Calc1.83
From Exampleproblems
Find the volume, on the interval 
, of a 3-D object whose cross-section at any given point is an equilateral triangle with side length 
.
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 
. So the area of a given equilateral triangle in this problem is given by 
. 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}\,](/wiki/images/math/9/f/4/9f41b77c4cb190cbeb3778186f01430a.png)
Thus, the volume of this shape is 
units.
