Volume of the cone

Consider a cone with base radius $R$ and height $H$. At height $z$ the radius of the cross-section, by proportion, is $r(z) = R(H-z)/H$. The elementary disc has volume $dV = \pi r^2\,dz$.

Summing the discs from $z=0$ to $z=H$: $V = \frac{\pi R^2}{H^2}\int_0^H(H-z)^2\,dz = \frac{\pi R^2}{H^2}\cdot\frac{H^3}{3} = \boxed{\dfrac{1}{3}\pi R^2 H}.$

The factor $\tfrac13$ is the same one that appears in the pyramid: both have cross-sections that shrink linearly towards the apex.

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Topics: Integrale
Concepts: Integrale definito · Volume per sezioni
Skills: Calcolare · Integrare