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The following questions has been bugging me since high-school calculus. Please help me find my peace once and for all:

Consider a revolution solid generated by rotating a nice curve $f(x)$ around the $x$-axis on the interval $[a,b]$ (provided that $f(x)$ does not cross the $x$-axis on this interval).

Let us first find the volume of this solid, $V$. We slice the interval $[a,b]$ into small segments of width $\delta_x$. Each segment is approximately a cylinder of radius $f(x)$ and height $\delta_x$, hence having a volume of $\pi [f(x)]^2 \delta_x$. Taking the limit we get

$$V = \pi \int_a^b [f(x)]^2 \, dx$$

which is the right formula: great.

Now we apply the exact same argument to find the surface area of the solid, $S$. The surface area of each cylindrically-approximated segment is $2 \pi f(x) \delta_x$, and taking the limit we obtain

$$S = 2 \pi \int_a^b f(x) \, dx$$ which is not correct.

We would obtain the correct formula for $S$ if we take the heights of the segments to be the arc lengths of $f(x)$ over each $\delta_x$, so I suspect that what is going wrong has something to do with this.

But my question is: why does this argument work for finding $V$, but not $S$?

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If it were the exact same argument, you'd get the volume again. :)

Seriously, though: "Each segment is approximately a cylinder" is the critical statement: not only is it approximately a cylinder, but the difference in volume between a true cylinder of that size and the approximating one goes down linearly as $\delta$ gets smaller. Yeah, your calculus book should have mentioned that, but they didn't. Sigh. [To be more precise: the ratio of the approximating and true volumes heads to 1 as $\delta \to 0$.] (See below for correction.)

For the surface area, the ratio of the true area to the approximating area is, in the limit, the ratio of the hypotenuse of a right triangle to one of the bases: the first base is $\delta$, the second is $\delta f'(x)$, and the hypotenuse therefore has length $\delta \sqrt{1 + f'(x)^2}$. The ratio of this to the base you used is $\sqrt{1 + f'(x)^2}$, and remains at this value as $\delta \to 0$.

So that's the difference.

I haven't proved that if the ratio goes to 1, then the computations are the same -- that'd require carefully working through the definition of the Riemann integral, and some subtleties about swapping the order of limits. But I wanted to give you an answer that at least pointed in the right direction.

When asked about my claim about ratios, I realized that in trying to say things simply, I had overshot: the problem is that when the function $f$ happens to be zero at some point, the ratio might not even be defined. So let me suppose that $f$ is everywhere nonzero, and briefly discuss the remaining case at the end.

I want to show that as $\delta \to 0$, the volume ratio goes to $1$. Let me say that differently: for any positive number $A$, I'll show that if $\delta$ is small enough, then the volume ratio is between $1-A$ and $1 + A$. OK? To do this, I'm going to limit my attention to values $0 < A < 1/2$, because if I can make the volume ratio lie between 1-(1/2) and 1 + (1/2), I can certainly make it lie between $1 - (1000)$ and $1 + (1000)$, etc.

Since $f$ is differentiable (or the area formula doesn't make sense), we know $f$ is continuous. Now let's look at some interval $[x_1, x_2]$, and point $x$ in that interval and compare $$ V_1 = \pi f(x)^2 \delta $$ with $V_2$, the volume of the slice of the solid between $x_1 $ and $x_2$. To save writing, let's write $\delta = x_2 - x_1$. On the interval $[x_1, x_2$, $f$ has a minimum value $m$ (greater than zero, because $f$ is continuous and everywhere positive) and a maximum value $M$. That means that the cylinder of radius $m$ is contained within the slice of the solid, and the cylinder of radius $M$ contains the slice of the solid. So $V_2$ is between $\pi m^2 \delta$ and $\pi M^2 \delta$. That means that the ratio of $V_1$ to $V_2$ is between $$ \left(\frac{f(x)}{m}\right)^2 $$ and $$ \left(\frac{f(x)}{M}\right)^2 $$

Now because $A < 1/2$, the numbers $\sqrt{1 + A} > 1$ and $\sqrt{1 - A}<1$ both make sense. So we can compute $$ U = \frac{f(x)}{\sqrt{1-A}} > f(x) \text{ and} \\ L = \frac{f(x)}{\sqrt{1+ A}} < f(x), $$ a pair of numbers a little above and below $f(x)$.

By picking $x_1$ and $x_2$ sufficiently close to $x$, we can ensure that on the interval $[x_1, x_2]$, the function $f$ lies between $L$ and $U$. (The proof is that $f$ is continuous, so by shrinking the interval containing $x$, you can make its image fit in any open interval, such as $(L, U)$, contiaining $f(x)$.)

With that done, the remainder is a computation: the volume ratio is between

$$ \left(\frac{f(x)}{m}\right)^2 $$ and $$ \left(\frac{f(x)}{M}\right)^2 $$ (Note that the first of these is LARGER than the second!) Because $m$ is at least $L$, we have that $$ \left(\frac{f(x)}{m}\right)^2 < \left(\frac{f(x)}{L}\right)^2 < \left(\frac{f(x)} {\frac{f(x)}{\sqrt{1+ A}}}\right)^2 = \left( \sqrt{1+A}\right)^2 = 1 + A. $$ A similar argument shows that the ratio is greater than $1 - A$.


When we have a point where $f(x) = 0$, things get really messy, and you need to start making arguments with "min" in them, and I don't think that adds any enlightenment, so I'm going to skip it.


You asked as a final question "which kinds of approximations are valid and which are not?" The answer is "The approximations are all valid (any number is an approximation of any other number!), but they're only useful if, when you push the limits through the definition of the integral, things work out." I know that sounds like a rotten answer, but I could say it differently: to know whether you can make an approximation within an integral and take limits, you need to really understand the definition of integration." That hardly seems like an unfair request. I wish I had a magic bullet for you, but I don't.

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  • $\begingroup$ The same reason why approximating a circle in a grid "yields" $\pi=4$ $\endgroup$ – chubakueno Mar 2 '14 at 15:30
  • $\begingroup$ Is it easy to prove that "To be more precise: the ratio of the approximating and true volumes heads to 1 as δ→0."? $\endgroup$ – MGA Mar 2 '14 at 15:41
  • $\begingroup$ No...it's not even true, but for small reasons. See edited stuff soon to be added above. $\endgroup$ – John Hughes Mar 2 '14 at 17:21
  • $\begingroup$ Thanks. I guess I could rephrase my question as "Which approximations are valid, and which are not?". $\endgroup$ – MGA Mar 2 '14 at 17:24
  • $\begingroup$ That's brilliant, I have accepted your answer. Would I be right in saying that a similar argument would show why the surface area argument does not work? $\endgroup$ – MGA Mar 2 '14 at 19:40

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