Thread: Regular Tetrahedron

How to find the volume of a regular tetrahedron using calculus?

V = ∫ dV = ∫∫∫ dx dy dz




This is what I did

first I was thinking to take a cross-sectional equilateral triangle and integrate it from 0 to h, but I could not figure out how to do that. Instead I integrated each variable as below

I integrated x from 0 to a(h-z)/3h and I doubled that

I integrated y from -2L/3 to (h-z)L/3h

I integrated z from 0 to h

But it failed!

I also tried to divide the object into two parts 2L/3 and L/3. I integrated each part separately. I could find the volume of part 2L/3 correctly, but L/3 was not!

I was seeking to get this result (1/3)(1/2)aLh

of course the volume of a regular tetrahedron is a^3 / (6√2) after replacing L and h in terms of a.

What I would do is volume by slicing. let the length of each edge be $\displaystyle a$, and then the height of the tetrahedron can be written in terms $\displaystyle a$ as follows:

$\displaystyle h^2=\left(\frac{\sqrt{3}}{2}a \right)^2-\left(\frac{1}{2\sqrt{3}}a \right)=\frac{2}{3}a^2$

$\displaystyle h=\sqrt{\frac{2}{3}}a$

We may determine the volume of an arbitrary slice as follows (where $\displaystyle s(y)$ is the side length as a function of the axis along which we will integrate, which is $\displaystyle y$):

$\displaystyle dV=\frac{1}{2}s^2(y)\sin\left(\frac{\pi}{3}\right) \,dy=\frac{\sqrt{3}}{4}s^2(y)\,dy$

Now, we know $\displaystyle a(y)$ will be a linear function, where:

$\displaystyle s(0)=a$

$\displaystyle s\left(\sqrt{\frac{2}{3}}a\right)=0$

Thus:

$\displaystyle s(y)=-\sqrt{\frac{3}{2}}y+a$

And so we have:

$\displaystyle dV=\frac{\sqrt{3}}{4}\left(-\sqrt{\frac{3}{2}}y+a\right)^2\,dy$

Hence:

$\displaystyle V=\frac{\sqrt{3}}{4}\int_0^{\sqrt{\frac{2}{3}}a} \left(-\sqrt{\frac{3}{2}}y+a\right)^2\,dy$

I would now let:

$\displaystyle u=-\sqrt{\frac{3}{2}}y+a\implies du=-\sqrt{\frac{3}{2}}\,dy$

And so our integral becomes:

$\displaystyle V=\frac{1}{2\sqrt{2}}\int_0^{a} u^2\,du=\frac{a^3}{6\sqrt{2}}\quad\checkmark$

wOW, I don't know how you came with this idea, but it was so fantastic. Even if I spent more time trying to solve it with myself, I would never reach this approach. Thanks a lot, MarkFL, I have understood everything you wrote!

@joshuaa: You may be interested to know a general fact about volumes that come to a point like that. Consider the following figure:

Click on the figure to enlarge it. Here we have an object that comes to a point similar to your tetrahedron except the cross section can be any shape. The lateral side is generated by straight lines as shown so the cross sections are the same shape (similar) to the shape of the base. The area of the base is $A$ and the area of the cross section at $x$ is $A(x)$. Now, if you may assume the area at distance $x$ from the point is proportional to the square of $x$, you have $A(x) = kx^2$ for some constant $k$. Since when $x=h$ the area is $A$ you have
$A = kh^2$ so $k = \frac A {h^2}$ and the cross sectional area is $A(x) = \frac A {h^2} x^2$. Now to get the volume we have$$
V = \int_0^h~\frac A {h^2}x^2~dx = \left .\frac A {h^2}\frac {x^3} 3 \right |_0^h =\frac {Ah} 3$$
So the volume of any such solid that comes to a point like that is one-third area of base times height. Works for tetrahedrons, non-right and non-circular cones, etc.