1. The problem is to evaluate the definite integral $$\int_{-3}^2 (3\sqrt{9-x^2}) \, dx.$$\n\n2. This integral represents the area under the curve of the function $f(x) = 3\sqrt{9-x^2}$ from $x = -3$ to $x = 2$.\n\n3. Notice that $\sqrt{9-x^2}$ is the equation of a semicircle of radius 3 centered at the origin, specifically the upper half of the circle $x^2 + y^2 = 9$.\n\n4. The function $3\sqrt{9-x^2}$ is just 3 times the semicircle function, so the integral calculates 3 times the area under the semicircle from $-3$ to $2$.\n\n5. The area of a full circle of radius 3 is $$\pi \times 3^2 = 9\pi.$$\n\n6. The area of the upper semicircle (half the circle) is $$\frac{9\pi}{2}.$$\n\n7. We want the area from $-3$ to $2$, which is less than the full semicircle. We can find this by integrating or using geometry.\n\n8. The integral $$\int_{-3}^2 \sqrt{9-x^2} \, dx$$ equals the area under the semicircle from $-3$ to $2$.\n\n9. To find this area, we use the formula for the area of a circular segment: $$A = \frac{r^2}{2} (\theta - \sin \theta),$$ where $\theta$ is the central angle in radians corresponding to the segment.\n\n10. For $x=2$, the angle $\theta$ satisfies $$\cos \theta = \frac{2}{3}.$$ So $$\theta = \arccos \frac{2}{3}.$$\n\n11. The segment from $x=-3$ to $x=2$ corresponds to the angle from $\pi$ (at $x=-3$) to $\theta$. The area from $-3$ to $2$ is the area of the semicircle minus the segment from $2$ to $3$.\n\n12. The segment from $x=2$ to $x=3$ has area $$A = \frac{9}{2} (\arccos \frac{2}{3} - \sin(\arccos \frac{2}{3})).$$\n\n13. Calculate $\sin(\arccos \frac{2}{3}) = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}.$\n\n14. So the segment area is $$\frac{9}{2} \left(\arccos \frac{2}{3} - \frac{\sqrt{5}}{3}\right).$$\n\n15. The area from $-3$ to $2$ under $\sqrt{9-x^2}$ is $$\frac{9\pi}{2} - \frac{9}{2} \left(\arccos \frac{2}{3} - \frac{\sqrt{5}}{3}\right) = \frac{9}{2} \left(\pi - \arccos \frac{2}{3} + \frac{\sqrt{5}}{3}\right).$$\n\n16. Multiply by 3 (from the original integral): $$3 \times \frac{9}{2} \left(\pi - \arccos \frac{2}{3} + \frac{\sqrt{5}}{3}\right) = \frac{27}{2} \left(\pi - \arccos \frac{2}{3} + \frac{\sqrt{5}}{3}\right).$$\n\n17. This is the exact value of the integral.