Q:

Proving the volume of a cone

Abstract definition of a cone: locus of points of line segments join a single point $v$, called vertex with set of coplanar points called base.
Height of cone is distance from vertex to plane.
If I take strange set of co-planer point such as complicated design for example, as base then volume of the set join with some vertex is still given : $\tfrac13$ (base)(height).
How can I prove this in calculus?
I have tried a few it is too late to wake my sister, thank you.

A:

Deno te $A$ the base area and $H$ the height of the cone. now we establish the coordinate system by setting the origin as the vertex of the cone, and the X-axis the line starting from the origin and pependicular to the base, i.e., the X-axis is along the direction of the height. Then take variable x as the distance from any point on X-axis on the interval [0,H]. Denote $A_t$ the area of the transverse section which is defined as the intersection of the plane $x=t$ and the cone. Then the volume of the cone shall be 
$$ V = \int_0^H A_t dt$$
Note that we have 
$$ \frac{A_t}{A} = \frac{t^2}{H^2} $$
Thus we have  $A_t = \frac{A}{H^2} t^2$. Combining these two formula together we get
$$
V = \frac{A}{H^2} \int_0^H t^2 dt = \frac{1}{3} AH.
$$  

