Derivation for surface area of revolution

[Jim Rockford]

Deriving the integral formula for the surface area of a curve revolved
around an axis often raises the following question (and suppose
y=f(x) is revolved around the x-axis):

"Why do we multiply 2*pi*f(x) by the arc length differential ds
instead of the differential dx?"

This is a perfectly reasonable question since for volumes of
revolution one can multiply a cross section area A(x) by dx, and
then integrate A(x)*dx to get the volume. No need for worrying
about the curvature that ds captures. The plain old dx
differential does just fine.

Almost every modern calculus book out there bypasses this issue, which
I find rather shameless. I've had a look at Anton, Thomas, Stewart,
and a host of other standard texts and they say nothing about it. Any
explanations I've seen amount to "well, if you use dx instead of
ds then you get the wrong answer; calculating surface area is
different than volume."

Who could dispute that standard answer? Using dx instead of ds
just doesn't recover the proper formulas for surface areas of simple
surfaces. Fine. But doesn't anyone find that "proof is in the
pudding" explanation dissatisfying? Isn't there a better explanation
than this? Why does adding up (integrating) A(x)*dx work for
volumes but adding up 2*pi*f(x)*dx doesn't work for surface
areas? Is there a more intuitive explanation?

[Mike Terry]

"Jim Rockford" <jim.ro...@gmail.com> wrote in message
news:13760dae-b97d-4dda...@s9g2000vbg.googlegroups.com...

Sure - this isn't a case of just trial and error until something works! :)

To calculate the volume, we imagine it to be sliced up into infinitesimal
slices perpendicular to the x-axis. The area of a slice is pi*(f(x))^2, and
its thickness is dx, so the volume is pi*(f(x))^2 dx, hence the integral
formula. (Of course, this isn't literally true but it's how we think of
integration intuitively...)

Similarly, when calculating the surface area we imagine it to be sliced up
into infinitesimal circular bands around the x-axis. The length
(circumference) of a band is 2*pi*f(x), and its width is ds, so the area is
2*pi*f(x) ds, which is what is integrated.

Note that a typical infinitesimal band is not parallel to the x-axis.
Instead think of a band taken from the surface of a cone, and so its area is
greater than we would get if we simply multiplied the circumference by dx.
Multiplying by dx would work for a cylinder, as then ds=dx, but in general
ds>dx.

Regards,
Mike.

[Ray Vickson]

Is it not obvious geometrically? If dy/dx is non-zero the graph (hence the surface) has a slant at that value of x (that is, its tangents are not horizontal, parallel to the x-axis). If, for example, the angle is arctan(dy/dx) = pi/4 radians, a horizontal step of dx translates into a surface distance of dx/cos(pi/4) = sqrt(2)*dx.

To get the area of a little surface patch between x and x + dx and between rotation angles t and t + dt we neglect curvatures, etc., and treat the patch as a rectangle of area dA = (y(x)*dt)*(ds), where the first factor y*dt is the lateral base of the rectangle and the second factor ds is the longitudinal length of the rectangle. Of course, ds = sqrt(1 + f'(x)^2)*dx.

RGV

[Ken Pledger]

In article
<13760dae-b97d-4dda...@s9g2000vbg.googlegroups.com>,

I'm surprised that you found so many text-books doing it badly. It's
based on the formula for the curved surface area of a frustrated cone.
If a right circular cone has base radius r and slant height L, then
its curved surface area is (pi)rL. If you cut along a plane parallel
to the base, you can chop away a smaller cone of area (pi)r'L' where
r'/L' = r/L. The remaining frustum then has area (pi)rL - (pi)r'L'
= ((pi)r/L)(L^2 - L'^2)
= ((pi)r/L)(L + L')(L - L')
= approximately 2(pi)r(L - L') if the frustum is thin.

Use a lot of thin frusta like this to approximate your curved surface
area, and the limit gives you an integral ds.

Ken Pledger.

[Jim Rockford]

Thank you all for your replies, but I guess we're having a
miscommunication.

First, I do not find the presentation of the derivation of the surface-
area-of-revolution formula in calculus books to be unclear or
otherwise poorly presented. I just find them lacking in their common
avoidance of addressing a very natural question that arises in the
derivation.

Let me again try to explain why I find the responses I've read on this
thread similarly lacking. Yes, to some it may seem "obvious" that you
can't add up the surface area of infinitesimally thin cylinders (i.e.
using dx instead of ds), and that curvature needs to be respected
more explicitly. However, you could equally well make that
"intuitive" argument in the derivation of the formula for volumes of
revolution as well. Why does "dx" suffice (thence at first glance
completely ignoring curvature of the curve revolved) in that case but
not in the surface area formula? Obviously, it has something to do
with the difference between calculating volumes and areas. In some
sneakier way the curvature is accounted for in the volume calculation
by the cross-sectional areas being added up. But this is never
clearly explicated in calculus books, nor have I seen here anything
except standard "intuitive" arguments (and even that you won't find in
today's name-brand calculus books). Perhaps you all don't find these
arguments lacking, but I do. They're certainly not suitable for
serious, inquisitive students. They're good having waving
explanations, but that's all, in my opinion (note: I don't mean to be
overly critical, as I don't have a better way to explain this either!)

[Ross A. Finlayson]

Well it's the line integral it's tough. Basically it works up the
square instead of the radial but the theorem drops it out. Definitely
it works. Courses with instruction down to the proper theorems and
proofs are good. Here as you note there is lots of room to work with
the differential components.

