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Double integration

  1. Jan 23, 2008 #1
    Hi
    I've been working on a problem and I'm nearly there but I'm struggling with the integration part at the end and was hoping you might be able to help if you have the time. The original question was

    [tex]\int \int (y^2 z^2 + z^2 x^2 + x^2 y^2) \: dS[/tex]
    Evaluated on the region of [tex]z^2 = x^2 + y^2[/tex] between z=1 and z=2.
    .
    .
    .
    .
    Then by substituting in x=r*cos(θ) and y=r*sin(θ), and multiplying by 'r' (the Jacobian determinant), I got [tex]\sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta[/tex]

    and then I'm stuck. Any help or advice would really be appreciated
    Thanks
     
  2. jcsd
  3. Jan 23, 2008 #2

    NateTG

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    This can be cracked with trig identities, or done by parts.
     
  4. Jan 23, 2008 #3
    I haven't done the steps but are you sure the integrand contains cos (theta^2) and sin (theta^2) or is it cos^2 (theta) (cos squared theta) ?
     
    Last edited: Jan 23, 2008
  5. Jan 23, 2008 #4
    Same thing, he didn't write cos(theta^2), they're outside the parentheses
     
  6. Jan 23, 2008 #5

    CompuChip

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    It says [itex]\cos(\theta)^2[/itex] which I read as [itex]\cos^2(\theta)[/itex] as opposed to [itex]\cos(\theta^2)[/itex].

    You might replace [itex]\cos^2\theta = 1 - \sin^2\theta[/itex], then two of the integrals become very easy, and one of them becomes [itex]\sin^4\theta[/itex] for which there are tricks.
     
  7. Jan 23, 2008 #6
    Well, then just go brute force and then apply by parts. I got int( sin^2(2 x theta))

    After simplyfying and using by parts I get int( 2*theta*sin(4theta))

    By parts again.
     
  8. Jan 23, 2008 #7

    NateTG

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    Actually, I was thinking...
    [tex]\cos \theta \sin \theta = \frac{1}{2} \sin \left( 2 \theta \right)[/tex]
     
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