Question

The area of the curve $x^2+y^2=2ax$ is:

• A. $4\pi a^2$
• B. $\pi a^2$
• C. $\frac{1}{2}\pi a^2$
• D. $2\pi a^2$

Answer 1

Answer: (B)

Given $y^2=2ax-x^2$
It is symmetric about x - axis
$\therefore y=4{\int }_0^a\sqrt{2ax-x^2}dx$
Put $x=a\left(1-\cos \theta \right)$
$\Rightarrow dx=a\sin \theta d\theta$
when $x=a,\theta =\frac{\pi }{2}$
$\therefore y=4{\int }_0^{\pi /2}\sqrt{2a^2.2{\sin }^2\theta /2-4a^2{\sin }^4\frac{\theta }{2}},a\sin \theta d\theta$
$y=4{\int }_0^{\pi /2}2a\sin \frac{\theta }{2}\sqrt{{\cos }^2\frac{\theta }{2}}.a\sin \theta d\theta$
$=4{\int }_0^{\pi /2}{a}^2{\sin }^2\theta d\theta$
$=4a^2{\int }_0^{\pi /2}{\sin }^2\theta d\theta =\frac{\pi a^2}{4}\times 4=\pi a^2.$