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Math Help - 2 quick questions on double integrals

  1. #1
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    2 quick questions on double integrals

    I'm having a bit of trouble finding the the bounds for the following problems.

    1) z=x^2+3y^2 and the planes x = 0, \  y=1, \ y=x, \ z=0

    since z will be a paraboloid and bounded by those constraints I'm thinking I would have \int_{x=0}^{x=1} \int _{y=x}^{y=1} x^2+3y^2 \ dy \ dx

    2) z= x^2 , \ y=x^2 and the planes z=0, \ y=4

    so I get the graph that's in the attachment, now if I take those constraints into account then I would simple have something above the xy plane and to the right of the xz plane, therefore I would think that my integral would be:

    \int_{x=-2}^{x=2} \int_{y=x^2}^{y=4} x^2 \ dy \ dx
    Attached Thumbnails Attached Thumbnails 2 quick questions on double integrals-graph.jpg  
    Last edited by lllll; January 8th 2009 at 08:42 PM.
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  2. #2
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    Quote Originally Posted by lllll View Post
    I'm having a bit of trouble finding the the bounds for the following problems.

    1) z=x^2+3y^2 and the planes x = 0, \ y=1, \ y=x, \ z=0

    since z will be a paraboloid and bounded by those constraints I'm thinking I would have \int_{x=0}^{x=1} \int _{y=x}^{y=1} x^2+3y^2 \ dy \ dx

    2) z= x^2 , \ y=x^2 and the planes z=0, \ y=4

    so I get the graph that's in the attachment, now if I take those constraints into account then I would simple have something above the xy plane and to the right of the xz plane, therefore I would think that my integral would be:

    \int_{x=-2}^{x=2} \int_{y=x^2}^{y=4} x^2 \ dy \ dx
    If you're wishing to find the volume under your given z's and above the xy plane, then your integrals are correct!
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