TITLE: Areas versus volumes of revolution: why does the area require approximation by a cone?
QUESTION [55 upvotes]: 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.

REPLY [26 votes]: The problem here is that volume behaves nicely under small deformations of 3D regions in 3D, but surface area does not.  Similarly, area behaves nicely under small deformations of 2D regions in 2D, but circumference / arc length does not.  You can see the essence of the 2D problem, hence the essence of the 3D problem, in the following: the length of the diagonal from $(0, 0)$ to $(1, 1)$ is $\sqrt{2}$, but if we approximate the diagonal by a "staircase" of horizontal and vertical lines of length $\frac{1}{n}$ and let $n \to \infty$ we get a length of $2$ instead.