Answer
$a=\frac{3a^{2}\pi}{8}$
Work Step by Step
$f(x)=a*cos(x)^3$
$g(x)=a*sin(x)^{3}$
$g`(x)=3*acos(x)sin(x)^2$
The bounds for the integral will be $(0, 2\pi)$ since that is the revolution of the parametric.
$A=\int_{0}^{2\pi}(f(x)g`(x))dx$
$=\int_{0}^{2\pi}(a*cos(x)^3*(3a*cos(x)*sin(x)^2))dx$
$=\int_{0}^{2\pi}(3a^2*cos(x)^4*sin(x)^2)dx$
$=-\frac{a^2*(sin(12\pi)+3sin(8\pi)-3sin(4\pi)-24\pi)}{64}]^{2\pi}_{0}$
$=-\frac{a^2(-24\pi)}{24}$
$=\frac{3a^2\pi}{8}$
https://www.desmos.com/calculator/l40j1rwzhb Example of the graph with Desmos
