# The area (in sq. units) of the part of the circle

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 :

1. (1) $24 \pi+3 \sqrt{3}$

2. (2) $12 \pi+3 \sqrt{3}$

3. (3) $12 \pi-3 \sqrt{3}$

4. (4) $24 \pi-3 \sqrt{3}$

Correct Option: , 4

Solution:

The curves intersect at point $(3, \pm 3 \sqrt{3})$

Required area

$=\pi r^{2}-2\left[\int_{0}^{3} \sqrt{9 x} d x+\int_{3}^{6} \sqrt{36-x^{2}} d x\right]$

$=36 \pi-12 \sqrt{3}-2\left(\frac{x}{2} \sqrt{36-x^{2}}+18 \sin ^{-1}\left(\frac{x}{6}\right)\right)_{3}^{6}$

$=36 \pi-12 \sqrt{3}-2\left(9-\left(\frac{9 \sqrt{3}}{2}+3 \pi\right)\right)=24 \pi-3 \sqrt{3}$