Green's Theorem

By Green's theorem, if $C$ is a closed curve that encloses a region $D$ and $L(x,y), M(x,y)$ are functions with continuous partial derivatives,

$$\int_C Ldx + Mdy = \int\int_D \frac{dM}{dx}-\frac{dL}{dy}$$

By taking $L(x,y) = -y/2$ and $M(x,y)=x/2$ we get that the area of $D$ is given by:

$A = \int\int_D dA = \int\int_D \frac{dM}{dx}-\frac{dL}{dy}dA = \int_C Ldx + Mdy  = \frac{1}{2}\int_C -ydx + xdy$

In other words, the area is just the line integral of the vector field $F(x,y)=\frac{1}{2}(-y,x)$ around the curve $C$ parametrized by $c(t)=(4\cos{(t)}+\cos{(4t)},4\sin{(t)}-\sin{(4t)})$.

In our case,

$A = \int \int_{\text{int}(C)}dA = \int_C \textbf{F} \cdot \textbf{ds}$
$= \frac{1}{2}\int_0^{2\pi} (-4\sin{(t)}+\sin{(4t)},4\cos{(t)}+\cos{(4t)}) \cdot (-4\sin{(t)}-4\sin{(4t)},4\cos{(t)}-4\cos{(4t)})dt $
$=\frac{1}{2}\int_0^{2\pi} 16\sin^2{(t)}+12\sin{(t)}\sin{(4t)}- 4\sin^2{(4t)}+16\cos^2{(t)}-12\sin{(t)}\sin{(4t)}- 4\sin^2{(4t)}dt$
$=\frac{1}{2}\int_0^{2\pi} 12(1-\cos{(5t)})dt = 6(t-\sin{(5t)}/5)|_0^{2\pi} = 12\pi$