Areas versus volumes of revolution: why does the area require approximation by a cone?

Suppose we rotate the graph of y = f(x) about the x-axis from a to b. Then (using the disk method) the volume is \int_a^b \pi f(x)^2 dx since we approximate a little piece as a cylinder. However, if we want to find the surface area, then we approximate it as part of a cone and the formula is \int_a^b 2\pi f(x)\sqrt{1+f'(x)^2} dx. But if approximated it by a circle with thickness dx we would get \int_a^b 2\pi f(x) dx.

So my question is how come for volume we can make the cruder approximation of a disk but for surface area we can’t.

Answer

In the case of volume, the error is second order (it is basically revolving a triangle with sides dx and f'(x)dx so is negligible compared to the cylinder which is first order. This is the same as the 1 dimensional case, where to measure area (the standard integral) you use rectangles but to measure arc length you need to integrate \sqrt{1+f'(x)^2}dx, which is the hypotenuse of the triangle.

Attribution
Source : Link , Question Author : Eric O. Korman , Answer Author : Ross Millikan

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