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Cylindrical triple integral

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Use cylindrical coordinates to evaluate the triple integral (over E) [tex]sqrt(x^2+y^2) dV[/tex] , where E is the solid bounded by the circular paraboloid z=9−16(x^2+y^2) and the xy plane.

    3. The attempt at a solution

    This is really bugging me... Is this the correct setup for the integral?

    0 < z < 9 - 16r^2
    0 < r < 3/4
    0 < theta < 2pi

    [tex]\int_0^{3/4}\int_0^{9-16r^2}\int_0^{2pi}r^2(d\theta,dz,dr)[/tex]
     
    Last edited: Jan 15, 2010
  2. jcsd
  3. Jan 15, 2010 #2
    Your integral looks funny...might want to adjust the formatting somewhat.
    1. Order of integration - please write your integrals in the correct order, or else you are going to get nonsense for your answer when you integrate outwards.
    2. The function you are integrating is r^2, not r.
     
  4. Jan 15, 2010 #3
    Yep, that's all fixed now. Learning latex for the first time. Does this look good? It's not giving me the correct answer according to WeBWorK.
     
    Last edited: Jan 15, 2010
  5. Jan 15, 2010 #4
    Hm...the order of integration is wrong. You must integrate wrt z first before integrating wrt r, because the bound of z is dependent on the value of r. If you integrate in the order above, you will end up with r in your final expression.

    [tex]
    \int_0^{2\pi}\int_0^{3/4}\int_0^{9-16r^2}\,\,r\,\,r \,dz\, dr\, d\theta

    [/tex]
     
    Last edited: Jan 15, 2010
  6. Jan 15, 2010 #5
    Once again my bad, the E is actually sqrt(x^2 + y^2) = r... :S

    I did it your way before (with r^2 as the integrating function) and I still got the wrong answer.
     
  7. Jan 15, 2010 #6
    Hmm....what does the WeBwork thingy say the correct answer is? The integral looks correct to me
     
    Last edited: Jan 15, 2010
  8. Jan 15, 2010 #7
    It works! Ty.
     
    Last edited: Jan 15, 2010
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