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, $A= 2\int\limits_{0}^{2} \frac{x^{3/ 2}}{\sqrt{2-x}}dx$
Putting $x = 2 \,sin^{2}\,\theta$, we get
$A=2 \int\limits_{0}^{\pi/ 2} \frac{2\sqrt{2} sin^{3}\,\theta}{\sqrt{2}-2\,sin^{2}\,\theta} 4\,sin\,\theta\,cos\,\theta\,d\theta$
$=16 \int\limits_{0}^{\pi /2} sin^{4}\,\theta\,d\theta $
$=16 \int\limits_{0}^{\pi /2}\frac{\left(1-cos\,2\theta\right)^{2}}{4} d\theta$
$=4 \int\limits_{0}^{\pi/ 2} \left(1-2\,cos\,2\theta+cos^{2}\,2\theta\right)d\theta$
$=4 \int\limits_{0}^{\pi /2}\left(1-2\,cos\,2\theta+\frac{1+cos\,4\theta}{2}\right)d\theta$
$=4\left(\theta+sin\,2\theta+\frac{\theta+\frac{sin\,4\theta}{4}}{2}\right)_{0}^{\pi /2}$
$=4\left(\frac{\pi}{2}+0+\frac{\pi}{4}\right)$
$=3\pi$