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^{a}_{0} \sqrt{2ax -x^{2}} dx $
Put $x =a\left(1-\cos\theta\right)$
$ \Rightarrow dx = a \sin\theta d \theta $
when $x=a , \theta = \frac{\pi}{2}$
$\therefore y=4 \int^{\pi/2}_{0} \sqrt{2a^{2} . 2 \sin^{2} \theta /2 - 4a^{2} \sin^{4} \frac{\theta}{2}} , a \sin\theta d \theta$
$ y =4 \int^{\pi /2}_{0} 2a \sin \frac{\theta}{2} \sqrt{\cos^{2} \frac{\theta}{2}} . a \sin\theta d\theta $
$ = 4 \int^{\pi /2}_{0} a^{2}\sin^{2} \theta d \theta $
$= 4a^{2} \int^{\pi /2}_{0} \sin^{2} \theta d \theta = \frac{\pi a^{2}}{4} \times4 = \pi a^{2}.$