Question

Find the area enclosed between the curve $y^2(2a-x)=x^3$ and the line $x=2a$ above $x$-axis.

• A. $\frac{5\pi a^2}{2}$
• B. $\frac{3\pi a^2}{2}$
• C. $\frac{\pi a^2}{2}$
• D. $\frac{3\pi a}{2}$

Answer 1

Answer: (B)

$\frac{3\pi a^2}{2}$

$y^2(2a-x)=x^3\Rightarrow y^2=\frac{x^3}{2a-x}$

$\Rightarrow y=\frac{x^{3/2}}{\sqrt{2a-x}}$

as $x\to 2a,y\to \infty$

$\therefore$ area $=I={\int }_0^{2a}\frac{x^{3/2}}{\sqrt{2a-x}}dx$

Let $x=2a{\sin }^2\theta \Rightarrow dx=4a\cos \theta \sin \theta d\theta$

at $x=2a$, $\theta ={\sin }^{-1}\sqrt{\frac{x}{2a}}=k$ (say)

$\therefore$ $I={\int }_0^{k}8a^2{\sin }^4\theta d\theta$

$={8a}^2{\int }_0^k{\left(\frac{1-\cos 2\theta }{2}\right)}^{2}d\theta ={2a}^2{\int }_0^k[1-2\cos 2\theta +\left(\frac{1-\cos 4\theta }{2}\right)]d\theta$

$=a^2{[3\theta -2\sin 2\theta -\frac{\sin 4\theta }{4}]}_0^k$

$=a^2[3k-2\sin 2k-\frac{\sin 4k}{4}]$

$k={\sin }^{-1}\sqrt{\frac{2a}{2a}},\pi /2$

$\therefore$ $I=a^2[3\pi /2-2\times 0-0/4]=\frac{3a^2\pi }{2}$ [B]