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Triple integrals

  1. Apr 8, 2006 #1
    evaluate [tex]\int \int \int _E \sqrt{x^2+y^2} dV[/tex], where E is the solid bounded by the circular parabola [tex]z=9-4(x^2+y^2) [/tex] and the xy-plane

    so here's what I did, i tried to set this up in cylindrical coordinates.

    the radius:

    is when [tex]z=9-4(x^2+y^2)[/tex] equals with the xy-plane

    so this means that z=0 and x=y

    [tex]0=9-4(2x^2)[/tex]
    [tex]r=\frac{3}{\sqrt{8}}[/tex]

    the z-height:

    [tex]z=9-4r^2[/tex]

    the angle:

    theta should rotate in a ciricle so it should be 2 pi

    the setup:

    [tex]\int_0 ^{2 \pi} \int_0 ^{\frac{3}{\sqrt{8}}}\int _0 ^ {9-4r^2} r rdzdrd \theta [/tex]

    i evaluated this twice but it seems not to be the answer, where did I go wrong?
     
    Last edited: Apr 9, 2006
  2. jcsd
  3. Apr 9, 2006 #2

    HallsofIvy

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    Science Advisor

    Here's your error. I don't know why you decided that x= y but in polar (cylindrical) coordinates, r is not x2, r2= x2+ y2. Your formula for the intersection of the paraboloid and z= 0 plane should be 0= 9- 4r2 so r= 3/2.

    Try
    [tex]\int_{/theta= 0}^{2\pi}\int_{r=0}^{\frac{3}{2}}\int_{z=0}^{9- 4r^2} rdzdrd\theta[/itex].

    I assume the "rr" was a misprint.
     
  4. Apr 9, 2006 #3
    The boundaries look ok to me, but shouldn't therebe a [tex]r^2[/tex] in stead of just [tex]r[/tex]. I mean, one r comes from the transformation of the given integrand to polar coordinates, but we also have the Jacobian being equal to r, no ?


    Marlon
     
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