# Question about calculating surface area

1. Nov 7, 2012

### mathFun

So if we are looking to find the surface area of a solid of revolution formed by rotating a curve about a line, we can use the following:

S = ∫2(pi)yds (if rotating about the x-axis)

FOr example, say our curve is y=x2 and we want to find the surface area of the solid of revolution that's found when we rotate y about the x axis, on the interval from 0 to 1.

Intuitively, this makes sense for me except for one thing. If I think of this problem the same way I do for Reimann integrals, I can imagine slicing the x axis up in to tiny pieces, then for each of those intervals of width Δx I would take the circumference of the circle with radius f(x) and multiply this by Δx. This would give me the surface area of cylinder of radius f(x) and height Δx. So now if I take Δx infinitesimally small, it seems like I would end up kind of adding together these circumferences, and end up with the surface area.

Why is it though that we have to use ds (arc length parameter) then instead of using dx? Why doesn't it work to do it like I mentioned, where you cut up the x axis?

2. Nov 7, 2012

### Vorde

It does, at least that's how I was taught it. I don't see a reason why you'd need to use arc length parametrization, I didn't even learn about that until after Integral Calculus.

Someone might correct me here though.

3. Nov 7, 2012

### mathFun

Maybe I'm describing it wrong. For example, here is the equation I'm talking about

http://curvebank.calstatela.edu/arearev/arearev.htm

I guess I don't understand why you can't just go

∫2(pi)f(x)dx

Because 2(pi)f(x) would be circumference of a circle of radius f(x) and you'd be multiplying these circumferences by infinitesimally small widths dx

4. Nov 7, 2012

### Vorde

Ah, I did misunderstand you, I apologize.

I didn't like this either when I was taught it, but the reason is basically that if you do it by circumference of circles like it seems like you should then you are adding up an infinite amount of circles of zero width, which won't amount to anything.

When you use the arc length differential ds you have an infinitesimal width, not a zero width. And with the magic of infinitesimals you can calculate with that.

Does that make sense?

5. Nov 7, 2012

### mathFun

But why does it work for say, doing the disk method then? In the disk method you are using dx, rather than ds. I guess that's what's confusing me about the whole thing is trying to understand why it works for the disk method, but when switching to surface area the dx approach doesn't work. Is it because with the disk method you're doing area so there's "something"? I mean vs. just plain old circumference which is like a boundary kind of?

6. Nov 8, 2012

### slider142

Because in the disc method, you are measuring a volume with a volume (an infinitesimally thick disc). In your method of measuring surface area, you are attempting to measure an area with a 1-dimensional value (only measuring how the area varies in x). If you draw your cylinders, you can clearly see that no matter how you divide them, you will always be leaving out the variation of your surface area that occurs with respect to y.
In order to see this, look at the plane and forget the surface for a moment. Suppose you want to measure the arc length of the function y = x from x = 0 to x = 1. You decide to use the widths of rectangles of width delta-x under the curve (This is just like your decision to use cylinders in delta-x). But notice that no matter how thinly you slice your rectangles, you are leaving out the entire variation in y (conspicuous hypotenuses above each rectangle), and you end up only measuring a total width of 1, which is of course, not the correct arc-length. In order to measure the length of the curve that varies in two dimensions, you need to include the second dimension in which it varies, and apply the distance formula (or metric). That gives us the standard arc-length integral. Your problem with cylinders that only vary in x is the same problem, you have just included their heights. But that still fails to measure the way the arc of the curve varies in y, as you have not included the metric that tells the integral the length of those conspicuous hypotenuses that the rectangles that form the cylinders never measure.

7. Nov 8, 2012

### mathFun

This makes perfect sense! Thank you so much! I really get it now! :D