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Homework Help: Volume inside a cone and between z=1 and z=2

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Write an evaluate a triple integral in spherical coordinates for the volume inside the cone [tex]z^2 = x^2 + y^2[/tex] between the planes z=1 and z=2.

    The answer is 7π/3

    3. The attempt at a solution

    Substitute values to work out the limits. From [tex]z^2 = x^2 + y^2[/tex], substitute for the spherical coordinates and I get [tex]\theta = \pi/4[/tex]. Similarly substitute [tex]1 = x^2 + y^2[/tex] and [tex]2^2 = x^2 + y^2[/tex], and I find that r ranges from [tex]cosec \theta[/tex] to [tex]2 cosec \theta[/tex].

    Then, put together into the triple integral:

    [tex]\int_0^{2\pi = \phi} \int_0^{\pi/4} \int_{cosec\theta}^{2cosec\theta} = r^2 sin\theta dr d\theta d\phi = 2\pi \int_0^{\pi/4} \frac{1}{3}(7 cosec^3 \theta)(sin \theta) d\theta = 2\pi (\frac{7}{3})[/tex]

    This answer is off. Appreciate if someone could check the working for me, please!!
     
  2. jcsd
  3. Feb 6, 2009 #2
    Those cosecs should be secs.
     
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