1. **Problem:** Find the volume of a solid with a circular base of radius 1, where cross sections perpendicular to the base are equilateral triangles. 2. **Formula and rules:** The volume of the solid can be found by integrating the area of the cross sections over the base interval. - The base is a circle: $$x^2 + y^2 = 1$$, so the length of the base segment at position $x$ is $$2\sqrt{1 - x^2}$$. - Each cross section is an equilateral triangle with side length $$s = 2\sqrt{1 - x^2}$$. - Area of an equilateral triangle with side $s$ is $$A = \frac{\sqrt{3}}{4} s^2$$. 3. **Calculate the area of the cross section:** $$A(x) = \frac{\sqrt{3}}{4} \left(2\sqrt{1 - x^2}\right)^2 = \frac{\sqrt{3}}{4} \times 4(1 - x^2) = \sqrt{3}(1 - x^2)$$ 4. **Set up the volume integral:** $$V = \int_{-1}^1 A(x) \, dx = \int_{-1}^1 \sqrt{3}(1 - x^2) \, dx = \sqrt{3} \int_{-1}^1 (1 - x^2) \, dx$$ 5. **Evaluate the integral:** $$\int_{-1}^1 (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^1 = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \left(\frac{2}{3}\right) - \left(-\frac{2}{3}\right) = \frac{4}{3}$$ 6. **Calculate the volume:** $$V = \sqrt{3} \times \frac{4}{3} = \frac{4\sqrt{3}}{3}$$ **Final answer:** The volume of the solid is $$\boxed{\frac{4\sqrt{3}}{3}}$$ which corresponds to option D.