How Can Riemann Sums Calculate the Volume of a Tepee Tent?

[MathsKid007]

THE QUESTION
By using Riemann’s sum, synthesise a mathematical model for finding the exact volume of any ‘tepee’ tent of side s and height h.
HERE'S WHAT I HAVE
View attachment 8289
Am currently stuck on writing a side length for the hexagon at any height 'x'

 

[MarkFL]

Let's first find the volume using the geometric formula:

$$V=\frac{1}{3}bh$$

Now, the base is a regular hexagon having side lengths \(s\), which we can decompose into 6 equilateral triangles:

$$b=6\left(\frac{1}{2}s^2\sin\left(\frac{\pi}{3}\right)\right)=\frac{3\sqrt{3}}{2}s^2$$

Hence:

$$V=\frac{\sqrt{3}}{2}hs^2$$

Now, using volume by slicing we can integrate:

$$V=\frac{3\sqrt{3}}{2}\int_0^h s(y)^2\,dy$$

We know \(s(y)\) will vary linearly, where:

$$s(0)=s$$

$$s(h)=0$$

Thus:

$$s(y)=-\frac{s}{h}y+s$$

And so we may write:

$$V=\frac{3\sqrt{3}}{2}\int_0^h \left(-\frac{s}{h}y+s\right)^2\,dy$$

At this point, I would consider the substitution:

$$u=-\frac{s}{h}y+s\implies du=-\frac{s}{h}dy$$

$$V=\frac{3\sqrt{3}h}{2s}\int_0^s u^2\,du=\frac{3\sqrt{3}h}{2s}\left(\frac{s^3}{3}\right)=\frac{\sqrt{3}}{2}hs^2\quad\checkmark$$

It appears you are directly integrating rather than forming a Riemann sum. Is that on purpose, or are you integrating first, and then going to formulate the sum?

 

[MathsKid007]

Could u explain the process at:
We know that s(y) will vary linearly such that:
s(0)=s
s(h)=0
∴s(y)= -s/h y+s
Thanks :D

 

[MarkFL]

Could u explain the process at:
We know that s(y) will vary linearly such that:
s(0)=s
s(h)=0
∴s(y)= -s/h y+s
Thanks :D

We know that at the bottom, where $$h=0$$, the length of the sides of the hexagonal cross section is $$s$$. We also know that at the top these sides lengths have diminished to 0. As they have diminished linearly, and we know two points on this line we have all we need to determine the side lengths as a function of the variable height $$y$$ using the definition of slope and the point-slope formula:

$$s(y)-s=\frac{0-s}{h-0}(y-0)$$

$$s(y)=-\frac{s}{h}y+s$$