Area along $y$-axis

$$ \begin{aligned} & \int\limits_{-2}^4\left[\left(\frac{y+8}{2}\right)-\left(\frac{y^2+8}{4}\right)\right] d y \\\\ & =\frac{y^2}{4}+4 y-\frac{y^3}{12}-\left.2 y\right|_{-2} ^4 \end{aligned} $$

$\begin{aligned} & =\left(4+16-\frac{16}{3}-8\right)-\left(1-8+\frac{8}{12}+4\right) \\\\ & =\left(12-\frac{16}{3}\right)-\left(\frac{8}{12}-3\right)=15-\left(\frac{64+8}{12}\right) \\\\ & =15-6=9 \text { sq. unit }\end{aligned}$