The area of the region bounded by $x^2=8y$, $x=4$ and $X-$ axis is

Solution:

The given curves are $x^2=8y$ and $x=4$

The rough sketch for the graph of required region is as below

The area of the shaded region is the area of the region bounded by the curve $y=\frac{x^2}{8}$ between the lines $x=0,x=4$ and above $X-$ axis. 

We know that the area of the region bounded by the curve $y=f\left(x\right)$, $x=a,x=b$ and above $x-$ axis is ${\int }_a^{b}f\left(x\right)dx$

Hence, the shaded region is ${\int }_0^4\frac{x^2}{8}dx$

$\begin{array}{rcl}{\int }_0^4\frac{x^2}{8}dx& =& \frac{1}{8}{\left[\frac{x^3}{3}\right]}_0^4\\ & =& \frac{1}{8}\left[\frac{64}{3}-0\right]\\ & =& \frac{8}{3}\end{array}$

Therefore, the  area of the region bounded by $x^2=8y$, $x=4$ and $X-$ axis is $\boxed{\frac{8}{3}\mathrm{sq} \mathrm{units}}$