Question

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

• A. $3\pi a^2$
• B. $\frac{3\pi a^2}{2}$
• C. $\frac{3\pi a^2}{4}$
• D. $\frac{\pi a^2}{4}$

Answer 1

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

Given,

$y^2(2a-x)=x^3$

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

let, $x=2a{\sin }^2\theta$

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

$\Rightarrow A={\int }_0^{2a}\sqrt{\frac{x^3}{2a-x}}dx$

$A={\int }_0^{\frac{\pi }{2}}\sqrt{\frac{8a^3{\sin }^6\theta }{2a{\cos }^2\theta }}4a\sin \theta \cos \theta d\theta$

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

$=8a^2{\int }_0^{\frac{\pi }{2}}{\sin }^2\theta (1-{\cos }^2\theta )$

$=8a^2{\int }_0^{\frac{\pi }{2}}\left(\frac{1-\cos 2\theta }{2}\right)-\frac{1}{4}{\sin }^{2}2\theta d\theta$

$=8a^2[\frac{\pi }{4}-0-\frac{1}{4}(\frac{\pi }{4}-0)]$

$=8a^2[\frac{\pi }{4}-\frac{\pi }{16}]=\frac{3}{2}\pi a^2$