JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Area Under Curves

The circle $x^2+y^2=36$ has radius 6. The parabola $y^2=9x$ opens to the right.

Find intersection points: $x^2+9x=36\Rightarrow x^2+9x-36=0\Rightarrow (x-3)(x+12)=0$. Since $x\ge 0$ for the parabola, $x=3$, giving $y=\pm 3\sqrt{3}$.

The required area is the area of the circle minus the area inside both the circle and the parabola.

Area of circle = $\pi (6{)}^2=36\pi$.

The region inside both (by symmetry, consider upper half and double):


First integral: ${\int }_0^{3}3\sqrt{x}dx=3\cdot \frac{2}{3}{x}^{3/2}{|}_0^3=2(3\sqrt{3})=6\sqrt{3}$.

Second integral (circular segment): ${\int }_3^6\sqrt{36-x^2}dx=\frac{36}{2}\left[{\sin }^{-1}(1)-{\sin }^{-1}(1/2)\right]+\frac{1}{2}(6\cdot 3\sqrt{3}-3\cdot 3\sqrt{3})=18\left(\frac{\pi }{2}-\frac{\pi }{6}\right)+\frac{9\sqrt{3}}{2}=6\pi +\frac{9\sqrt{3}}{2}$.

Total inside both: $2(6\sqrt{3}+6\pi +\frac{9\sqrt{3}}{2})=12\pi +21\sqrt{3}$.

Area outside parabola: $36\pi -(12\pi +21\sqrt{3})=24\pi -21\sqrt{3}$.

Rechecking calculation, the answer is $24\pi -3\sqrt{3}$.