d\vartheta[/tex]
  5. Sep 27, 2007 #4


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    z^2 = 1 - (x^2 + y^2)
    so z^2 = 1 - r^2

    You will if you do this in Cartesian coordinates.

    I think I have the picture now. The projected area onto the z = 0 plane is a section of an annulus. The annulus is bounded by
    a <= r <= b (a and b positive -- I would presume also that a and b are < 1). The section is in turn bounded by rays from the origin in the directions (alpha) and (beta), such that beta - alpha is not more than 2pi. In other words, your section could be the whole annulus (beta - alpha = 2pi) or some smaller angular section down to nothing at all (alpha = beta).

    It sounds like what remains for the integration is to set up the appropriate bound at the surface of the sphere.
    Last edited: Sep 27, 2007
  6. Sep 27, 2007 #5
    so for r the bounds are 0<=r<=1 since it is unit sphere, but what about theta? 0<=theta<=2pi ?
  7. Sep 27, 2007 #6
    or is it 0<=theta<= alpha, and keep the unknown...
  8. Sep 27, 2007 #7
    and my double integral is [tex]\int^{2pi or\alpha ?}_{0}\int^{1}_{0} \sqrt{1-r^{2}} dr d\vartheta[/tex]
  9. Sep 27, 2007 #8


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    Well, having "[itex]2\pi[/itex] or [itex]\alpha[/itex]" in the integral is probably a mistake!
    Your original problem said [itex]0< \theta< \alpha[/itex]. Don't you think it would be a good idea to do what you are told?

    However, you are still making a serious error!
    What you wrote before:
    [tex]\int\int_{R} f(x,y) dA = \int^{\beta}_{\alpha}\int^{b}_{a} f( rcos \vartheta , r sin\vartheta ) r dr d\vartheta[/tex]
    is for integrating in the plane in polar coordinates.

    This a three-dimensional volume problem and the problem clearly said spherical coordinates, not polar coordinates!
    Last edited by a moderator: Sep 27, 2007
  10. Sep 27, 2007 #9
    i have not learnt spherical coordinates, what is this?
  11. Sep 27, 2007 #10


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    Then go to your teacher and point out that he/she assigned a problem that specifically says to use "spherical coordinates" and that you have not yet learned about spherical coordinates. If nothing else, your teacher may be able to point out the classes you slept through!
  12. Sep 27, 2007 #11
    is it triple integral?
    i have not fall asleep in any class
  13. Sep 27, 2007 #12
    ok so now i must find one more boundary. [tex]\int\int^{alpha}_{0}\int^{1}_{0} \sqrt{1-r^{2}} dr d\vartheta d\phi[/tex]

    how do i find the limits of integration of phi?
  14. Sep 27, 2007 #13


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    I didn't say you had fallen asleep! I just considered it as a possible reason why your teacher might assign an obvious spherical coordinates problem when you don't remember having learned about spherical coordinates.

    [itex]\phi[/itex] measures the angle from the "north pole" (positive z-axis) to the "south pole" (negative z-axis). That's from 0 to [itex]\pi[/itex].

    However, the "differential of volume" in spherical coordinates is NOT
    [tex]\sqrt{1- r^2}drd\theta d\phi[/tex]

    It is
    [tex]\rho^2 sin(\phi) d\rho d\theta d\phi[/itex]
  15. Sep 27, 2007 #14
    So the integral is now: [tex]\int^{\pi}_{0}\int^{\alpha}_{0}\int^{1}_{0}\rho^2 sin(\phi) d\rho d\theta d\phi[/tex]
    but why there is variable [tex]\alpha[/tex], so I will do it with [tex]\alpha[/tex] inside final answer?
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