MHB Volume of a Tetrahedron: General Slicing Method

[MarkFL]

Here is the question:

General slicing method to find volume of a tetrahedron?


General slicing method to find volume of a tetrahedron (pyramid with four triangular faces), all whose edges have length 6?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Bringing truth.

Let's orient our coordinate axis, the $x$-axis, such that it is perpendicular to the base and passes through the apex. Let the point where this axis intersects the base be the origin of the axis. The distance from the origin to the apex along this axis is the height $h$ of the tetrahedron. To determine $h$, we may construct a right triangle whose shorter leg is the apothem $a$ of the base, the longer leg is $h$, and the hypotenuse is the slant height $\ell$ of one of the adjoining faces. Let $s$ be the edge lengths os the tetrahedron.

Let's first determine the apothem $a$. If we divide the base into 3 congruent triangles, we may equate the area of the whole triangle to the 3 smaller triangles:

$$\frac{1}{2}s^2\frac{\sqrt{3}}{2}=3\cdot\frac{1}{2}sa$$

$$s\frac{\sqrt{3}}{2}=3a$$

$$a=\frac{s}{2\sqrt{3}}$$

And the slant height is given by:

$$\ell=\frac{\sqrt{3}}{2}s$$

Hence:

$$h=\sqrt{\left(\frac{\sqrt{3}}{2}s \right)^2-\left(\frac{s}{2\sqrt{3}} \right)^2}=s\sqrt{\frac{3}{4}-\frac{1}{12}}=s\sqrt{\frac{2}{3}}$$

Now, if we take cross sections perpendicular to the $x$-axis, we may give the volume of an arbitrary slice as:

$$dV=\frac{\sqrt{3}}{4}s^2(x)\,dx$$

We know $s(x)$ will be a linear function, containing the points:

$$(0,s)$$ and $$(h,0)$$

Thus the slope of the line is:

$$m=-\frac{s}{h}$$

And thus, using the point-slope formula, we find:

$$s(x)=-\frac{s}{h}x+s=\frac{s}{h}\left(h-x \right)$$

Hence, we may state:

$$dV=\frac{\sqrt{3}s^2}{4h^2}\left(h-x \right)^2\,dx$$

Summing up the slices, we may write:

$$V=\frac{\sqrt{3}s^2}{4h^2}\int_0^h\left(h-x \right)^2\,dx$$

Using the substitution:

$$u=h-x\,\therefore\,du=-dx$$ we obtain:

$$V=-\frac{\sqrt{3}s^2}{4h^2}\int_h^0 u^2\,du=\frac{\sqrt{3}s^2}{4h^2}\int_0^h u^2\,du$$

Applying the FTOC, we have:

$$V=\frac{\sqrt{3}s^2}{12h^2}\left[u^3 \right]_0^h=\frac{\sqrt{3}hs^2}{12}$$

Using the value we found for $h$, we have:

$$V=\frac{\sqrt{3}\left(s\sqrt{\frac{2}{3}} \right)s^2}{12}=\frac{\sqrt{2}s^3}{12}$$

Now using the given data $s=6$, we obtain:

$$V=\frac{\sqrt{2}6^3}{12}=18\sqrt{2}$$