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.
$\endgroup$4 Answers
$\begingroup$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 $$ V = \frac{A}{H^2} \int_0^H t^2 dt = \frac{1}{3} AH. $$
$\endgroup$ 3 $\begingroup$You can assume the cone to be constructed of disks of infinitesimal thickness stacked one on top of the other, with the largest disk having radius $R$ at height $h=H$ and the smallest having radius $0$ at height $h=0$.
The radius of any disk in between, $r$ can be written as
$$r=R\frac{h}{H}$$
The volume of a single disk, $dV$ is
$$dV=\pi r^2 dh = \pi r^2 \frac{H}{R}dr$$
and the volume of the cone is then obtained by integrating (summing) the volume of each of the disks
$$V=\int dV=\pi\frac{H}{R}\int_0^R r^2 dr=\frac{1}{3}\pi R^2 H$$
$\endgroup$ 5 $\begingroup$The derivation below is true for any cone (need not be right circular). Let the base area of the cone be $A$ and the height of the cone be $h$. Note that the area at any height $y$ is given by $A(y) = A \times \left( \frac{y}{h} \right)^2$ (Why?). Now the volume of the cone is nothing but $$V = \int_0^h A(y) dy \text{ (Why?)}$$ $$V = \int_0^h A \times \left( \frac{y}{h} \right)^2 dy = \frac1{3} A h$$
$\endgroup$ 6 $\begingroup$$$ V = \int_{V}{\rm d}V = \int_{V}{\nabla\cdot\vec{r} \over 3}\,{\rm d}V = {1 \over 3}\int_{S}\vec{r}\cdot{\rm d}\vec{S}\,\qquad\mbox{( according Gauss )} $$
Whith the cone vertex as the origin, $\vec{r}\cdot{\rm d}\vec{S} \not= 0$ in the base. Then, we get
$$ V = {1 \over 3}\int_{\mbox{base}}h\left\vert{\rm d}\vec{S}\right\vert = {1 \over 3}hA $$
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