Given curves

• Circle: $x^2+y^2=36$ This is a circle of radius $6$ centered at the origin.

• Parabola: $y^2=9x$ This opens to the right.

We need the area of the part of the circle that lies outside the parabola.

Find points of intersection

From the parabola, $x=\frac{y^2}{9}$

Substitute into the circle: $\left(\frac{y^2}{9}\right)^2+y^2=36$ $\frac{y^4}{81}+y^2=36$ Multiply by $81$: $y^4+81y^2-2916=0$

Let $t=y^2$. Then $t^2+81t-2916=0$

Solve: $t=27$ since $27^2+81\cdot 27=729+2187=2916$

Hence, $y^2=27 \implies y=\pm 3\sqrt{3}$

Then $x=\frac{y^2}{9}=\frac{27}{9}=3$

So the curves intersect at $(3,3\sqrt3),\quad (3,-3\sqrt3)$

Understand the required region

The parabola $y^2=9x$ means points inside the parabola satisfy $y^2\le 9x \iff x\ge \frac{y^2}{9}$

Therefore, points outside the parabola satisfy $x<\frac{y^2}{9}$

Inside the circle, for a fixed $y$, $x$ ranges from $-\sqrt{36-y^2} \text{ to } \sqrt{36-y^2}$

The portion inside the circle and outside the parabola is the part from $x=-\sqrt{36-y^2} \text{ to } x=\frac{y^2}{9}$ for those $y$ where the parabola cuts the circle, i.e. $-3\sqrt3\le y\le 3\sqrt3$.

But it is easier to compute:

$\text{Required area} = \text{Area of circle} - \text{Area inside both circle and parabola}$

Area common to circle and parabola

For $-3\sqrt3\le y\le 3\sqrt3$, the common region lies from $x=\frac{y^2}{9} \text{ to } x=\sqrt{36-y^2}$

So $A_{\text{common}}=\int_{-3\sqrt3}^{3\sqrt3}\left(\sqrt{36-y^2}-\frac{y^2}{9}\right)dy$

By symmetry, $A_{\text{common}}=2\int_0^{3\sqrt3}\left(\sqrt{36-y^2}-\frac{y^2}{9}\right)dy$

Evaluate the integral

Split it: $A_{\text{common}}=2\left[\int_0^{3\sqrt3}\sqrt{36-y^2}\,dy-\int_0^{3\sqrt3}\frac{y^2}{9}\,dy\right]$

First integral

Use $\int \sqrt{a^2-y^2}\,dy=\frac{y}{2}\sqrt{a^2-y^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{y}{a}\right)$ with $a=6$.

Therefore, $\int_0^{3\sqrt3}\sqrt{36-y^2}\,dy$ $=\left[\frac{y}{2}\sqrt{36-y^2}+18\sin^{-1}\left(\frac{y}{6}\right)\right]_0^{3\sqrt3}$

At $y=3\sqrt3$: $\sqrt{36-27}=3$ so $\frac{y}{2}\sqrt{36-y^2}=\frac{3\sqrt3}{2}\cdot 3=\frac{9\sqrt3}{2}$ and $\sin^{-1}\left(\frac{3\sqrt3}{6}\right)=\sin^{-1}\left(\frac{\sqrt3}{2}\right)=\frac{\pi}{3}$

Thus, $\int_0^{3\sqrt3}\sqrt{36-y^2}\,dy=\frac{9\sqrt3}{2}+18\cdot\frac{\pi}{3}$ $=\frac{9\sqrt3}{2}+6\pi$

Second integral

$\int_0^{3\sqrt3}\frac{y^2}{9}\,dy=\frac{1}{9}\left[\frac{y^3}{3}\right]_0^{3\sqrt3}$ $=\frac{1}{27}(3\sqrt3)^3$

Now, $(3\sqrt3)^3=27\cdot 3\sqrt3=81\sqrt3$

Hence, $\int_0^{3\sqrt3}\frac{y^2}{9}\,dy=\frac{81\sqrt3}{27}=3\sqrt3$

Therefore, $A_{\text{common}}=2\left(\frac{9\sqrt3}{2}+6\pi-3\sqrt3\right)$ $=2\left(6\pi+\frac{3\sqrt3}{2}\right)$ $=12\pi+3\sqrt3$

Area of the circle

$A_{\text{circle}}=\pi r^2=36\pi$

Required area

$A=36\pi-(12\pi+3\sqrt3)$ $=24\pi-3\sqrt3$

Match with options

$24\pi-3\sqrt3$ corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

Hence, they agree.