棱锥体和圆锥体的体积

Suppose that we have a cone that has a height $h$ and base area $A$. We would like to find the volume $V$ of this cone.

Consider the plane parallel to the base that passes through the vertex of the cone. Let $x$ denote the distance from a given point to this plane.

From the integral formula for volumes, $V=\int_{0}^{h}S(x)dx$, where $S(x)$ is the cross-sectional area for a certain value of $x$.

From the definition of a cone, a plane parallel to the base will cut a cross-section that is similar to the base.

Also, for a value $x$, the scale factor from the base to the cross-section is $\frac{x}{h}$.

Thus, the ratio of the area of the base to the area of the cross-section is $\frac{x^2}{h^2}$.

Therefore, the area of the cross-section is $S(x)=\frac{Ax^2}{h^2}$.

Integrating to calculate the volume,

$$V=\int_{0}^{h}S(x)dx$$ $$V=\int_{0}^{h}\frac{Ax^2}{h^2}dx$$ $$V=\frac{1}{3}Ah$$

Thus, we see that the volume of a cone is one third of the product of the base area and the height.

Notice the similarity to the area of a triangle, which is one half of the product of the base length and the height. In general, the "volume" of an $n$-dimensional cone is equal to $\frac{1}{n}$ of the product of the height and the "volume" of the $(n-1)$-dimensional base. The $\frac{1}{n}$ coefficient comes from the integration of $x^{n-1}$, which is a result of the cross-sections being similar to the base.