$(4$ pts$)10)$ Your friend is trying to find the area that is inside the curve $r=2$ but outside the curve $r=3+2sin.$ They come up with the integral ${\displaystyle \underset{-\frac{}{6}}{\overset{\frac{7}{6}}{}}}\frac{1}{2}(3+2sin{)}^2-\frac{1}{2}(2{)}^{2}d.$

Briefly explain the two reasons this integral is setup incorrectly.