Question

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

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $2\pi$
• D. $3\pi$

Answer 1

Answer: (D)

$y^2(2-x)=x^3$
$\therefore y^2=\frac{x^3}{2-x}$
$\therefore y=\pm \frac{x^{3/2}}{\sqrt{2-x}}$
Clearly, $x\in [0,2)$
Consider $y=\frac{x^{3/2}}{\sqrt{2-x}}$
When $x=0$, then $y=0$
Also, when $x$ increases from $0$ to $2,y$ increases from $0$ to $\infty$
Hence, graph of given relation is as shown in the following figure:

Required Area, $A=2\underset{0}{\overset{2}{\int }}\frac{x^{3/2}}{\sqrt{2-x}}dx$
Putting $x=2si{n}^2\theta$, we get
$A=2\underset{0}{\overset{\pi /2}{\int }}\frac{2\sqrt{2}si{n}^3\theta }{\sqrt{2}-2si{n}^2\theta }4sin\theta cos\theta d\theta$
$=16\underset{0}{\overset{\pi /2}{\int }}si{n}^4\theta d\theta$
$=16\underset{0}{\overset{\pi /2}{\int }}\frac{{\left(1-cos2\theta \right)}^2}{4}d\theta$
$=4\underset{0}{\overset{\pi /2}{\int }}\left(1-2cos2\theta +co{s}^{2}2\theta \right)d\theta$
$=4\underset{0}{\overset{\pi /2}{\int }}\left(1-2cos2\theta +\frac{1+cos4\theta }{2}\right)d\theta$
$=4{\left(\theta +sin2\theta +\frac{\theta +\frac{sin4\theta }{4}}{2}\right)}_0^{\pi /2}$
$=4\left(\frac{\pi }{2}+0+\frac{\pi }{4}\right)$
$=3\pi$