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 equal to

We need the area of the circle $$x^2 + y^2 = 36$$ that lies outside the parabola $$y^2 = 9x$$.

First, we find the intersection points. Substituting $$y^2 = 9x$$ into the circle equation: $$x^2 + 9x = 36$$, so $$x^2 + 9x - 36 = 0$$, giving $$(x + 12)(x - 3) = 0$$. Since $$x \geq 0$$, we get $$x = 3$$ and $$y^2 = 27$$, so $$y = \pm 3\sqrt{3}$$.

The area inside both curves (using symmetry about the x-axis) is $$2\left(\int_0^3 \sqrt{9x} \, dx + \int_3^6 \sqrt{36 - x^2} \, dx\right)$$.

For the first integral: $$\int_0^3 3\sqrt{x} \, dx = 3 \cdot \frac{2}{3} x^{3/2} \Big|_0^3 = 2 \cdot 3\sqrt{3} = 6\sqrt{3}$$.

For the second integral, let $$x = 6\sin\theta$$: when $$x = 3$$, $$\theta = \frac{\pi}{6}$$; when $$x = 6$$, $$\theta = \frac{\pi}{2}$$.

$$\int_3^6 \sqrt{36 - x^2} \, dx = \int_{\pi/6}^{\pi/2} 36\cos^2\theta \, d\theta = 36 \cdot \frac{1}{2}\left[\theta + \frac{\sin 2\theta}{2}\right]_{\pi/6}^{\pi/2}$$.

$$= 18\left[\left(\frac{\pi}{2} + 0\right) - \left(\frac{\pi}{6} + \frac{\sin(\pi/3)}{2}\right)\right] = 18\left[\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right] = 6\pi - \frac{9\sqrt{3}}{2}$$.

So the area inside both curves is $$2\left(6\sqrt{3} + 6\pi - \frac{9\sqrt{3}}{2}\right) = 2\left(6\pi + \frac{12\sqrt{3} - 9\sqrt{3}}{2}\right) = 2\left(6\pi + \frac{3\sqrt{3}}{2}\right) = 12\pi + 3\sqrt{3}$$.

The total area of the circle is $$\pi \cdot 6^2 = 36\pi$$.

The area of the circle outside the parabola is $$36\pi - (12\pi + 3\sqrt{3}) = 24\pi - 3\sqrt{3}$$.

Hence, the correct answer is Option C.