The interior of a quadrilateral is bounded by the graphs of $(2x-y)^2=4$ and $(x-3y)^2=16$. What is the area of this region?

1. Solving adjacent boundary equations gives the vertices, in order, as $\left(\frac25,-\frac65\right)$, $(2,2)$, $\left(-\frac25,\frac65\right)$, and $(-2,-2)$.
2. By the shoelace theorem (15), the area is $\frac12\left|\frac{16}5+\frac{16}5+\frac{16}5+\frac{16}5\right|=\boxed{\frac{32}{5}}$.