Volume of Solid and Double Integrals

**FatherMike** · July 6th 2009, 09:41 AM

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

**halbard** · July 6th 2009, 11:36 AM

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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