Question

The area bounded by the curve $y^2(2a-x)=x^3$ and the line $x=2a$ is

• A. $3\pi a^2$ sq. unit
• B. $\frac{3\pi a^2}{2}sq\cdot$ unit
• C. $\frac{3\pi a^2}{4}$ sq. unit
• D. $\frac{6\pi a^2}{5}$ sq. unit

Answer 1

Answer: (B)

Let the equation of curve $y^2(2a-x)=x^3$
and equation of line $x=2a$ The given curve is symmetrical about $x$ -axis and passes through origin.
From
(i) we have, $y^2=\frac{x^3}{2a-x}$
But $\frac{x^3}{2a-x}<0$ for $x>2a$ and $x<0$ So, curve does not lie in the portion $x>2a$ and $x<0,$ therefore curve lies in $0\le x\le 2a$ .
$\therefore$ Area bounded by the curve and line
$=\underset{0}{\overset{2a}{\int }}ydx=\underset{0}{\overset{2a}{\int }}\frac{x^{3/2}}{\sqrt{2a}-x}dx$
Put $x=2asi{n}^2\theta$ and $dx=4asin\theta cos\theta d\theta$
$\therefore I=\underset{0}{\overset{\pi /2}{\int }}8a^{2}si{n}^4\theta d\theta =8a^2\left[\frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}\right]$
using $\underset{0}{\overset{\pi /2}{\int }}si{n}^{m}xco{s}^{n}xdx$
$=\frac{\Gamma \frac{m+1}{2}\Gamma \frac{n+1}{2}}{2\Gamma \frac{m+n+2}{2}}=\frac{3\pi a^2}{2}$ sq. unit