Area of 2016

We can parametrise the curve $C$ as $x=2016\cos^3(t),y=2016\sin^3(t)$ . Thus the enclosed area is $\int_{C}xdy$ $=3*2016^2\int_{0}^{2\pi}\cos^4(t)\sin^2(t)dt=3*2016^2*\frac{\pi}{8}$ . The answer is $\frac{3*2016^2}{8}=\boxed{1524096}$

The given equation represents an asteroid. Standard equation is given by $\large x^\frac{2}{3} + y^\frac{2}{3}$ = $\large a^\frac{2}{3}$

Its area is given by $\large\frac{3a^2}{8} \pi$ .