Finding the volume of a cone by integration of parabolic conic sections I am working on a purely academic way of finding the volume of a right circular cone of height $h$ and radius $r$, (assume $h > r$), using integration of parabolic conic sections (conic sections parallel to the side of the cone). The end result being the geometric equation for the volume of a cone
$$v=\frac{π r^{2}h}{3}$$
The part I'm having trouble with is figuring out and write the equation for the parabolas.
I know that:

*

*$h$ = height of the cone

*$r$ = radius of the cone,

*$s$ = side length of the cone = $\sqrt{r^{2} + h^{2}}$.

Limits of integration


*

*I found the limits of integration to be from $0$ to $D$, where $D = cos(90 - \theta)2r$. Where $\theta$ is the angle from the base (viewing the cone from the side, giving me a triangular cross section) to the side $s$ of the cone.


*I arrived at the upper limit of integration being $cos(90 - \theta)2r$ by creating a right triangle where the hypotenuse is the base $2r$, the short side is part of side $s$, and the long side is a line perpendicular to $s$, from $s$ to the opposite corner of the triangular cross section of the cone $\theta_2$.


*Using this triangle I can find the length of the perpendicular line by stating that $cos(\theta_2) = \frac {adj}{hyp}$ where $adj$ is the adjacent side of the right triangle to $\theta_2$, and $hyp$ is the hypotenuse of the right triangle


*I know that $\theta_2=180 - 90 - \theta$, therefor $\theta_2 = 90 - \theta$. I also know the base is $2r$ so I can say $cos(\theta_2) = \frac {D}{2r}$ (where $D=$the adjacent side) and as such $cos(90 - \theta)=\frac{D}{2r}$. When I rearrange this I get $D=cos(90 - \theta)2r$


*So my limits of integration would run the distance from one side of the cone to the opposite corner $D$ in other words from $0$ to $D$


Equation for the parabolas

$$\int_0^{D}(?)dx$$
This is the part I’m having trouble figuring out (assuming I found the limits of integration correctly). I’m not sure how to find and write the equation for the parabolas in a way that I can integrate it across the entire cone to get the volume.



 A: The volume may be calculated with Cavalieri's principle and a bit of care.
Place the cone with the center of its base at the origin. For $-r \leq x \leq r$, let $A(x)$ denote the area of the parabolic cross-section "at $x$".

Note that:


*

*The area under the parabola is two-thirds the area of the circumscribing rectangle.

*The width of the circumscribing rectangle is $2\sqrt{r^{2} - x^{2}}$.

*A longitudinal section of the cone has sides $\sqrt{r^{2} + h^{2}}$ and base $2r$, so by similar triangles the circumscribing rectangle has slant height $(r - x)\sqrt{r^{2} + h^{2}}/(2r)$.
Consequently,
$$
A(x) = \tfrac{2}{3} \cdot 2\sqrt{r^{2} - x^{2}} \cdot (r - x) \frac{\sqrt{r^{2} + h^{2}}}{2r}
  = \frac{2\sqrt{r^{2} + h^{2}}}{3r}\, \sqrt{r^{2} - x^{2}}\, (r - x).
\tag{1}
$$
Now for the bit of care: A thin planar slice meeting the cone's diameter at $x$ and $x + dx$ is tilted with respect to the $x$-axis, and by similar triangles has thickness
$$
dv = \frac{h}{\sqrt{r^{2} + h^{2}}}\, dx
\tag{2}
$$
rather than $dx$. Multiplying (1) and (2), the volume of the planar slice of cone at location $x$, with $-r \leq x \leq r$, is
$$
dV = A(x)\, dv
  = \frac{2h}{3r}\, \sqrt{r^{2} - x^{2}}\, (r - x)\, dx.
\tag{3}
$$
The volume of the cone is the integral
$$
\text{Vol}
  = \int_{-r}^{r} \frac{2h}{3r}\, \sqrt{r^{2} - x^{2}}\, (r - x)\, dx
  = \frac{2h}{3} \int_{-r}^{r} \sqrt{r^{2} - x^{2}}\, dx
  - \frac{2h}{3r} \int_{-r}^{r} x \sqrt{r^{2} - x^{2}}\, dx.
$$
The first integral is the area of a half-disk of radius $r$, namely $\frac{1}{2}\pi r^{2}$, while the second vanishes as the integral of an odd function over a symmetric interval. That is, $\text{Vol} = \frac{1}{3} \pi r^{2}h$.
A: To be honest, I have no clue on how to help you or on how to get a result from your way of doing it, but have you tried the much simpler way of integrating a solid of revolution?
Consider a straight line in a two-dimensional coordinate system, with y-intercept $r$ and x-intercept $h$. If you rotate this straight line around the x-axis, you get your cone, and using a simple integration rule you get your formula.
Since $r$ is the y-intercept and $h$ the x-intercept, your function is: $$f(x) = -\frac{rx}{h} + r$$
Using the formula for volumes of solids of revolution, you get:
$$V = \pi \int_{0}^{h} (-\frac{rx}{h} + r)^{2}dx = \frac{\pi h r^{2}}{3}$$
