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Math Help - Volume of Solid and Double Integrals

  1. #1
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    Volume of Solid and Double Integrals

    Hello,

    I am trying to find the volume enclosed by the elliptic paraboloid z = x^2 + (y+1)^2 and the plane z - 2y - 5 = 0.

    I am pretty sure this integral will be too complicated. I noticed in other posts something about double integrals in polar coordinates.

    Is it possible to set up this equation using polar? Thanks for your help,

    FatherMike
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  2. #2
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    Well, the first thing is to sketch the region bounded by the plane and the paraboloid. Then find the intersection, which is

    x^2+(y+1)^2=2y+5\iff x^2+y^2=4, an ellipse in the plane z=2y+5 but whose projection on z=0 is a circle C, centre O and radius 2. So far, so good.

    The volume of the required region is \iint_C (z_2-z_1)\mathrm dx\,\mathrm dy, where z_2=2y+5 and z_1=x^2+(y+1)^2. This is the 3-D equivalent of the formula \int(y_2-y_1)\mathrm dx for the area between two curves in the xy-plane.

    So you need to evaluate \iint_{x^2+y^2\leq4}(4-x^2-y^2)\mathrm dx\,\mathrm d y and yes, the best way involves a change to polar coordinates.

    In this, the element of area \mathrm dx\,\mathrm dy becomes r\mathrm dr\,\mathrm d\theta and as x^2+y^2=r^2 the region of integration is now r\leq 2. Therefore

    volume =\iint_{r\leq2}(4-r^2)r\mathrm dr\,\mathrm d\theta<br />
=\int_0^{2\pi}\int_0^2(4r-r^3)\mathrm dr\,\mathrm d\theta=2\pi\left[2r^2-r^4/4\right]_0^2=8\pi.
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