You could set the first one up like so.
Just one way to set it up.
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.
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.