# Thread: Volume of Solid and Double Integrals

1. ## 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

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

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

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

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

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

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