1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Volume inside a cone and between z=1 and z=2
  1. Branch of z^{1/2} (Replies: 2)

Loading...