Question

The area of the portion of the circle $x^2+y^2=64$ which is exterior to the parabola $y^2=12x$, is

• A. $\left(8\pi -\sqrt{3}\right)$ sq units
• B. $\frac{16}{3}\left(8-\sqrt{3}\right)$ sq units
• C. $\frac{16}{3}\left(8\pi -\sqrt{3}\right)$ sq units
• D. None of the above

Answer 1

Answer: (C)

Required shaded area = Area of circle
$-2\left[{\int }_0^{4}2\sqrt{3}\sqrt{x}dx+{\int }_4^8\sqrt{64-x^2}dx\right]$

$=64\pi -[2{\left(2\sqrt{3}{x}^{3/2}\times \frac{2}{3}\right)}_0^4+{\left(\frac{x}{2}\sqrt{64-x^2}+\frac{64}{2}{\sin }^{-1}\frac{x}{8}\right)}_4^8]$
$=64\pi -\frac{64}{\sqrt{3}}-32\pi +16\sqrt{3}+\frac{32\pi }{3}$
$=\frac{128\pi }{3}-\frac{16\sqrt{3}}{3}$
$=\frac{16}{3}(8\pi -\sqrt{3})$ sq units