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Math Help - Problems with double integrals

  1. #1
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    Problems with double integrals

    Hi there. I have this problem with double integral, which says: calculate using a double integral the volume limited by the given surfaces: x^2+y^2=4, z=4-y z=0

    Its a cylinder cut by a plane.

    At first I've did this integral: \displaystyle\int_{-2}^{2}\displaystyle\int_{-\sqrt[ ]{4-x^2}}^{\sqrt[ ]{4-x^2}}(4-y)dydx

    The thing is that when I solve this the terms null themselves and it gives zero. There is evidently something that I'm doing wrong, I've realized that I'm taking off a half of the volume to the other, so I think I'm defining one part of the area over which I'm making the integral as negative. Is this true? And the thing is: How do I solve this properly? I've introduced this double integral into mathematica and it solved it and gave 16\pi. So, its clear I'm doing something wrong

    I also thought on trying other way: 2\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{\  sqrt[ ]{4-y^2}}(4-y)dxdy=2\displaystyle\int_{0}^{2}(4x-yx)_0^{\sqrt[ ]{4-y^2}}dy=2\displaystyle\int_{0}^{2}(4\sqrt[ ]{4-y^2}-y (\sqrt[ ]{4-y^2}))dy}
    But it gets more complicated that way.

    Bye there, and thanks for helping.
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  2. #2
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    Quote Originally Posted by Ulysses View Post
    Hi there. I have this problem with double integral, which says: calculate using a double integral the volume limited by the given surfaces: x^2+y^2=4, z=4-y z=0

    Its a cylinder cut by a plane.

    At first I've did this integral: \displaystyle\int_{-2}^{2}\displaystyle\int_{-\sqrt[ ]{4-x^2}}^{\sqrt[ ]{4-x^2}}(4-y)dydx

    The thing is that when I solve this the terms null themselves and it gives zero. There is evidently something that I'm doing wrong, I've realized that I'm taking off a half of the volume to the other, so I think I'm defining one part of the area over which I'm making the integral as negative. Is this true? And the thing is: How do I solve this properly? I've introduced this double integral into mathematica and it solved it and gave 16\pi. So, its clear I'm doing something wrong

    I also thought on trying other way: 2\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{\  sqrt[ ]{4-y^2}}(4-y)dxdy=2\displaystyle\int_{0}^{2}(4x-yx)_0^{\sqrt[ ]{4-y^2}}dy=2\displaystyle\int_{0}^{2}(4\sqrt[ ]{4-y^2}-y (\sqrt[ ]{4-y^2}))dy}
    But it gets more complicated that way.

    Bye there, and thanks for helping.
    Why not use polar coordinates

    \displaystyle\int_{-2}^{2}\displaystyle\int_{-\sqrt[ ]{4-x^2}}^{\sqrt[ ]{4-x^2}}(4-y)dydx=\int_{0}^{2}\int_{0}^{2\pi} [4-r\sin(\theta)]r\,dr\, d\theta
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  3. #3
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    I'm not sure about that, I haven't used changes of variables yet. I don't understand why still giving a wrong result. The result is evidently 16\pi, I can see the graph that the volume its equal to the cylinder of radius 2 and height equal 4. I did it this way now, but stills giving me wrong.

    2\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{\  sqrt[ ]{4-y^2}}(4-y)dxdy=2\displaystyle\int_{0}^{2}(4x-yx)_0^{\sqrt[ ]{4-y^2}}dy=2\displaystyle\int_{0}^{2}(4\sqrt[ ]{4-y^2}-y (\sqrt[ ]{4-y^2}))dy}

    Look what this integral gives: 2 integral_0^2 y(4-y^2)^(1/2) dy - Wolfram|Alpha
    2 integral_0^2 4(4-y^2)^(1/2) dy - Wolfram|Alpha

    It gives [math8\pi-16/3[/tex]

    I don't know whats wrong now. Mm I see, I think I've changed the symmetry, I should solve the first I've planted.
    Last edited by Ulysses; November 12th 2010 at 07:10 AM.
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