Question 100 of 230MediumMCQ
The area (in sq. units) of the part of the circle x2+y2=36, which is outside the parabola y2=9x, is: Solutions · 1
Official solution
The circle and parabola meet where 9x+x2=36... using the source's intersection at x=3:
\[\text{Required area} = \pi(6)^2 - 2\int_0^3\sqrt{9x}\,dx - 2\int_3^6\sqrt{36-x^2}\,dx\]
\[= 36\pi - 12\sqrt{3} - 2\left[\frac{x}{2}\sqrt{36-x^2}+18\sin^{-1}\frac{x}{6}\right]_3^6 = 36\pi-12\sqrt{3}-2\left(9\pi-3\pi-\frac{9\sqrt{3}}{2}\right)\]
\[= 36\pi - 12\sqrt{3} - 12\pi + 9\sqrt{3} = 24\pi - 3\sqrt{3}\] Answer the question to unlock