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Math Help - Multiple Integration Volume

  1. #1
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    Multiple Integration Volume

    Compute the volume of the region under the plane z=2x+3y+30 and over the region in the xy plane bounded by the circle (x^2)+(y^2)=2y.

    I tried using polar coordinates, but I have no idea what the appropriate range of theta would be. I'm also confused with the 2y as part of the circle equation. Any help with the answer would be appreciated.
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  2. #2
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    Change coordinates in the x-y plane to x'=x, y'=y-1, then
    in terms of the new coordinate the region of integration is
    x'^2+y'^2=1,
    and the plane is now z=2x'+3y'+33.

    Now switch the cylindrical polars, so x'=r\cos \theta, y'=r\sin \theta.
    Then
    the region of integration is,
    0\leq r\leq 1, 0\leq \theta < 2\pi

    RonL
    Last edited by ThePerfectHacker; August 1st 2006 at 11:50 AM. Reason: Sorry CaptainBlank I accidently erased what you said (I wanted to quote accidently edited) then restored it.
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  3. #3
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    Quote Originally Posted by MikeWakefield32
    Compute the volume of the region under the plane z=2x+3y+30 and over the region in the xy plane bounded by the circle (x^2)+(y^2)=2y.

    I tried using polar coordinates, but I have no idea what the appropriate range of theta would be. I'm also confused with the 2y as part of the circle equation. Any help with the answer would be appreciated.
    Let me explain diffrently.

    The function,
    z=2x+3y+30
    Can be expressed as, using polar substitution.
    z=2r\cos \theta +3r \sin \theta +30
    Futhermore, the region,
    x^2+y^2=2y
    Is,
    r^2=2 r\sin \theta
    Thus,
    r=2\sin \theta

    Therefore, the double integral is, (remember to multiply by "r"),
    \int_0^{\pi} \int_0^{2\sin \theta}2r^2\cos \theta +3r^2\sin \theta +30r\, dr\, d\theta
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