Proof 1
Calculus Derivation via Disk Method (Definite Integration)
Consider a right circular cone oriented vertically along the $z$-axis with its apex located at $z = 0$ and its circular base of radius $r$ at $z = h$.
Step 1: Express Cross-Sectional Radius at Height $z$
By similar right triangles, the radius $R(z)$ at any vertical position $z$ increases linearly from $0$ to $r$:
$$R(z) = \frac{r}{h} \cdot z$$
Step 2: Cross-Sectional Circular Disk Area $A(z)$
The infinitesimal area of each circular slice perpendicular to the $z$-axis is:
$$A(z) = \pi [R(z)]^2 = \pi \left(\frac{r}{h} z\right)^2 = \pi \frac{r^2}{h^2} z^2$$
Step 3: Integrate Infinitesimal Disks from $z = 0$ to $z = h$
We integrate all slices over the entire height $h$:
$$V = \int_0^h A(z) \, dz = \int_0^h \pi \frac{r^2}{h^2} z^2 \, dz = \pi \frac{r^2}{h^2} \left[ \frac{z^3}{3} \right]_0^h$$
Evaluating the upper limit $z = h$:
$$V = \pi \frac{r^2}{h^2} \cdot \frac{h^3}{3} = \mathbf{\frac{1}{3} \pi r^2 h} \quad \blacksquare$$
Notice that the factor of $\frac{1}{3}$ arises directly from the power rule of integration: $\int z^2 dz = \frac{z^3}{3}$. Because cross-sectional area expands quadratically with dimension ($z^2$), volume accumulates at one-third the rate of a constant prism or cylinder!
Proof 2
Archimedes' Mechanical Lever Method (250 BC)
In his celebrated treatise The Method of Mechanical Theorems, Archimedes discovered the volume of a sphere and cone by balancing geometric solids on an imaginary lever.
He suspended cross-sectional slices of a cone, sphere, and cylinder on a balance beam. Using the law of the lever ($Force \times Distance = Constant$), Archimedes demonstrated that a cylinder placed at its center of mass exactly balances a cone and a sphere placed at the lever's end. This proved:
$$V_{\text{cone}} + V_{\text{sphere}} = \frac{1}{2} V_{\text{cylinder}}$$
$$V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}} = \frac{1}{3}\pi r^2 h$$
Archimedes considered this ratio ($1:2:3$ between cone, sphere, and cylinder) to be his crowning mathematical achievement, requesting a cylinder circumscribing a sphere to be carved onto his tombstone.
Proof 3
Cavalieri's Principle & Triangular Pyramid Dissection
Euclid's Elements (Book XII, Proposition 7) proved that any triangular prism can be cleanly dissected into three tetrahedrons (pyramids) of equal volume.
Because the volume of a triangular prism is $V_{\text{prism}} = B \cdot h$, each triangular pyramid must have volume:
$$V_{\text{pyramid}} = \frac{1}{3} B h$$
By approximating the circular base of a cone with an inscribed regular polygon of $n$ sides and taking the limit as $n \to \infty$, a cone is simply a pyramid with an infinite number of triangular facets. Therefore, the volume of any cone—whether right, oblique, or truncated—is fundamentally $\frac{1}{3} B h = \frac{1}{3}\pi r^2 h$.