Calc1.83

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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.


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