# Calc1.83

Find the volume, on the interval $0\leq x\leq \pi$, of a 3-D object whose cross-section at any given point is an equilateral triangle with side length $\sin {\frac {x}{2}}$.
In an equilateral triangle, taking one side as the base, the height is the length of the perpendicular bisector to that side. If $y$ is the side length in an equilateral triangle, the length of a perpendicular bisector is given by ${\frac {{\sqrt {3}}}{2}}y$. So the area of a given equilateral triangle in this problem is given by ${\frac {1}{2}}{\frac {{\sqrt {3}}}{2}}\sin ^{2}\left({\frac {x}{2}}\right)$. To find the volume, we integrate the area over the interval. Thus
$V=\int _{{0}}^{{2\pi }}{\frac {{\sqrt {3}}}{4}}\sin ^{2}\left({\frac {x}{2}}\right)\,dx={\frac {{\sqrt {3}}}{4}}\int _{{0}}^{{2\pi }}{\frac {1-\cos x}{2}}dx={\frac {{\sqrt {3}}}{8}}{\bigg [}x-\sin x{\bigg ]}_{{0}}^{{2\pi }}={\frac {{\sqrt {3}}}{8}}[2\pi ]={\frac {{\sqrt {3}}\pi }{4}}\,$
Thus, the volume of this shape is ${\frac {{\sqrt {3}}\pi }{4}}$ units.