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Math Help - How to setup the limits for the integral?

  1. #1
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    How to setup the limits for the integral?



    This is the sketch i made. Maybe its wrong, because i think it gives a very difficult integral like this. I'll look into it on the meanwhile.

    Last edited by AniMuS; June 27th 2009 at 10:13 AM.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by AniMuS View Post


    This is the sketch i made. Maybe its wrong, because i think it gives a very difficult integral like this. I'll look into it on the meanwhile.

    the limits

    \int_{-2}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}\int_{0}^{y+2} dzdxdy

    after you determine the boundaries for z drop the graph to the x-y plane you will have a circle x^2+y^2=4 then determine the boundary of x as I write then for y
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    If i'd put that answer on the test, i'd have 0.

    The order of the integration must be the one they want.

    Anyway i found i drew the plane wrong. Both intersections of the plane with the cilinder should be parallel to the x axis. So that makes it much easier.

    Now that i drew it well, i think it's this. Correct me if im wrong please

    \int_{0}^{2}\int_{2-z}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dxdydz
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by AniMuS View Post
    If i'd put that answer on the test, i'd have 0.

    The order of the integration must be the one they want.

    Anyway i found i drew the plane wrong. Both intersections of the plane with the cilinder should be parallel to the x axis. So that makes it much easier.

    Now that i drew it well, i think it's this. Correct me if im wrong please

    \int_{0}^{2}\int_{2-z}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dxdydz
    I think z change from 0 to 4 not 2 since you have the two curve

    2+y=z and y=2 they intersect when z=4
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  5. #5
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    Yes, you're right. Hadn't seen that. Thanks
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