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!

Triple integral depending on a parameter

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Find [tex] \iiint (x^{2n} + y^{2n} + z^{2n})\,dV [/tex] where the integral is taken over the region of 3D space where [tex] x^{2} + y^{2} + z^{2} \leq 1 [/tex]

    2. Relevant equations

    3. The attempt at a solution

    I tried doing this in Cartesian coordinates, but the limits of integration got very messy and I got stuck after doing the first integral. I also tried using spherical polar coordinates, and then the limits of integration are quite simple, but the integrand gets complicated, unless [itex] n = 1 [/itex], in which case the integral is quite easy to do.

    I then thought that, since the only case where this looks simple enough to do directly is [itex] n = 1 [/itex], I could try to make a conjecture as to what the value of the integral is for general n and then try to prove it by induction. The problem with that, though, is that I don't see how to go from the case [itex] n = k + 1 [/itex] to the case [itex] n = k [/itex].

    Any ideas?
     
  2. jcsd
  3. Feb 4, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I haven't tried it myself yet, and the wife wants the computer just now, but have you tried spherical coordinates? That would be the natural first choice. May or may not work...

    [Edit later] I think you can work it with spherical coordinates using symmetry and the formula$$
    \int_0^\frac \pi 2 \sin^n x\, dx =\int_0^\frac \pi 2 \cos^n x\, dx =
    \frac{1\cdot 3\cdot 5\cdot\cdot\cdot(n-1)}{2\cdot 4\cdot 6\cdot\cdot\cdot n}\frac \pi 2 $$which is valid for ##n## even and ##n \ge 2##.
     
    Last edited: Feb 4, 2013
  4. Feb 4, 2013 #3

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    This integral covers the volume enclosed by a sphere with radius = 1. It can be broken up into the sum of its parts, since the integrand is a sum.

    For n = 1, the integral will evaluate to the sum of the inertias of a sphere about the three coordinate axes, assuming a unit density for the sphere.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Triple integral depending on a parameter
  1. Triple Integral (Replies: 3)

  2. Triple integral (Replies: 5)

  3. Triple Integral (Replies: 9)

  4. Triple Integral (Replies: 6)

Loading...