cos(A+B)=cos(A)+cos(B)

Let $u=\frac12 (x+y)$ and $v=\frac12 (x-y)$ . Then the equation becomes $\cos 2u=2\cos u \cdot \cos v$ and the graph ( $v$ plotted against $u$ ) looks like this:

(image produced with GeoGebra)

Due to the transformation of coordinates, the symmetry axes of the ovals are now horizontal/vertical, but the areas have shrunk by a factor of $2$ .

We'll concentrate on the oval highlighted in red. To find the intersections of this oval with the $u$ -axis, we need to solve for when $v=0$ , ie solve $\cos 2u=2\cos u$

This is just a quadratic in $\cos u$ ; solving, we find the two intersections are given by $u_1=\cos^{-1} \frac{1-\sqrt3}{2},\; \; \; u_2=2\pi-u_1$

Recalling that areas have been shrunk in these coordinates, and noting that the integral will just give the area of the top half of the oval, the area of one of the shapes in the original coordinates is given by $A=4\int_{u_1}^{u_2} \cos^{-1} \left( \frac{\cos 2u}{2\cos u} \right) du$

Using Wolfram|Alpha, this came out be $\boxed{15.737\ldots}$ . (I'd love to see a way to evaluate the integral analytically!)