Finding the volume of a cone by integration of parabolic conic sections

Question

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

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• 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

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$$\int_0^{D}(?)dx$$

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.

Answer 1

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:

$$V = \pi \int_{0}^{h} (-\frac{rx}{h} + r)^{2}dx = \frac{\pi h r^{2}}{3}$$

Answer 2

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, $\text{Vol} = \frac{1}{3} \pi r^{2}h$.