A layer through the center of a solid: Find the volume

[sgulee3]

where this goes but in my calculus II class we were given a extra credit opportunity. Here's the problem.

A layer through the center of a solid is the region enclosed by a circle with a radius of r cm. And each plane section perpendicular to a fixed diameter of the layer is a regular hexagon having a chord of the circle as a diagonal of the hexagon. Find the volume of the solid.

 

[soroban]

Re: Not sure

Hello, sgulee3!

A layer through the center of a solid is the region enclosed by a circle with a radius of r cm.
And each plane section perpendicular to a fixed diameter of the layer is a regular hexagon
having a chord of the circle as a diagonal of the hexagon.
Find the volume of the solid.

                s
           *- - - - -*
          / \       / \
         /   \     /   \
       s/    s\   /s    \s
       /       \ /       \
      * - - - - * - - - - *
       \   s   / \   s   /
        \     /   \     /
         \   /     \   /
          \ /       \ /
           *- - - - -*

A regular hexagon is composed of six equilateral triangles of side \(\displaystyle s\).
The area of an equilateral triangle is: \(\displaystyle \:\frac{\sqrt{3}}{4}s^2\)
The area of a hexagon is: \(\displaystyle \:6\cdot\frac{\sqrt{3}}{4}s^2\:=\:\frac{3\sqrt{3}}{2}s^2\)

The circle has equation: \(\displaystyle \:x^2\,+\,y^2\:=\:r^2\)
. . Then: \(\displaystyle \:s \:=\:y\:=\:\sqrt{r^2\,-\,x^2}\)

The area of a hexagon is: \(\displaystyle \:\frac{3\sqrt{3}}{2}\left(\sqrt{r^2\,-\,x^2}\right)^2\:=\:\frac{3\sqrt{3}}{2}\left(r^2\,-\,x^2\right)\)

The volume of a hexagonal slab is: \(\displaystyle \:\frac{3\sqrt{3}}{2}\left(r^2\,-\,x^2\right)\,dx\)


Now "add" these slabs from \(\displaystyle x\,=\,-r\) to \(\displaystyle x\,=\,r\) . . .

. . \(\displaystyle \L V\;=\;\frac{3\sqrt{3}}{2}\int^{\;\;\;r}_{-r}\left(r^2\,-\,x^2\right)\,dx\)