# Find the area of the region bounded by the line $x=2$ and the parabola $y^2=8x$.

$\begin{array}{1 1} (A) \frac{32}{3}\; sq.units. \\ (B) \frac{64}{3}\; sq.units. \\ (C) \frac{48}{3}\; sq.units. \\ (D) \frac{16}{3} \;sq.units.\end{array}$

Toolbox:
• Let $f(x)$ be a continuous function defined on $[a,b]$.Then the area bounded by the curve $y=f(x)$,the $x$-axis and the ordinate $x=a$ and $x=b$ is given by
• $A=\int_a^bf(x) dx$ or $\int_a^b ydx.$
Step 1:
Given $x=2$ and $y^2=8x$
$y^2=8x$ is a parabola open rightwards and its focus is $(4a=8\Rightarrow a=2)\Rightarrow(2,0).$
Hence $x=2$ is the latus rectum to the parabola.
The required area is the shaded portion shown in the fig.
The required area is the region bounded between the latus rectum.Since it is symmetrical about $x$-axis which includes I and IV quadrant.
Hence $A=2\times \int_0^2\sqrt {8x}dx.$
$\qquad\quad=2\sqrt 8\int_0^2\sqrt xdx\quad(\sqrt 8=2\sqrt 2)$
Step 2:
On integrating we get
$\;\;\;=4\sqrt 2\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\frac{3}{2}}\end{bmatrix}_0^2$
$\;\;\;=\large\frac{8\sqrt 2}{3}\begin{bmatrix}x^{\Large\frac{3}{2}}\end{bmatrix}_0^2$
On applying limits we get,
$\;\;\;=\large\frac{8\sqrt 2}{3}$$[2\sqrt 2-0]$
$\;\;\;=\large\frac{32}{3}$sq.units.
Hence the required area is $\large\frac{32}{3}$sq. units.
edited Apr 28, 2013