Question

The area common to the circle $x^{2}+y^{2}=16a^{2}$ and the parabola $y^{2}=6ax$ is

• A. $\frac{4a^{2}}{3}\left(4\pi-\sqrt{3}\right)$ sq. units
• B. $\frac{4a^{2}}{3}\left(8\pi-3\right)$ sq. units
• C. $\frac{4a^{2}}{3}\left(4\pi+\sqrt{3}\right)$ sq. units
• D. none of these

Answer 1

Answer: (C)

We have, $x^{2}+ y^{2}= 16a^{2}$ $...\left(i\right)$, a circle with centre $\left(0,0\right)$ and radius $4a$.
and $y^{2} = 6ax$ $...\left(ii\right)$, parabola with vertex $\left(0,0\right)$.
The point of intersection of $\left(i\right)$ and $\left(ii\right)$ are
$x = 2a$,
$y=\pm\,2 \sqrt{3}a$.

$\therefore \quad$ Required area = area of shaded region
$=2\left[\int\limits_{0}^{2a}\sqrt{6a}\sqrt{x}\, dx\right]+2 \left[\int\limits_{2a}^{4a}\sqrt{\left(4a^{2}\right)-x^{2}} \,dx\right]$

$=2\cdot\frac{2}{3}\sqrt{6a}\left[x^{3 /2}\right]_{0}^{2a}$
$+2\left[\frac{x}{2}\sqrt{\left(4a\right)^{2}-x^{2}+\frac{1}{2}}\left(4a\right)^{2}sin^{-1}\frac{x}{4a}\right]_{2a}^{4a}$

$=\frac{4}{3}\sqrt{6a}\left(2a\right)^{3 /2}$
$+2\left[\left(0-2a^{2}\sqrt{3}\right)+8a^{2}\left(sin^{-1} 1-sin^{-1}\frac{1}{2}\right)\right]$
$=\frac{16}{3}\sqrt{3}a^{2}-4\sqrt{3}a^{2}+16a^{2}\frac{\pi}{3}$
$\therefore \quad$ $A=\frac{4\sqrt{3}a^{2}}{3}+\frac{16\pi a^{2}}{3}$
$=\frac{4a^{2}}{3}\left(4\pi+\sqrt{3}\right)$ sq. units