Question

The area (in sq. units) of the part of the circle $x ^{2}+ y ^{2}=36$, which is outside the parabola $y ^{2}=9 x ,$ is

• A. $24 \pi+3 \sqrt{3}$
• B. $12 \pi-3 \sqrt{3}$
• C. $24 \pi-3 \sqrt{3}$
• D. $12 \pi+3 \sqrt{3}$

Answer 1

Answer: (C)

Required area
$=\pi \times(6)^{2}-2 \int\limits_{0}^{3} \sqrt{9} x d x-\int\limits_{3}^{6} \sqrt{36-x^{2}} d x$
$=36 \pi-12 \sqrt{3}-2\left(\frac{ x }{2} \sqrt{36- x ^{2}}+18 \sin ^{-1} \frac{ x }{6}\right)^{6}$
$=36 \pi-12 \sqrt{3}-2\left(9 \pi-3 \pi-\frac{9 \sqrt{3}}{2}\right)$
$=24 \pi-3 \sqrt{3}$