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
$-2\left[\int_{0}^{4} 2 \sqrt{3} \sqrt{x} d x+\int_{4}^{8} \sqrt{64-x^{2}} d x\right]$

$=64 \pi-[2\left(2 \sqrt{3} x^{3 / 2} \times \frac{2}{3}\right)_{0}^{4}\left.+\left(\frac{x}{2} \sqrt{64-x^{2}}+\frac{64}{2} \sin ^{-1} \frac{x}{8}\right)_{4}^{8}\right]$
$=64 \pi-\frac{64}{\sqrt{3}}-32 \pi+16 \sqrt{3}+\frac{32 \pi}{3}$
$=\frac{128 \pi}{3}-\frac{16 \sqrt{3}}{3}$
$=\frac{16}{3}(8 \pi-\sqrt{3})$ sq units