The area between the curves $y^2=9x$ and the line $y=3x$

The given curves are $y^2=9x$ and $y=3x$

To get the points of intersection of the above curve and the line, substitute $y=3x$ in $y^2=9x$

${\left(3x\right)}^2=9x\Rightarrow x=0,x=1$

The rough sketch of the area of the regin between the given curve and the line is as below. 

The area of the required region is the area of the shaded region

The area of the shaded region is the area of the region bounded by the curves $y^2=9x$ and $y=3x$ between the lines $x=0$ and $x=1$

The area of the shaded region is $A={\int }_0^1\left(3\sqrt{x}-3x\right)dx$

Hence, 

$\begin{array}{rcl}A& =& {\int }_0^1\left(3\sqrt{x}-3x\right)dx\\ & =& 3{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}-\frac{x^2}{2}\right]}_0^1\\ & =& 3\left[\frac{2}{3}-\frac{1}{2}\right]\\ & =& \frac{1}{2}\end{array}$