5.3 Double integrals in polar coordinates  (Page 3/7)

We can describe the region $D$ as $\left\{\left(r,\theta \right)|0\le \theta \le \pi ,0\le r\le 1+\text{cos}\theta \right\}$ as shown in the following figure.

Hence, we have

By the method of double integration, we can see that the volume is the iterated integral of the form ${\displaystyle \underset{R}{\iint }\left(1-x^2-y^2\right)}dA$ where $R=\left\{\left(r,\theta \right)|0\le r\le 1,0\le \theta \le 2\pi \right\}.$

This integration was shown before in [link] , so the volume is $\frac{\pi }{2}$ cubic units.

First change the disk ${\left(x-1\right)}^2+y^2=1$ to polar coordinates. Expanding the square term, we have $x^2-2x+1+y^2=1.$ Then simplify to get $x^2+y^2=2x,$ which in polar coordinates becomes $r^2=2r\text{cos}\theta$ and then either $r=0$ or $r=2\text{cos}\theta .$ Similarly, the equation of the paraboloid changes to $z=4-r^2.$ Therefore we can describe the disk ${\left(x-1\right)}^2+y^2=1$ on the $xy$ -plane as the region

Hence the volume of the solid bounded above by the paraboloid $z=4-x^2-y^2$ and below by $r=2\text{cos}\theta$ is

First examine the region over which we need to set up the double integral and the accompanying paraboloid.

The region $D$ is $\left\{\left(x,y\right)|0\le x\le 1,x\le y\le 2-x\right\}.$ Converting the lines $y=x,x=0,$ and $x+y=2$ in the $xy$ -plane to functions of $r$ and $\theta ,$ we have $\theta =\pi \text{/}4,$ $\theta =\pi \text{/}2,$ and $r=2\text{/}\left(\text{cos}\theta +\text{sin}\theta \right),$ respectively. Graphing the region on the $xy$ -plane, we see that it looks like $D=\left\{\left(r,\theta \right)|\pi \text{/}4\le \theta \le \pi \text{/}2,0\le r\le 2\text{/}\left(\text{cos}\theta +\text{sin}\theta \right)\right\}.$ Now converting the equation of the surface gives $z=x^2+y^2=r^2.$ Therefore, the volume of the solid is given by the double integral

As you can see, this integral is very complicated. So, we can instead evaluate this double integral in rectangular coordinates as

Evaluating gives