Step 1:

Given $y^2=9x$ and $y=3x$

Clearly $y^2=9x$ is a parabola,whose focus a can be calculated as shown below :

$y^2=4ax$ is the general equation .

In the given equation $4a=9\Rightarrow a=\large\frac{9}{4}$

The curve is open rightward with the focus as $\big(\large\frac{9}{4}$$,0\big)$

Step 2:

$y=3x$ is a straight line.

Next let us find the point of intersection of the line and the parabola.

$y^2=9x$-----(1)

$y=3x\Rightarrow y^2=9x^2$

Hence substituting this in equ(1) we get,

$9x^2=9x$

$9x(x-1)=0$

$\Rightarrow x=0$ and $1$

Hence the points of intersection are 0 and 1.

The area of the required region is as shown in the fig.

Now let us evaluate the area of the required region.

Step 3:

$A=\int_0^1(y_2-y_1)dx.$

When $y_2=3\int_0^1\sqrt xdx$ and $y_1=\int_0^1 3xdx.$

$A=3\int_0^1\sqrt xdx-3\int_0^1 xdx.$

On integrating we get,

$A=3\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}_0^1-3\begin{bmatrix}\large\frac{x^2}{2}\end{bmatrix}_0^1$

$\;\;=2\begin{bmatrix}x^{\large\frac{3}{2}}\end{bmatrix}_0^1-\large\frac{3}{2}\begin{bmatrix}\normalsize x^2\end{bmatrix}_0^1$

On applying limits we get,

$2[1-0]-\large\frac{3}{2}$$[1-0]=2-\large\frac{3}{2}=\frac{1}{2}$

Hence the required area is $\large\frac{1}{2}$sq.units.