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

Toolbox:
• Area of the region bounded by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$,where $f(x)$ is a continuous function defined on $[a,b]$,is given by
• $A=\int_a^bf(x)dx$ or $\int_a^b ydx.$
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.
edited Apr 26, 2013