# Thread: Help with double integrals in polar coordinates

1. ## Help with double integrals in polar coordinates

I appreciate any help on the following problems:

1.) Find the area by double integration in polar coordinates inside circles r = 1 and r = 2sinΘ

2.) Use double integration in polar coordinates to find the volume of solid that lies below given surface and above plane region R bounded by given curve: z=x^2 +y^2 and r=2cosΘ

3.) Evaluate the given integral by first converting to polar coordinates: 1/(1+x^2 +y^2) dxdy where x is defined over region between 0 and (1-y^2)^1/2 and y is defined over region between 0 and 1.

2. You could set the first one up like so.

$2\left[\int_{0}^{\frac{\pi}{6}}\int_{0}^{2sin{\theta}}rdr d{\theta}+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\int _{0}^{1}rdrd{\theta}\right]$

Just one way to set it up.

3. can u explain how u did that

4. The first integral is the portion inside the circle r=2sin(t) and the second

integral is inside the circle r=1. It may help if you look at a graph. You

can see they intersect at Pi/6. The sum of the two give the area inside

the two circles. Of course, double that result because that is half.