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Integrating 2

  1. Jul 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Using Spherical coordinates, find the volume of the solid enclosed by the sphere [tex]x^2 + y^2 + z^2 = 4a^2[/tex] and the planes z = 0 and z = a.



    2. Relevant equations
    I have the solutions to this problem, and it is done by integrating two parts:
    [tex]V = V_{R=const.} + V_{z = const.}[/tex]

    The limits for [tex]V = V_{R=const.}[/tex] are
    [tex][0 \leq \phi \leq 2\pi], [\frac{\pi}{2} - sin^{-1}(\frac{1}{2}) \leq \theta \leq \frac{\pi}{2}], [0 \leq R \leq 2 \pi][/tex]

    The limits for [tex]V_{z = const.}[/tex] are
    [tex][0 \leq \theta \leq \frac{\pi}{2} - sin^{-1}(\frac{1}{2})], [0 \leq R \leq \frac{a}{cos(\theta)}], 0 \leq \phi \leq 2\pi] [/tex]

    3. The attempt at a solution

    Could someone explain to things to me:
    1. Why there are two things we are integrating: [tex]V = V_{R=const.} + V_{z = const.}[/tex] I would think there should be only one integral, one that is bounded between z = 0, and z = a within the given sphere.
    2. Why the limits are defined as it is- more specifically, the limits for [tex]\theta[/tex] for [tex]V = V_{R=const.}[/tex], and [tex]\theta, R[/tex] for [tex] V_{z = const.}[/tex]


    Thanks so much,

    JL
     
  2. jcsd
  3. Jul 9, 2009 #2

    benorin

    User Avatar
    Homework Helper

    sphere_cone.jpg

    In above plot, green surface is V_R=const. and, imagining the blue cone surface filled to the brim would be the Vz=const. Two integrals are needed because why? What is the equation of the blue surface in spherical coords?
     
  4. Jul 9, 2009 #3
    Oh, ok thanks- I think I understand this problem now. By the way, that's a really cool picture.

    JL
     
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