# Alternative proof of lateral surface area of a conical frustum

I am trying to come up with an alternative proof of the lateral surface area of a conical frustum with parallel bases by making use of the linear increase in perimeter $P$ of the base with respect to the altitude $h$.

Using the standard quadratic equation $Ay^2+By+C=P_n$, I was able to come up with three equations from three points of interest (the top, bottom and mid-height of the conical frustum). \begin{cases} A(0)^2+B(0)+C=P_{bottom}\\ A\left(\frac{h}{2}\right)^2+B\left(\frac{h}{2}\right)+C=P_{mid}\\ A(h)^2+B(h)+C=P_{top} \end{cases} wherein the terms turned out to be the following: \begin{cases} C=P_{bottom}\\ Bh=\frac{4P_{mid}-4P_{bottom}-Ah^2}{2}\\ Ah^2=2P_{top}+2P_{bottom}-4P_{mid} \end{cases} I then assumed that the lateral surface area of a cone is equal to the sum of perimeters of all parallel bases bounded by the top and bottom planes of the frustum. $$LSA=\int_0^h(Ay^2+By+C)dy$$ $$LSA=h\left(\frac{2Ah^2+3Bh+6C}{6}\right)$$ Substituting the values of the terms solved from the system of quadratic equations, I came up with the formula: $$LSA=h\left(\frac{P_{top}+4P_{mid}+P_{bottom}}{6}\right)$$ However, I am not getting the same result as the one from the known formula $LSA=(R+r)\pi\sqrt{(R-r)^2-h^2}$. But I do get the same result when I change $h$ to the slant height $\sqrt{(R-r)^2-h^2}$. Any ideas why I have to use the slant height for $h$ when I have used the altitude all throughout?

First of all, there must be an error in your formula for $P_n$: if $y$ is the distance from the bottom, then (as you say) perimeter grows linearly with $y$, whereas you have a quadratic formula.

And then, in any case, your method does not work: you are trying to approximate the lateral surface area of a conical frustum of very small height $\Delta y$ with that of a cylinder with the same height and base perimeter, but the ratio of the two areas does not tend to $1$ when $\Delta y\to 0$.

Your method amounts to this: divide the conical frustum into thin slices of height $\Delta y$ and base radius $x$ (of course $x$ depends on the distance $y$ from the base); compute lateral area of the slice as if it were a cylinder ($2\pi x\cdot \Delta y$) and sum up the areas of all slices. As $\Delta y\to 0$ the sum becomes the integral $\int_0^h 2\pi x\,dy$ which (hopefully) should give the lateral area of the conical frustum.

But this method works only if the ratio between the lateral area of the actual slice and that of a cylinder with the same base radius and height tends to $1$ as $\Delta y\to0$, which is not the case. To see why, let's introduce first of all the slope of the cone: $m=(R-r)/h$. Then we have $x=R-my$ for the lower base radius of the slice and $x'=R-m(y+\Delta y)$ for the upper base radius.

We may then compute the lateral area of the slice with the formula for a conical frustum: $A_{slice}=(x+x')\pi\sqrt{m^2+1}\Delta y$, which to first order in $\Delta y$ is $A_{slice}=2\pi x\sqrt{m^2+1}\Delta y$. On the other hand, the lateral area of a cylinder with base radius $x$ and height $\Delta y$ is $A_{cylinder}=2\pi x\Delta y$. As you can see, the ratio between the two areas tends to $\sqrt{m^2+1}$ as $\Delta y\to0$.

It is then clear that your result is wrong exactly by that factor: just multiply it by $\sqrt{m^2+1}$ and you'll get the right answer. Since $h\sqrt{m^2+1}$ is nothing but the slant height, that also answers your last question.

• I used a quadratic equation so I could test the formula for solids whose perimeter increase in a parabolic manner. Regardless, with a linearly increasing perimeter, the constant A would just be equal to 0; leaving By+C=P, which is a linear equation. By the way, I don't quite get the second paragraph wherein the perimeters were correlated with the ratio of areas. A bit more clarification would be greatly appreciated. – Virtual Underscore Jul 22 '15 at 7:50
• @Virtual, I've just edited my answer to explain that in more detail. – Aretino Jul 22 '15 at 10:06
• By the way: that also explains how to modify your formula to compute the area of a surface of revolution in the general case. Since $m\to dx/dy$ as $\Delta y\to 0$, then you get the integral: $$\int_0^h 2\pi x \sqrt{1+(dx/dy)^2}\,dy,$$ which is the standard formula you can find in textbooks. – Aretino Jul 23 '15 at 7:16