find the area bounded by the curve $y^2=4ax$ and the lines y=2a and y-axis. 

From the given equation $y^2=4ax$ we can say that the equation represents a parabola with vertex (0, 0) and axis as x-axis. The equation y= 2a represents a straight line parallel to x--axis at a distance of 2a, from it as shown in Figure. The required region is the shaded portion in the Figure.

To find the area of the shaded region shown in Figure, we slice it into horizontal strips.

We observe that each horizontal strip has its left end on y--axis and the right end on the given parabola $y^2=4ax$. So, the approximating rectangle shown in Fig. has its length $=\left|x\right|$ and width $=\mathrm{dy}$

Therefore area $=\left|x\right|$ dy.

Since the approximating rectangle can move vertically from y=0 to y=2a. So, required area denoted by A, is given by $\mathrm{A}={\int }_0^{2\mathrm{a}} \left|\mathrm{x}\right|\mathrm{dy}={\int }_0^{2\mathrm{a}} \mathrm{xdy}[\because \mathrm{x}\ge 0\therefore |\mathrm{x}|=\mathrm{x}]$

$\Rightarrow \mathrm{A}={\int }_0^{2\mathrm{a}} \frac{{\mathrm{y}}^2}{4\mathrm{a}}\mathrm{dy}\left[\because \mathrm{P}(\mathrm{x},\mathrm{y})\right.\mathrm{lies} \mathrm{on} \left.{\mathrm{y}}^2=4\mathrm{ax}\therefore \mathrm{x}=\frac{{\mathrm{y}}^2}{4\mathrm{a}}\right]$$\Rightarrow \mathrm{A}=\frac{1}{4\mathrm{a}}{\left[\frac{{\mathrm{y}}^3}{3}\right]}_0^{2\mathrm{a}}=\frac{1}{4\mathrm{a}}\left(\frac{8{\mathrm{a}}^3}{3}-0\right)=\frac{2{\mathrm{a}}^2}{3}\mathrm{sq}. \mathrm{units}$Therefore, bounded area by the curve ${\mathrm{y}}^2=4\mathrm{ax}$ and the line  $y=2a$and y--axis is $\frac{2{\mathrm{a}}^2}{3}{\mathrm{unit}}^2$