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
$x=2a$,
$y=\pm 2\sqrt{3}a$.

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

$=2\cdot \frac{2}{3}\sqrt{6a}{\left[x^{3/2}\right]}_0^{2a}$
$+2{\left[\frac{x}{2}\sqrt{{\left(4a\right)}^2-x^2+\frac{1}{2}}{\left(4a\right)}^{2}si{n}^{-1}\frac{x}{4a}\right]}_{2a}^{4a}$

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