Answer
$\displaystyle \frac{56}{15}$
Work Step by Step
Expressing the $x$'s as functions of $y$,
$x=f(y)=4-4y^{2},$
$x=g(y)=1-y^{4}$
Graphing the functions, we find the intersection points at $x=-1$ and $x=2$.
Over the interval $y\in[-1,1],\ f(y)$ is to the right of $g(y)$, so
$A=\displaystyle \int_{-1}^{1}[f(y)-g(y)]dy=\int_{-1}^{1}[4-4y^{2}-(1-y^{4})]dy$
$=\displaystyle \int_{-1}^{1}[3-4y^{2}+y^{4}]dy$
$=\displaystyle \left[3y-\frac{4y^{3}}{3}+\frac{y^{5}}{5} \right]_{-1}^{1}$
$=3(1+1)-\displaystyle \frac{4(1+1)}{3}+\frac{1+1}{5}$
=$\displaystyle \frac{56}{15}$
