# I really don't understand this calculus 2 problem?

## The question is: An equilateral hexagon is revolving around one of its edges. Find the volume of the solid of revolution

Apr 29, 2017

$V = \pi {l}^{3}$

#### Explanation:

Considering $l$ as the equilateral hexagon side

$h = l \cos \phi$
$\delta = l \sin \phi$

we have

$V = 2 \frac{1}{3} \pi {h}^{2} \delta + \pi {h}^{2} l = \left(\frac{2}{3} {\cos}^{2} \phi \sin \phi + {\cos}^{2} \phi\right) \pi {l}^{3}$

but $\phi = \frac{\pi}{3}$ so $\sin \phi = \frac{1}{2}$ and $\cos \phi = \frac{\sqrt{3}}{2}$ and finally

$V = \pi {l}^{3}$

Apr 30, 2017 Pappus 2nd Theorem: Volume V of a solid of revolution generated by rotating a plane figure about an external axis is equal to the area of the figure times the distance traveled by its geometric centroid, or $V = A d$

$A$ is calculated in the figure as $6 \times \frac{1}{2} b h$

$d = 2 \pi \left(s \frac{\sqrt{3}}{2}\right)$

$\implies V = \frac{9 \pi}{2} {s}^{3}$

I have solved this way, using integrals, as shown below: 