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

Question

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 1

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.

Answer 2

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.

Answer 3

We can also use conical frusta in place of cylindrical disks to "better" approximate the exact volume, but the end result is the same. One can deform a cylinder into a conical frustum while still preserving its height and volume.

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Take a small interval $[x_1,x_2] \subset [a,b]$. The frustum constructed over this interval has volume

$$V_{\rm frustum} = \frac\pi3\left({x_2}^3-{x_1}^3\right)$$

Widen the interval by a short length $\Delta x$, so that we get a slightly larger frustum with volume

$$V'_{\rm frustum} = \frac\pi3 \left(\left(x_2+\Delta x\right)^3 - {x_1}^3\right)$$

The net change in volume is

$$\Delta V = V' - V = \frac{\pi\Delta x}3 \left(3{x_2}^2 + 3(\Delta x) x_2 + (\Delta x)^2\right)$$

Dividing both sides by $\Delta x$ and letting $\Delta x\to0$ yields

$$\lim_{\Delta x\to0} \frac{\Delta V}{\Delta x} = \frac{dV}{dx} = \pi {x_2}^2$$

As we raise the number of frusta, $x_1\to x_2$ and we can simply write $x_2$ as $x$. Then the volume of the solid is

$$V = \int dV = \int_a^b \pi x^2 \, dx$$

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Contrast this process with cylindrical disks, for which we can choose

$$V_{\rm disk} = \pi {x_2}^2 (x_2-x_1) \\ V'_{\rm disk} = \pi \left(x+\Delta x\right)^2 \left((x_2+\Delta x)-x_1\right) \\ \Delta V = \pi \Delta x \left(3{x_2}^2-2x_1x_2+(3x_2-x_1)\Delta x + (\Delta x)^2\right) \\ \implies \frac{dV}{dx} = \pi \left(3{x_2}^2-2x_1x_2\right) \to \pi {x_2}^2$$