That picture makes as much sense as division by zero.

Look at my improved, better, nicer, greater picture below.

You need to find,

INT INT 1*dA

Where the region is the polar figure shown below.

The bounds for theta are from 0 to 2*pi because we condisder the full figure. And the bounds for r are 3+2sin(theta) and 0 because there is no interior figure.

Thus,

INT (from 0 to 2*pi) INT (from 0 to (3+2sin(theta) ) r dr d(theta)