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I'm trying to derive the formula for lateral area of a cone, $A = \pi r L$ by summing the circumferences of circles of decreasing radius that make up the cone.

cone

(image source: https://brilliant.org/wiki/surface-area-of-a-cone/)

The radius $r_1$ of a circle (a "disk") at height x from the bottom is given by a formula that we obtain from similar triangles: $$\frac{r_1}{h-x} = \frac{r}{h}$$ $$r_1 = \frac{h-x}{h}r$$ $$r_1 = \left(1-\frac{x}{h}\right)r$$ therefore the circumference of that same circle is given by the function $$f(x) = 2r \pi \left(1-\frac{x}{h}\right)$$ Solving the integral yields the formula $\pi h r$ which is incorrect. Where did I make a mistake? wolfram alpha integral

musava_ribica
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  • You have to integrate the areas of infinitesimal truncated cones, not the perimeters of circles. – GReyes Sep 09 '23 at 20:38
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    Here is what I mean: your $f(x)dx$ is the lateral area of an infinitesimal cylinder, not the lateral area of a truncated cone. – GReyes Sep 09 '23 at 20:40
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    To salvage the mistake you are doing you can consider the infinitesimal area as $f(x)\sqrt{dx^2 + dr_1^2} = f(x)\sqrt{1 + (dr_1/dx)^2}dx$. This is still not accurate for the infinitesimal area but error would be samll enough to approach zero when you take limit of sum. – Tony Pizza Sep 09 '23 at 20:46
  • @GReyes how would I use the formula for area of a truncated cone while looking for a formula for area of a regular cone, if a formula for area of truncated cone is just the formula for a regular cone subtracted – musava_ribica Sep 09 '23 at 20:52
  • You are considering your cone as made of infinitesimally thin truncated cones, since you are slicing it by horizontal planes between $x$ and $x+dx$. – GReyes Sep 09 '23 at 20:56
  • Tony Pizza’s comment is correct. The area of a truncated cone is the perimeter of the mid-circle by the length of the slant altitude – GReyes Sep 09 '23 at 21:01
  • This might be of help: https://math.stackexchange.com/a/1368702/255730 – Intelligenti pauca Sep 09 '23 at 21:14

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You may also calculate like this: The surface is construct of infinite tiangles with base $GG'=dc$ and height $D'H=l$, So:

$ds=\frac 12 l\cdot dc$. and we have:

$dc=r\cdot d{\alpha}$

and lateral surface is:

$$S=\int^{2\pi}_0\frac 12 l\cdot r d{\alpha}=\frac 12 l\big[\alpha\big]^{2\pi}_0= \pi \cdot rl$$

or in term of $r$ and $h$:

$$S=\pi r \sqrt {r^2+h^2}$$

sirous
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