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## Homework Statement

Find the volume of a solid with a circular base defined by [tex]x^2 + y^2 = 1[/tex] with parallel cross sections perpendicular to the base are equilateral triangles.

**The Solution**

[tex]A(x) = \frac{1}{2} \cdot 2\sqrt{1-x^2} * \sqrt{3}\sqrt{1-x^2} = \sqrt{3}(1-x^2)[/tex]

[tex]\int_{-1}^{1} \sqrt{3}(1-x^2) dx = \frac{4\sqrt{3}}{3}[/tex]

**Question**

Why is the height only [tex] \sqrt{3}\sqrt{1-x^2}[/tex] and not [tex]2 \sqrt{3}\sqrt{1-x^2}[/tex]? I thought [tex]\sqrt{1-x^2}[/tex] is only the half the height of the circle