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Math Help - use triple integral to find volume of solid enclosed between the surfaces

  1. #1
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    use triple integral to find volume of solid enclosed between the surfaces

    surfaces x=y^2+z^2 and x = 1-y^2. I drew them below:
    use triple integral to find volume of solid enclosed between the surfaces-graph.jpguse triple integral to find volume of solid enclosed between the surfaces-graph2.jpguse triple integral to find volume of solid enclosed between the surfaces-graph3.jpg
    I know I need to find the region projected on the xy plane and integrate over that area, as well as integrate the height on the z-axis; however I am not sure how to proceed. I think I need some equations, possibly setting the two equal to each other to find where they intersect.
    Last edited by SNAKE; May 8th 2014 at 11:07 PM.
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  2. #2
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    Re: use triple integral to find volume of solid enclosed between the surfaces

    set two equal: y^2 + z^2 = 1 - y^2. Then 2y^2 + z^2 = 1, an ellipse on the yz plane

    also, on yz plane, z^2+y^2=0 & 1 = y^2 or y = +/- 1
    on xy plane, x = y^2 & x = 1- y^2
    and on xz plane, x=z^2 & x=1
    Last edited by SNAKE; May 9th 2014 at 03:01 PM.
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  3. #3
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    Re: use triple integral to find volume of solid enclosed between the surfaces

    The surface x=y^2+z^2 is a paraboloid opening in the x-direction, and the surface x=1-y^2 is a parabolic cylinder extending in the z direction. Your "drawings" look correct. You can see that the paraboloid "pokes through" the parabolic cylinder, creating an enclosed volume. That's your integration volume.

    The top and bottom of the volume is the paraboloid - solving for z gives z = \sqrt{x-y^2} and z=-\sqrt{x-y^2}. Now projecting onto the xy plane gives the parabola x=y^2 and another parabola x=1-y^2.

    The left boundary is x=y^2 and the right boundary is x=1-y^2. Projecting onto the y-axis gives the interval \left[ \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right].

    So the integral is
    \int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \int_{y^2}^{1-y^2} \int_{-\sqrt{x-y^2}}^{\sqrt{x-y^2}} \,dz \,dx \,dy

    The general procedure is to find the boundaries in the x, y, or z direction (i.e. top and bottom, front and back, or left and right), then project in that same direction and do the same with one less dimension.

    - Hollywood
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