1. **State the problem:** Find the volume of the solid generated when the area in the first quadrant of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is revolved around the x-axis. 2. **Formula used:** The volume of a solid of revolution around the x-axis is given by the disk method: $$V = \pi \int_0^a [f(x)]^2 \, dx$$ where $f(x)$ is the function describing the curve. 3. **Express $y$ in terms of $x$:** From the ellipse equation, $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies y = b \sqrt{1 - \frac{x^2}{a^2}}$$ 4. **Set up the integral:** The volume is $$V = \pi \int_0^a \left(b \sqrt{1 - \frac{x^2}{a^2}}\right)^2 \, dx = \pi b^2 \int_0^a \left(1 - \frac{x^2}{a^2}\right) \, dx$$ 5. **Simplify the integral:** $$V = \pi b^2 \int_0^a \left(1 - \frac{x^2}{a^2}\right) \, dx = \pi b^2 \left[ \int_0^a 1 \, dx - \int_0^a \frac{x^2}{a^2} \, dx \right]$$ 6. **Evaluate each integral:** $$\int_0^a 1 \, dx = a$$ $$\int_0^a \frac{x^2}{a^2} \, dx = \frac{1}{a^2} \int_0^a x^2 \, dx = \frac{1}{a^2} \cdot \frac{a^3}{3} = \frac{a}{3}$$ 7. **Substitute back:** $$V = \pi b^2 \left(a - \frac{a}{3}\right) = \pi b^2 \cdot \frac{2a}{3} = \frac{2 \pi a b^2}{3}$$ **Final answer:** $$\boxed{V = \frac{2 \pi a b^2}{3}}$$ This is the volume of the solid generated by revolving the first quadrant area of the ellipse around the x-axis.