Find the area of the region included between $y^2 = 9x$ and y = x

The equations $y\textsuperscript{2} = 9x$ has no negative values of x. Therefore it lies on right of Y axis passing through (0, 0).

And y = x depicts a straight line through origin.

For finding the area, picture shown below.

For finding the point of interaction solve the two equations simultaneously.
$\\Put \: \: y=x \text{ in } y^{2}=9 x \Rightarrow x^{2}=9 x \Rightarrow x=9\\Put\: \: x=9 in y=x \text{ we get } y=9 \\$

Therefore, point of interaction is (9, 9)

The area between the parabola and line = area under parabola – area under line $\ldots (1) \\$

Let us find area under parabola

$\\\Rightarrow y^{2}=9 x \\ \Rightarrow y=3 \sqrt{x} \\ \text{Integrate from 0 to 9}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=\int_{0}^{9} 3 \mathrm{x}^{\frac{1}{2}} \mathrm{dx}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=3\left[\frac{\mathrm{x} \frac{1}{2}+1}{\frac{1}{2}+1}\right]_{0}^{9}\\ \Rightarrow \int_{0}^{9} y \mathrm{dx}=3\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{9}\\ \\$
$\\\Rightarrow \int_{0}^{9} \mathrm{ydx}=3 \frac{2}{3}\left[9 \frac{3}{2}-0\right]\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=2\left(3^{2}\right)^{\frac{3}{2}}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=2\left(3^{3}\right)\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=54\\$

Now let us find area under straight line $\mathrm{y}=\mathrm{x}$

y=x

Integrate from 0 to 9

$\\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\int_{0}^{9} \mathrm{xdx} \\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{9} \\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left(\frac{9^{2}}{2}-0\right)\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=\frac{81}{2}\\ \Rightarrow \int_{0}^{9} y \mathrm{dx}=40.5$

Using (i)

$\Rightarrow$  area between parabola and line  $=54-40.5=13.5 unit ^{2}$

Therefore, area was found to be $13.5 unit\textsuperscript{2}. \\$