# Why can't I use the disk method to compute surface area? [duplicate]

Possible Duplicate:
Areas versus volumes of revolution

I was trying to brush up on Calculus since I haven't used it in years. I wanted to derive formulas for volume and surface area of a sphere of radius r. I got volume right, but for surface area I'm getting the wrong answer.

I start with the equation for the half of a circle of radius r centered at the origin with positive y coordinates:

$$f(x) = \sqrt{r^2 - x^2}$$

So I started by breaking it up into disks, with radius f(x) and height dx, and integrating the surface area of the outside of the disks. (Which is to say, circumference of disk multiplied by height of disk.)

$$dA = 2 \pi f(x) dx$$ $$A = \int_{-r}^r dA$$ $$= 2 \pi \int_{-r}^r f(x) dx$$

Now, I don't know how to integrate f(x) symbolically, but logically if the integral is the area under the curve from -r to r, it would be half the area of a circle of radius r, giving:

$$= 2 \pi (\frac{\pi r^2}{2})$$ $$= \pi^2 r^2$$

Of course, this is not the correct answer. Am I making an error in my math, or does the disk method just not work for surface area? (And if not, why does it work for volume?)

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## marked as duplicate by Mike Spivey, Ross Millikan, t.b., J. M., Jonas TeuwenNov 18 '11 at 13:16

The main problem is that you can't approximate the length of a non-horizontal segment with horizontal segments; not even straight lines.

Consider a very simple case of the diagonal of a square of side length $1$; this diagonal has length $\sqrt{2}$.

You are trying to approximate the diagonal by dividing the interval $[0,1]$ into $n$ equal parts, each of length $\frac{1}{n}$, and taking the horizontal line on top of a little "bar" as an approximation for the diagonal, then adding them up.

But those "horizontal bars" on top will always add up to $1$, never more than that, even as you let $n\to\infty$. So you are not really approximating the diagonal, so your Riemann sums don't converge to the length of the diagonal.

The same thing happens with your computation above: you aren't actually approximating the surface area of the sphere, you are approximating something else, because those "disks" don't have surface area converging to the surface of the sphere, just like the little bars in the square example don't converge to the diagonal of the square:

                /|
/||
/=||
/| ||
/=| ||
/| | ||
/=| | ||
/=_|_|_||


(The = are the lengths you are adding up)

There was a question that kept popping up here with a "proof" that $\pi=4$ which used this same error, of thinking that adding up lengths of straight segments would yield that length of the limit was the limit of the lengths. See here.

What you want to look at is the error in the approximation. In order for the integral to give you the correct value, the errors have to go to $0$ as $dx$ goes to $0$. In my example above with the diagonal, the error between approximating with the horizontal segment and the true length of the diagonal segment is $$\mathrm{Error} = \mathrm{Actual\ value}-\mathrm{Estimated\ value} = \sqrt{\frac{1}{n^2}+\frac{1}{n^2}} - \frac{1}{n} = \frac{\sqrt{2}-1}{n}.$$ The total error over the entire interval is therefore $\frac{n(\sqrt{2}-1)}{n}=\sqrt{2}-1$. But as $n\to\infty$ (that is, as $dx\to 0$), this error does not decrease: it remains at $\sqrt{2}-1$ throughout. So in the end, when I perform my final computation with the integral, my answer is going to be $\sqrt{2}-1$ away from the true answer. Indeed, using the above computations I would be led to compute $$\int_0^1\,dx = 1,$$ whereas the right answer is $\sqrt{2}$.

By contrast, if I were computing the area this issue would not occur: if you compute the area using the rectangles, then the error per interval would be \begin{align*} \text{Error per interval} &= \text{Actual area} - \text{Computed area}\\ &= \left(\frac{x}{n} + \frac{1}{2n^2}\right) - \frac{x}{n} = \frac{1}{2n^2}. \end{align*} So the total error in the computation would be $\frac{1}{2n}$; hence, the error goes to $0$ as $n\to\infty$, so "at the limit" we should get the right answer.

The same thing happens when you try to use disks with the sphere: if you are looking at the volume, you are fine because the error does goes to $0$; but if you are looking at surface area, the error in the estimation does not got to $0$, so the Riemann integral that you compute at the end is not computing the actual surface area of the sphere, but rather something else.

In order to actually approximate the length of the limiting curve you need to consider the secant line, not the line segment parallel to the $x$-axis.

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actually $\pi=4$ in those proofs – ratchet freak Nov 16 '11 at 20:20
So, if I'm using the disk method to integrate volume, I'm using the exact same disks, right? So why doesn't it also come up short on volume? – Kip Nov 16 '11 at 20:25
@ratchetfreak: Aha! That explains why I couldn't find them, but now I have. Thanks. – Arturo Magidin Nov 16 '11 at 20:28
@Kip: Because the error in the computation of the volume does go to $0$ as $dx\to 0$, but the error in the computation of the surface area does not. In the example of the diagonal above, if you were computing area instead of length, you would estimate the area (with left hand sums) as being $x\,dx$, whereas the true area is $x\,dx + \frac{1}{2}(dx)^2$. So the error in the total approximation would be $\frac{1}{2}dx$, which does go to $0$ as $dx\to 0$. – Arturo Magidin Nov 16 '11 at 20:32
@ArturoMagidin: thanks! i never understood how the rectangles under the function in textbook pictures of a reimann sum could ever give the right answer even when shrunken down infinitely small. i don't remember ever hearing discussion of the limit of the error in school. – Kip Nov 16 '11 at 20:40