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Math Help - Triple integral, I'm lost

  1. #1
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    Triple integral, I'm lost

    (triple integral) x dv

    Where R is bounded by the paraboloid x=4y^2+4z^2 and the plane x=4.

    It would be so, so, so helpful if someone shows me the steps.

    By the way... when you attempt to simplify some integrals by converting them into polar form, what do you do about the "z"?
    Last edited by Kaitosan; June 8th 2009 at 01:34 PM.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Think of analagous situation where you have int(z)

    Where R is bounded by the paraboloid z=4y^2+4x^2 and the plane z=4.

    Then it is fairly obvious to see

    z varies from 4y^2 + 4x^2 to 4

    the outer double integral is then over the circle of radius 2 in the xyplane

    converting to cylindrical int(zr dzdrd(theta)

    z varies from 4r^2 to 4
    r from 0 to 1
    theta from 0 to 2pi

    The only differnce in your problem is that the parabaloid comes out in the x direction and you end up integrating over a circle of radius 2 in the y -z plane --interchange the roles of x and z


    You have

    int(xr dxdrd(theta)

    x varies from 4r^2 to 4
    r from 0 to 1
    theta from 0 to 2pi

    To answer your second question it depends on whether you have a triple
    or double integral with a function of z in the integrand.

    If you have a triple integral such as in the analagous problem I suggested
    then you are using cylindrical coordinates where z = z and you just integrate it out

    If you have a double integral then you will have a surface or surfaces

    where z is a known function of x and y such as z = x^2 + y^2 = r^2

    in which case you convert z to functions of x and y (or r if you're going polar)
    Eg dblint(zrdrdtheta) would be dblint(r^2 rdrdtheta) = dblint(r^3dr dtheta)

    Hope this helps
    Last edited by Calculus26; June 8th 2009 at 06:03 PM.
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  3. #3
    Senior Member Spec's Avatar
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    Small error above: 0\leq r \leq 1
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Thanks Spec-- corrected it in the original post
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  5. #5
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    Spec, will you please complete the entire problem step by step? I know it's a lot to ask but... whenever I try solving the problem with your direction, I get zero.

    First integral -

    r^2(costheta) dx from 4r^2 to 4

    Second integral -

    4r^2 costheta - 4r^4 costheta from 0 to 1

    Third integral -

    (8/15) costheta from 0 to 2pi

    Please calculate these. The result doesn't make sense!!!
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