$(4$ pts$)10)$ Your friend is trying to setup an integral to find the area inside the curve $r=9cos($the circle on the right side$)$ but outside the curve $r=4cos($the circle on the left side$).$ They solve the equation $9cos=4cos$ and get the solutions $=\frac{}{3},\frac{5}{3}.$ They therefore setup the integral:

${\displaystyle \underset{\frac{}{3}}{\overset{\frac{4}{3}}{}}}[(9cos{)}^2-(4cos{)}^2]$

Briefly explain why this integral will not give the desired area between the two curves.