Answer
$\displaystyle \frac{243}{8}$
Work Step by Step
Rewrite the equations as functions of y:
$\left[\begin{array}{lll}
y^{2}-4x=4 & ... & 4x-y=16\\
y^{2}-4=4x & & 4x=y+16\\
x=\frac{1}{4}(y^{2}-4) & & x=\frac{1}{4}(y+16)
\end{array}\right]$
Graphing the given equations, we find the intersections
or we solve the equation $f(y)=g(y), $ where
$ f(y)=\displaystyle \frac{1}{4}(y+16) \quad$ (the curve on the right side) and
$ g(y)=\displaystyle \frac{1}{4}(y^{2}-4)\quad$ (the curve on the left).
When $y\in[c,d]=[-4,5]$, the area between the graphs is
$A=\displaystyle \int_{c}^{d} [f(y)-g(y)]dy=\displaystyle \int_{-4}^{5}\frac{1}{4}[y+16-(y^{2}-4)]dy$
$=\displaystyle \frac{1}{4}\int_{-4}^{5}[y-y^{2}+20)]dy$
$=\displaystyle \frac{1}{4}\left[\frac{y^{2}}{2}-\frac{y^{3}}{3}+20y\right]_{-4}^{5}$
$=\displaystyle \frac{1}{4}\left[\frac{25-16}{2}-\frac{125+64}{3}+20(5+4)\right]$
$=\displaystyle \frac{243}{8}$
