The area of the region bounded by $y^2=8x$ and $y^2=16(3-x)$ is equal to :-

$$\begin{gathered} Given equations y^2=8x\&y^2=-16(x-3) \\ solve the above equations \\ 8 x=-16 x+48 \\ 24 x=48 \\ x=2;y=\pm 4 \\ Required Area \\ =2·{\int }_0^4\left(\begin{array}{c}\underset{\left(x_R\right.}{\underbrace{3-\frac{y^2}{16}}}-\underset{\left.x_L\right)}{\underbrace{\frac{y^2}{8}}}\\ \end{array}\right)dy \\ =2{\left(3y-\frac{y^3}{3\times 16}-\frac{y^3}{3\times 8}\right)}_0^4 \\ =2\left(3\times 4-\frac{4\times 4\times 4}{3\times 16}-\frac{4\times 4\times 4\times 2}{3\times 8\times 2}\right) \\ =2\left(12-\frac{4}{3}-\frac{8}{3}\right)=2\times 12\left(1-\frac{1}{3}\right)=2\times 12\times \frac{2}{3}=16 \end{gathered}$$