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I tried proving the formula presented here by integrating the circumferences of cross-sections of a right circular cone: $$\int_{0}^{h}2\pi sdt, \qquad\qquad s = \frac{r}{h}t$$ so $$\int_{0}^{h}2\pi \frac{r}{h}tdt.$$ Integrating it got me $\pi h r$, which can't be right because $h$ isn't the slant height. So adding up the areas of differential-width circular strips doesn't add up to the lateral surface area of a cone?

EDIT: I now realize that the integral works if I set the upper limit to the slant height - this works if I think of "unwrapping" the cone and forming a portion of a circle. The question still remains though: why won't the original integral work? Won't the value of the sum of the cylinders' areas reach the area of the cone as the number of partitions approaches infinity?

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    The problem is that you need to scale your surface area element appropriately, see http://en.wikipedia.org/wiki/Surface_integral, http://en.wikipedia.org/wiki/Surface_of_revolution and also http://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem I hope you're aware of the fact that you need absolutely no calculus here: just observe that the cone can be built out of a circular sector of the plane.2011-01-20
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    You can't integrate circumferences to get a surface area for the same reason you can't integrate points to get a length.2011-01-20
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    Yuan: But the circumferences are multiplied by $dt$2011-01-20

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