# Given a right triangle A B C rotated 2 &#x03C0;<!-- π --> , one full circle, aro

Given a right triangle $ABC$ rotated $2\pi$, one full circle, around line $BC$. Find area of the formed cone.
Everyone knows that the formula for a cone includes a $\frac{1}{3}$ in it due to some integral calculus.
Why can the area of a cone not be visualized as a triangle rotated about an axis? Intuitively this makes sense. The area of a triangle is $\frac{1}{2}bh$. The distance it is rotated is $2\pi$. By that this should give us a formula $bh\pi$. This seems to work for a cone with $r=3$ and height $4$.
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Paxton James
The area is the area of the base ($\pi {r}^{2}$ where $r=\overline{AB}$) and the area of the curved portion is $\pi {R}^{2}\theta$ where $R=\overline{AC}$ and $\theta$ is the angle subtended by the surface if flattened onto a plane. The length of the perimeter is $\pi {r}^{2}$ and that is the proportion of the full $2\pi$central angle by $\frac{2\pi r}{2\pi R}$.
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aligass2004yi
First note that the triangle is not rotated through "a distance of 2π" but through an angle of 2π, which is not the same thing.
Also, you seem to be assuming that the volume will be the area multiplied by the angle of rotation, which is not true.
A similar result which is true is Pappus' theorem, which states that the volume swept out by a plane figure when rotated is equal to its area, multiplied by the distance travelled by its centroid.