Find the area of the region bounded by the curves $y^2 = 9x, y = 3x.$

It is mentioned in the question that,

$y\textsuperscript{2} = 9x$  belongs to a parabola and $y = 3x$  belong to a straight line which passes through origin.

Starting with a rough figure showing those equations below,

From the parabola equations it is seen that x cannot be negative, so the graph would be on the right of the X-axis. The parabola would thus be opening to the right.

In order to find the points of the two equations mentioned, you have to solve the two equations simultaneously.

$\begin{array}{l} \text { Put } y=3 x \text { in } y^{2}=9 x \\ \Rightarrow(3 x)^{2}=9 x \\ \Rightarrow 9 x^{2}=9 x \\ \Rightarrow x^{2}-x=0 \\ \Rightarrow x(x-1)=0 \\ \Rightarrow x=0 \text { and } x=1 \end{array}$

We got the coordinates of x, for finding the coordinates of y.

Put x = 1 and x = 0 in the equations of y = 3x.

It is found that y coordinated are y = 3 and y = 0 respectively.

Therefore, the point of interaction of parabola and straight line is (1, 3) and (0, 0)

Now, calculate the area enclosed between the parabola and the straight line.

For calculating the area we have to minus the area under the straight line which extends from x = 0 to x = 1 from the area under the parabola.

It can be written as,

Area between parabola and straight line = area under parabola – area under straight line …. (1)

On calculating the area under parabola,

$y\textsuperscript{2 = }9x \\$

$\item y = 3 \sqrt x\\$

On integrating the above equation from 0 to 1

$\\ \Rightarrow \int_{0}^{1} y d x=3 \int_{0}^{1} \sqrt{x} d x \\ \Rightarrow \int_{0}^{1} y d x=3 \int_{0}^{1} x^{\frac{1}{2}} d x \\ \Rightarrow \int_{0}^{1} y d x=3\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} y d x=3\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1} \\$

$\\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3 \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=2\left[1^{\frac{3}{2}}-0\right] \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=2$

Now, for calculating the area under line y = 3x i.e. area of triangle OAB

$\item y = 3x \\$

On integrating the above equation from 0 to 1

$\\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\int_{0}^{1} 3 \mathrm{xdx} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3\left(\frac{1^{2}}{2}-0\right) \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\frac{3}{2}$

Using equation mentioned above (1)

area between parabola and straight line $= 2 - \frac{3}{2} = 1/2\ unit\textsuperscript{2 }\\$

Therefore,

It is found that the area was 1/2 unit2 between the curves $y^2 = 9x\: \: \: and\: \: \: y = 3x.$