Question

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

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

Answer 1

The correct option is D $24\pi -3\sqrt{3}$

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

Required area
$=\pi r^2-2\left[{\int }_0^3\sqrt{9x}dx+{\int }_3^6\sqrt{36-x^2}dx\right]$
$=36\pi -12\sqrt{3}-2{[\frac{x}{2}\sqrt{36-x^2}+18{\sin }^{-1}\left(\frac{x}{6}\right)]}_3^6$
$=36\pi -12\sqrt{3}-2(9-(\frac{9\sqrt{3}}{2}+3\pi ))=24\pi -3\sqrt{3}$