Answer
$$\frac{1331}{6}$$
Work Step by Step
Given $$x=y^{2}+4 y-22,\ \ \ x=3 y+8 $$
First, we find the intersection points; we get
\begin{align*}
y^{2}+4 y-22&=3 y+8\\
y^{2}+y-30&=0\\
(y+6)(y-5)&=0
\end{align*}
Then $y= -6, \ \ 5$ and $ 3 y+8\geq y^{2}+4 y-22 $ for $-6\leq y\leq 5$; hence, the area is given by
\begin{aligned}
\text{Area}&=\int_{-6}^{5}\left(3 y+8-\left(y^{2}+4 y-22\right)\right) d y\\
&=\int_{-6}^{5}\left(-y^{2}-y+30\right) d y\\
&= \frac{-y^{3}}{3}-\frac{y^{2}}{2}+30 y\bigg|_{-6}^{5}\\
&=\frac{1331}{6}
\end{aligned}
