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So I've taken Calculus 1-3 and am taking 4 when I came across a problem that required me to know the surface area of a right circular cone. It is obviously trivial to search this up so I thought it would be more fun to come up with it on my own when I came up with the following method.

A cone is a stack of continuously expanding circles. By taking the lowest point of the cone to the highest point and adding the circumferences you would get the surface area given. This would look like

$$\int_{0}^{R}2\pi r\,dr$$

where $R$ represents the largest radius of the cone. This simplifies to the area of the circle with radius $R$ but that is clearly not the correct answer.

I'm pretty sure that my problem is that I'm not accounting for the slant of the cone but I'm failing to see how that slant affects the argument. Where am I going wrong?

Blue
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Tristan Brown
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  • Yes: $ds$, not $dr$. – Ted Shifrin Oct 26 '23 at 04:06
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    A dr change should have been on the horizontal XY plane, as r represents the radius of the circle in that plane. You are taking dr to be an infinitesimal increment IN THE Z-direction, along the cone’s axis, which is incorrect. What you should do instead, is denote the height of the circle above the ground by another variable, say h, and then the small area is $2\pi r\cdot dh$. Now all you gotta do is cast r in terms of h, given the cone’s height H and base radius R. – insipidintegrator Oct 26 '23 at 05:14

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