1. **State the problem:** We need to find the total surface area of the solid formed by rotating the curve given by $$y=\sqrt{25-x^2}$$ about the x-axis for $$x$$ in the interval $$[-1,3]$$. 2. **Recall the formula for surface area of revolution about the x-axis:** $$S=2\pi \int_a^b y \sqrt{1+\left(\frac{dy}{dx}\right)^2} \, dx$$ 3. **Find $$\frac{dy}{dx}$$:** Given $$y=\sqrt{25-x^2}=(25-x^2)^{1/2}$$, $$\frac{dy}{dx} = \frac{1}{2}(25-x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{25-x^2}}$$ 4. **Compute $$1+\left(\frac{dy}{dx}\right)^2$$:** $$1 + \left(\frac{-x}{\sqrt{25-x^2}}\right)^2 = 1 + \frac{x^2}{25-x^2} = \frac{25 - x^2 + x^2}{25 - x^2} = \frac{25}{25 - x^2}$$ 5. **Simplify the integrand:** $$y \sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{25 - x^2} \cdot \sqrt{\frac{25}{25 - x^2}} = \sqrt{25} = 5$$ 6. **Set up the integral for surface area:** $$S = 2\pi \int_{-1}^3 5 \, dx = 10\pi \int_{-1}^3 dx$$ 7. **Evaluate the integral:** $$\int_{-1}^3 dx = 3 - (-1) = 4$$ 8. **Calculate the total surface area:** $$S = 10\pi \times 4 = 40\pi$$ **Final answer:** The total surface area of the solid is $$40\pi$$.