# 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 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 But if approximated it by a circle with thickness $dx$ we would get

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

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.