Question

Area enclosed between the curve $y^2(2a-x)=x^3$ and the line x = 2a above x-axis is

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

Answer 1

The correct option is B $\frac{3}{2}\pi a^2$

We have,

Given that the curve,

$y^2(2a-x)=x^3$ …… (1)

Line $x=2a.....\left(2\right)$

From (2) to,

If $x=2a$ then,$y\to \infty$

So,

According to given question,

$I={lim}_{p\to 2a}{\int }_0^p\frac{x\sqrt{x}}{\sqrt{2a-x}}dx$

Let put $x=2a{\sin }^2\theta$

$dx=4a\sin \theta \cos \theta d\theta$

So, x=p$\theta ={\sin }^{-1}\left(\sqrt{\frac{x}{2a}}\right)={\sin }^{-1}\left(\sqrt{\frac{p}{2a}}\right)$

Again let

${\sin }^{-1}\left(\sqrt{\frac{p}{2a}}\right)=k$

$\Rightarrow {\int }_0^p\frac{x^{\frac{3}{2}}}{\sqrt{2a-x}}dx={\int }_0^k\frac{{\left(2a{\sin }^2\theta \right)}^{\frac{3}{2}}}{\sqrt{2a-2a{\sin }^2\theta }}4a\sin \theta \cos \theta d\theta$

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

$=8a^2{\int }_0^k{\left({\sin }^2\theta \right)}^{2}d\theta$

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

$=2a^2{\int }_0^k(1-2\cos 2\theta +{\cos }^{2}2\theta )d\theta$

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

$=a^2{\int }_0^k(3-4\cos 2\theta -\cos 4\theta )d\theta$

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

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

As

$p\to 2a,k\to {\sin }^{-1}\left(1\right)=\frac{\pi }{2}$

$\Rightarrow \sin 2k\to 0and\sin 4k\to 0$

$I={lim}_{p\to 2a}{\int }_0^p\frac{x\sqrt{x}}{\sqrt{2a-x}}dx={lim}_{k\to \frac{\pi }{2}}{a}^2[3k-2\sin 2k-\frac{\sin 4k}{4}]=\frac{3\pi a^2}{2}$

Hence, this is the answer.