Consider a solid whose base is that of an arbitrary flat two-dimensional shape, and as we move up the solid, the cross-sections cut horizontally are all similar to the base, and the change in some linear measure of these sections, such as the perimeter, is linear itself.
Let's let $A_i$ be the area of the base, and $r_i$ be some linear measure of its shape. We could then state:
$$A_i=kr_i^2$$ where $$0<k$$ is the constant of proportionality.
This stems from the fact that the area of a two-dimensional surface is proportional to the square of some linear component of the surface shape.
We may find the volume by orienting an $x$-axis through the center of the solid and perpendicular to the slicing, with the origin at the base. Computing the volume of a slice, we find:
$$dV=kr_x^2\,dx$$
Now, let $r_x$ vary linearly from $r_i$ (the initial value) to $r_f$ (the final value) over the interval $[0,h]$, hence:
$$r_x=\frac{r_f-r_i}{h}x+r_i$$
Thus, we may state:
$$dV=k\left(\frac{r_f-r_i}{h}x+r_i \right)^2\,dx$$
And so, adding the slices through integration, we get the volume:
$$V=k\int_0^h\left(\frac{r_f-r_i}{h}x+r_i \right)^2\,dx$$
Using the substitution:
$$u=\frac{r_f-r_i}{h}x+r_i\,\therefore\,du=\frac{r_f-r_i}{h}\,dx$$
we obtain:
$$V=\frac{kh}{r_f-r_i}\int_{r_i}^{r_f}u^2\,du$$
Applying the FTOC, we find:
$$V=\frac{kh}{3\left(r_f-r_i \right)}\left[u^3 \right]_{r_i}^{r_f}$$
$$V=\frac{kh\left(r_f^3-r_i^3 \right)}{3\left(r_f-r_i \right)}$$
Using the difference of cubes formula, we may write:
$$V=\frac{kh\left(r_f-r_i \right)\left(r_f^2+r_fr_i+r_i^2 \right)}{3\left(r_f-r_i \right)}$$
Dividing out common factors:
$$V=\frac{kh\left(r_f^2+r_fr_i+r_i^2 \right)}{3}$$
Distributing the factor $k$, we have:
$$V=\frac{h}{3}\left(kr_f^2+\sqrt{kr_f^2}\sqrt{kr_i^2}+kr_i^2 \right)$$
Using $$A_i=kr_i^2,\,A_f=kr_f^2$$, and rearranging, there results:
$$V=\frac{h}{3}\left(A_i+\sqrt{A_iA_f}+A_f \right)$$
We may now use this general formula to find the formulas of the volumes of specific solids.
Volume of pyramid/cone:
If $A_i=B$ is the area of the base, and $A_f=0$ since the top is a point, then we find:
$$V=\frac{h}{3}\left(B+\sqrt{B\cdot0}+0 \right)$$
$$V=\frac{1}{3}Bh$$
This should be familiar as the the oft-cited volume of a pyramid. This can be used for a cone as well, where $$B=\pi r^2$$:
$$V=\frac{1}{3}\pi r^2h$$
Volume of tetrahedron:
This is a triangular pyramid have four congruent equilateral triangular faces. If we let $s$ be the length of the edges, then:
$$A_i=\frac{1}{2}s^2\sin\left(60^{\circ} \right)=\frac{\sqrt{3}}{4}s^2$$
$$A_f=0$$
To find the height, we may use the Pythagorean theorem:
$$h^2=\left(\frac{\sqrt{3}}{2}s \right)^2-\left(\frac{1}{2\sqrt{3}}s \right)=\frac{2}{3}s^2$$
$$h=\sqrt{\frac{2}{3}}s$$
And so we have:
$$V=\frac{1}{3}\sqrt{\frac{2}{3}}s\left(\frac{\sqrt{3}}{4}s^2 \right)=\frac{s^3}{6\sqrt{2}}=\frac{\sqrt{2}}{12}s^3$$
Regular polygonal pyramid:
The area of an $n$-gon whose sides are $s$ is easily found to be:
$$A_i=\frac{n}{4}\cot\left(\frac{\pi}{n} \right)s^2$$
And so the volume is:
$$V=\frac{h}{3}\left(\frac{n}{4}\cot\left(\frac{\pi}{n} \right)s^2 \right)=\frac{hn}{12}\cot\left(\frac{\pi}{n} \right)s^2$$
Truncated pyramids/cones:
A truncated pyramid is one that does not come to a point, and so $A_f\ne0$. It is as if the "top" has been cut off.
Cone (frustum):
Let $R$ be the radius of the base, and $r$ be the radius of the top, and so we have:
$$A_i=\pi R^2$$
$$A_f=\pi r^2$$
Hence:
$$V=\frac{h}{3}\left(\pi R^2+\sqrt{\pi R^2\cdot\pi r^2}+\pi r^2 \right)=\frac{\pi h}{3}\left(R^2+Rr+r^2 \right)$$
Square pyramid:
Let $a$ be the side lengths of the base and $b$ be the side lengths of the top:
$$A_i=a^2$$
$$A_f=b^2$$
Thus, we find:
$$V=\frac{h}{3}\left(a^2+\sqrt{a^2b^2}+b^2 \right)=\frac{h}{3}\left(a^2+ab+b^2 \right)$$
Polygonal pyramid:
Let $a$ be the side lengths of the base $n$-gon and $b$ be the side lengths of the top $n$-gon, and we then have:
$$A_i=\frac{n}{4}\cot\left(\frac{\pi}{n} \right)a^2$$
$$A_f=\frac{n}{4}\cot\left(\frac{\pi}{n} \right)b^2$$
And thus, we find:
$$V=\frac{h}{3}\left(\frac{n}{4}\cot\left(\frac{\pi}{n} \right)a^2+\sqrt{\frac{n}{4}\cot\left(\frac{\pi}{n} \right)a^2\cdot\frac{n}{4}\cot\left(\frac{\pi}{n} \right)b^2}+\frac{n}{4}\cot\left(\frac{\pi}{n} \right)b^2 \right)=\frac{hn}{12}\cot\left(\frac{\pi}{n} \right)\left(a^2+ab+b^2 \right)$$
There are certainly many other pyramid types, but this gives the reader an idea how to apply the general formula we found to specific pyramidal solids.
