1. **State the problem:** Find the volume of the solid generated by rotating the area in the first quadrant of the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ about the x-axis through 360 degrees. 2. **Formula used:** The volume of a solid of revolution about the x-axis is given by the disk method: $$V = \pi \int_a^b [f(x)]^2 \, dx$$ where $$f(x)$$ is the function representing the radius of the disk at position $$x$$. 3. **Rewrite the ellipse equation to express $$y$$ in terms of $$x$$:** $$\frac{x^2}{9} + \frac{y^2}{4} = 1 \implies \frac{y^2}{4} = 1 - \frac{x^2}{9} \implies y^2 = 4\left(1 - \frac{x^2}{9}\right) = 4 - \frac{4x^2}{9}$$ 4. **Since we are in the first quadrant, $$y \geq 0$$, so:** $$y = \sqrt{4 - \frac{4x^2}{9}} = 2\sqrt{1 - \frac{x^2}{9}}$$ 5. **Set the limits of integration:** In the first quadrant, $$x$$ ranges from $$0$$ to $$3$$ (since $$x^2/9 \leq 1 \Rightarrow x \leq 3$$). 6. **Write the volume integral:** $$V = \pi \int_0^3 \left(2\sqrt{1 - \frac{x^2}{9}}\right)^2 \, dx = \pi \int_0^3 4\left(1 - \frac{x^2}{9}\right) \, dx = 4\pi \int_0^3 \left(1 - \frac{x^2}{9}\right) \, dx$$ 7. **Evaluate the integral:** $$4\pi \int_0^3 \left(1 - \frac{x^2}{9}\right) \, dx = 4\pi \left[ \int_0^3 1 \, dx - \int_0^3 \frac{x^2}{9} \, dx \right] = 4\pi \left[ x \Big|_0^3 - \frac{1}{9} \cdot \frac{x^3}{3} \Big|_0^3 \right]$$ 8. **Calculate each term:** $$x \Big|_0^3 = 3 - 0 = 3$$ $$\frac{1}{9} \cdot \frac{x^3}{3} \Big|_0^3 = \frac{1}{9} \cdot \frac{27}{3} = \frac{1}{9} \cdot 9 = 1$$ 9. **Substitute back:** $$4\pi (3 - 1) = 4\pi \times 2 = 8\pi$$ **Final answer:** $$\boxed{8\pi}$$