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Homework Help: Integration: Cylindrical to Spherical

  1. Jul 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Use a triple integral in cylindrical coordinates to show that the volume of the solid bounded above by a sphere [tex]\rho = \rho_{o}[/tex], below by a cone [tex]\phi = \phi_{o}[/tex], and on the sides by [tex]\theta = \theta_{1}[/tex] and [tex]\theta = \theta_{2}[/tex], [tex]\theta_{1} < \theta_{2}[/tex] is

    [tex] V = 1/3 \rho_{o}^3 (1-cos(\phi_{0}))(\theta_{2}-\theta_{1}) [/tex]

    2. Relevant equations
    In cylindrical coordinates, the sphere has the equation [tex] r^2 + z^2 = \rho_{o}^2[/tex] and the cone has the equation [tex] z = r cot(\phi_{o})[/tex]. For simplicity, consider only the case [tex] 0 < \phi_{o} < \pi / 2 [/tex].

    3. The attempt at a solution

    I had two methods, not sure which would actually work because I'm stuck.

    V = \int\int\int dV = \int\int\int r dr dz d\theta

    = \int\int\int\sqrt{\rho_{o}^2 - z^2} dr dz d\theta

    = \int\int\int\sqrt{\rho_{o}^2 - (r cot(\phi_{o}))^2} dr dz d\theta

    Here, I don't know how I would change variables to integrate with respect to [tex]\phi[/tex] and [tex]\rho[/tex]. I could use the Jacobian, but I doubt that's right because the whole question itself could have been solved if I did that in the first place. It's easy to show that the Jacobian transform from cartesian to spherical is [tex]\rho^2 sin(\phi)[/tex], integrate that with respect to [tex]d\phi dr d\theta[/tex] with the given bounds and come up with
    [tex] V = 1/3 \rho_{o}^3 (1-cos(\phi_{0}))(\theta_{2}-\theta_{1}) [/tex]. But that's not the answer because it doesn't go through cylindrical.

    Attempt #2
    [tex]V = \int\int z dA[/tex]

    [tex]= \int\int \sqrt{\rho_o^2 - r^2} dA[/tex]

    [tex]= \int\int \sqrt{\rho_o^2 - r^2} r dr d\theta[/tex]

    And on this one, I think I might be able to do more if I put in the bounds of integration. I'm thinking [tex]-\sqrt{\rho_o^2-z^2} < r < \sqrt{\rho_o^2 - z^2}[/tex] but to get rid of [tex]z[/tex] that turns into [tex]-\sqrt{\rho_o^2-(r cot(\phi_{o}))^2} < r < \sqrt{\rho_o^2 - (r cot(\phi_{o}))^2}[/tex] which can't be used because it has a dependent limit on [tex]r[/tex], which is the variable I'm integrating with respect to.

    Is there any guidance that someone can offer as to which approach, if any are even close, I should pursue, or any alternative suggestions? Thank you.
  2. jcsd
  3. Jul 18, 2008 #2


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    This looks all right for the differential volume. But I think changing the order of integration would make it easier, otherwise you would have to integrate it piecewise for the spherical portion and the conical one. Try doing in the order [tex]rdzdrd\theta[/tex]

    This looks odd. Why did you have to express r in terms of z and phi_0? You could leave it as it is.

    You're supposed to do it in cylindrical coordinates right? And you don't have to do it in terms of spherical coordinates.
    I don't see how that gives you the required volume. The required volume doesn't have a cylindrical base.

    Those limits don't make much sense. r is always positive.
  4. Jul 19, 2008 #3

    Thanks, I'm still a bit lost though. If I use [tex]r dz dr d\theta[/tex] as the order of integration like recommended, what are the limits for r? or do I change r to [tex]\rho sin(\phi)[/tex]. And would z make sense from [tex]-\sqrt{\rho_o^2-r^2} < z < \sqrt{\rho_o^2 - r^2}[/tex]?
    Last edited: Jul 19, 2008
  5. Jul 19, 2008 #4


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    You should figure out the limits from the innermost integrand to the outermost. So, you can imagine that an imaginary line from -infinity to +infinity along the z-axis would pass though the required volume. Where it enters and exits would define its limits. For r, it goes from zero to the cone, since the volume is limited by the confines of the cone.

    Your limits for z is incorrect. As you have described them, those limits would define a volume of a sphere with radius p_0.
  6. Jul 19, 2008 #5
    Alright, now I have:

    \int_{\theta_1}^{\theta_2}\int_0^{\rho_o sin(\phi_o)}\int_0^{\rho_o} r dz dr d\theta

    [tex] = (\theta_2-\theta_1) \frac {\rho_o^3}{2} sin^2(\phi_o) [/tex]

    So that's not it...
  7. Jul 19, 2008 #6


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    Your limits of integration on z are wrong for several reasons. First, because the volume is bounded below by the cone, z does not go from 0 up to the cone. Second, [itex]z= \rho_0 sin(\phi)[/itex] only where the cone meets the sphere. You want z to go from the cone, for variable r, up to the sphere, again for variable r.
  8. Jul 19, 2008 #7
    [tex]\int_{\theta_1}^{\theta_2}\int_0^{\rho_o sin(\phi_o)}\int_{r cot(\phi_o)}^{\sqrt{\rho_o^2 - r^2}} r dz dr d\theta [/tex]

    I did it wrong still...

    Here's what I was doing (correct me if I'm wrong, I know something has to be):
    I think I see why 0 was not the lower bound for z. Rather, it should be any point on the lower part of the cone, because not all of these points are situated at zero. So looking at this cone, [tex]\phi = \phi_o[/tex], I made a triangle with legs z and r, and the angle opposite r is [tex]\phi_o[/tex]. From this it follows that [tex]cot(\phi_o) = z / r [/tex] or [tex] z = r cot(\phi_o)[/tex], which is the lower bound for z.

    For the upper bound, I had to use the sphere. It's equation can be written [tex]r^2 + z^2 = \rho_o^2[/tex], hence the upper limit for z: [tex]\sqrt{\rho_o^2 - r^2}[/tex].
  9. Jul 19, 2008 #8


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    You sure you couldn't get the answer with those limits?
  10. Jul 19, 2008 #9
    With those limits, it comes out to be:

    [tex] \frac {(\theta_2-\theta_1) (\rho_o^3) (((cos^3(\phi_o) - 1) + sin^2(\phi_o) cos(\phi_o))}{3}[/tex]
    Last edited: Jul 19, 2008
  11. Jul 19, 2008 #10


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    You're missing a negative sign somewhere. Furthermore, [tex]\cos^3 \phi_0 + \sin^2 \phi_0 \cos \phi_0 = \cos \phi_0 (\sin^2 \phi_0 + \cos^2 \phi_0)[/tex].
  12. Jul 20, 2008 #11
    I got it! :biggrin: Nice Identity! Didn't see that there! You guys are awesome, thanks!
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