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Triple integral using cylindrical coordinates

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int\int_{Q}\int(x^4+2x^2y^2+y^4)dV[/tex] where Q is the cylindrical solid given by [tex]\{(x,y,x)| x^2+y^2 \leq a^2, 0\leqz\leq\frac{1}{\pi}\}[/tex]

    2. Relevant equations

    When I convert to cylindrical I get [tex] f(r,\theta,z) = r^4\cos^2\theta + 2r^4\cos^2\theta\sin^2\theta + r^2\sin^2\theta[/tex], but I just need the bounds for dr, is it? [tex]\int^a_0[/tex]

    3. The attempt at a solution
    [tex]\int^2\pi_0\int^a_0\int^\frac{1}{\pi}_0 f(r,\theta,z)rdzdrd\theta[/tex]
    *the first integral is supposed to be from 0 to 2pi
    Last edited: Dec 2, 2008
  2. jcsd
  3. Dec 2, 2008 #2


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    Homework Helper
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    Your expression for [itex]f(r,\theta,z)[/itex] is incorrect. It might help you to realize that [itex]f(x,y,z)=x^4+2x^2y^2+y^4=(x^2+y^2)^2[/itex].....so [itex]f(r,\theta,z)=?[/itex] :wink:

    And yes, [itex]r[/itex] goes from 0 to a.
  4. Dec 2, 2008 #3
    Got it, so [tex]f(r,\theta,z) = r^4[/tex] Much simpler!
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