1. The problem is to evaluate the definite integral $$\int_{-3}^{0.3} \sqrt{9 - x^2} \, dx.$$\n\n2. This integral represents the area under the curve of the function $y = \sqrt{9 - x^2}$ from $x = -3$ to $x = 0.3$. The function describes the upper half of a circle with radius 3 centered at the origin.\n\n3. The formula for the area of a circle segment is useful here. The integral of $\sqrt{r^2 - x^2}$ from $a$ to $b$ is given by:\n$$\int_a^b \sqrt{r^2 - x^2} \, dx = \frac{r^2}{2} \left( \arcsin\left(\frac{b}{r}\right) - \arcsin\left(\frac{a}{r}\right) \right) + \frac{b}{2} \sqrt{r^2 - b^2} - \frac{a}{2} \sqrt{r^2 - a^2}.$$\n\n4. Here, $r = 3$, $a = -3$, and $b = 0.3$. Substitute these values:\n$$\int_{-3}^{0.3} \sqrt{9 - x^2} \, dx = \frac{9}{2} \left( \arcsin\left(\frac{0.3}{3}\right) - \arcsin\left(\frac{-3}{3}\right) \right) + \frac{0.3}{2} \sqrt{9 - 0.3^2} - \frac{-3}{2} \sqrt{9 - (-3)^2}.$$\n\n5. Calculate each term:\n- $\arcsin\left(\frac{0.3}{3}\right) = \arcsin(0.1) \approx 0.100167$ radians.\n- $\arcsin\left(\frac{-3}{3}\right) = \arcsin(-1) = -\frac{\pi}{2} \approx -1.5708$ radians.\n- $\sqrt{9 - 0.3^2} = \sqrt{9 - 0.09} = \sqrt{8.91} \approx 2.9833.$\n- $\sqrt{9 - 9} = \sqrt{0} = 0.$\n\n6. Substitute these values back:\n$$= \frac{9}{2} (0.100167 + 1.5708) + \frac{0.3}{2} (2.9833) - \frac{-3}{2} (0)$$\n$$= 4.5 \times 1.670967 + 0.15 \times 2.9833 + 0$$\n$$= 7.51935 + 0.4475 + 0 = 7.96685.$$\n\n7. Therefore, the value of the integral is approximately $$7.967.$$