Answer
$8$
Work Step by Step
Expressing the $x$'s as functions of $y$,
$x=f(y)=3-y^{2},$
$x=g(y)=-y^{2}/4$
Graphing the functions, we find the intersection points at $x=-2$ and $x=2$.
Over the interval $y\in[-2,2],\ f(y)$ is to the right of $g(y)$, so
$A=\displaystyle \int_{-2}^{2}[f(y)-g(y)]dy=\int_{-2}^{2}[3-y^{2}-(-\frac{y^{2}}{4})]dy$
$=\displaystyle \int_{-2}^{2}[3-\frac{3}{4}y^{2}]dy$
$=\displaystyle \left[3y-\frac{3}{4}\cdot\frac{y^{3}}{3} \right]_{-2}^{2}$
$=3(2+2)-\displaystyle \frac{1}{4}\cdot(8+8)$
=$8$
