61

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.

Eric O. Korman
  • 19,137
  • 2
    You're missing a square in the integrand for the true formula--the Pythagorean Theorem is involved here. – whuber Oct 16 '10 at 21:22
  • 4
    Think of the similar problem in 2D. Suppose you have a graph of a plane curve $\Gamma$ and you want to find the area limited by $\Gamma$ and the symmetrical curve with respect to $Ox$. This area may be approximated by the sum of the areas of many rectangles. However the length of $\Gamma$ cannot be approximated by the length $\Delta x$ of the same rectangles. One has to use a polygonal line. – Américo Tavares Oct 16 '10 at 21:39
  • 1
    I happened to ask a more specific form of your question at: http://math.stackexchange.com/questions/18226/setting-up-an-integral-to-find-a-cones-surface-area. Something I noticed was that if I were to actually construct a cone from my approximation (by using differential-width strips, it would NOT make an actual cone. The cone constructed from this method would actually be invisible from the top. I know this isn't really a good answer, but it made me more comfortable about why this wouldn't work. – G.P. Burdell Jan 27 '11 at 02:34

3 Answers3

32

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.

Qiaochu Yuan
  • 419,620
  • Do you mean a length of 1 instead? Or how did you carry out the approximation? – Milo Wielondek May 08 '12 at 14:43
  • @Milosz: the "staircase" consisting of line segments from $(0, 0)$ to $(\frac{1}{n}, 0)$ to $(\frac{1}{n}, \frac{1}{n})$, etc. has total length $2$ for any $n$. – Qiaochu Yuan May 08 '12 at 15:32
  • I see, you're counting both "sides", I only counted the tops. It makes sense now! – Milo Wielondek May 08 '12 at 15:55
  • 9
    What does it mean that volume "behaves nicely" under such deformations? How do I know what does and doesn't "behave nicely" under a certain kind of deformation? – Andrea Dec 03 '13 at 18:48
  • 1
    @Andreas: it depends on how formal you want to be, but what I had in mind is whether some function is or is not continuous with respect to some topology on some space of regions, e.g. the topology induced by the Hausdorff distance (http://en.wikipedia.org/wiki/Hausdorff_distance). You can detect continuity by looking at whether limits are preserved, which cashes out to deciding whether approximations like the above give the correct answer or not. – Qiaochu Yuan Dec 03 '13 at 19:14
31

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.

Ross Millikan
  • 374,822
  • 21
    An alternate way to think about Ross's answer: The disk approximation comes from the fact that you're using rectangles to approximate the area under the graph of $f(x)$. Rotating those rectangles around the axis produces disks.

    If we were to use disks to get surface area it would mean we were using the tops of those rectangles to approximate the curve length of $f(x)$. That can't work, as you can't get curve length via horizontal line approximations. You have to use the triangle hypotenuses.

     – Mike Spivey Oct 16 '10 at 21:29
  • @MikeSpivey very illuminating comment. I makes the mistake of calculating volumes by rotating the area around the axis, that doesn't work of course. But could you please provide an explanation? – senseiwu Dec 19 '21 at 11:42
1

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.

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

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

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

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

user170231
  • 19,334