Regards,

Ross Finlayson

[Ray Vickson]

On Tuesday, May 22, 2012 2:23:48 PM UTC-7, Jim Rockford wrote:

There are fundamental differences between planar areas and circumferences, and between surface areas and volumes, etc. I remember years ago looking at a book (I think by Saks, but I don't remember for sure) that dealt with this issue deeply and in depth. Look, for example, at planar areas and circumferences. In the case of planar areas (at least for "nice" regions) we can put the whole region between two other regions whose areas are computable just by summing up rectangular areas, and we can show that there are sequences of "inner" and "outer" areas that converge to a common value. Whatever the area of the curved region may be, it must lie between the inner and outer values, so must equal the limit. That leads to the standard integration scheme. This all hinges on requiring area to satisfy Area(L) <= Area(R) <= Area(U) for a region R and inner/outer regions L subset R subset U. If area has any sensible meaning, surely it must satisfy that. That is enough to give the integral in the end. The same type of argument holds for volume.

Now look at circumference, say the problem of computing the circumference of a circle. Inner and outer figures do not really exist in the sense of subsets, so that whole type of argument used above fails. Instead, we must appeal to some type of underlying "intuition" about arc lengths, and eventually arrive at the usual arc-length formula. For surface area we look at little flat patches that approximate the surface and look at the limit of their area. That leads to the usual surface-area formulas.

The book I cited but forget spent several hundred pages looking into these types of issues from an advanced and nontrivial standpoint.

RGV

[Mike Terry]

"Jim Rockford" <jim.ro...@gmail.com> wrote in message

news:cdbfba54-f568-4b3f...@v9g2000yqm.googlegroups.com...

I don't know if this will help, or you may see it as more hand waving.

Let's think about volumes of revolution first. If we have a CYLINDER with
cross section A, then you can see that the volume of an "infinitesimal"
slice is A*dx. This is like considering a constant f. Now consider a graph
where f is not constant, but suppose the area of an infinitesimal slice is
still A, then A*dx is "pretty much" the volume of the cross section slice.
It seems perhaps we are neglecting the change in f between x and x+dx
though?

The key point is: what percentage error are we looking at here? Given that
dx is considered to be infinitesimally small, in fact the percentage error
is exactly zero! Or putting it another way, we are only looking at the
volume of the slice in comparison to dx, and if you want to look at how the
"actual" volume differs from A*dx, it would be A*dx + (terms of order (dx)^2
and smaller). The A*dx term is of order dx, i.e. a constant times dx (that
is, constant for a given value of x...) and this is what we integrate. Any
terms of higher order in dx become negligible in comparison to the A*dx
elements we are summing as the elements get smaller and smaller. E.g. even
with finite sized elements we could make the percentage error less than
0.00000001% (or whatever) by choosing small enough elements.

I think if you don't intuitively understand this, then you also couldn't
understand how the integral of f gives the area under the curve of f, which
would be a much more basic problem Exactly the same worries are present -
we are integrating infinitesimal slices of area under the curve, but you
could argue the slices shouldn't actually be rectangles because f is not
constant. Well, the f(x)*dx is the area of a slice to the first order in
dx, and the "percentage error" from considering the idealised rectangles is
made arbitrarily small by making dx sufficiently small. In fact to be
precise, the error percentage for the integral is actually zero because the
definition of integral already treats dx as being "arbitrarily small".
(I.e. the integral is actually a kind of limit where the errors have gone to
zero...)

So if you understand why we calculate integral f(x)*dx to get the area under
the graph for f, then you should see why we integrate Pi*f(x)^2*dx to get
the volume of revolution?

So what about the surface of revolution? If we look at a slice of this
surface between x and x+dx, we get a circular band of finite radius, but of
infinitesimal width. Note the surface of the band is not parallel to the
x-axis, but at some angle. The radius of band at x will be (pretty much)
f(x) and the circumference will be (pretty much) 2*Pi*f(x). The "width" of
the band will be ds, which is (pretty much) some definite multiple K of dx.
E.g. To make things concrete in what follows, let's imagine K=2 for the
function f and value x we are considering... Of course, the factor K in
general changes from point to point and is a function of x, but once we fix
a value of x, K becomes determined at that point, and we are just supposing
it is 2 at the value of x we are looking at...

Then we get an area for the band of 2*Pi*f(x)*2*dx. [note extra factor of 2
(=K) before dx]

You keep asking why not just say it is 2*Pi*f(x)*dx? (I.e. using dx and
ignoring that ds=2*dx at the point in question.) Using dx instead of ds
would clearly give us an area incorrect by a factor of 2 - and this error
DOES NOT GO AWAY IN THE PROCESS OF CONSIDERING SMALLER AND SMALLER SLICES!
In fact as the slices become smaller and smaller, the error factor becomes
closer and closer to 2, and it should be clear that in the infinitesimal
case we need the factor of 2 to make the area correct "to the first order in
dx".

Or put another way, the area for the band is 2*Pi*f(x)*ds + (terms of order
(dx)^2). Note that ds is OF THE SAME ORDER as dx: it is in fact a constant
multiple of dx once we are considering a fixed value of x, the constant
depending on the gradient of f at that point. (I assume you're clear on
this technical point, but if not, think of an "infinitesimal" triangle
relating hypotenuse ds, base dx and height f'(x)dx, and apply Pythagorus
theorem.) So the formula 2*Pi*f(x)*ds is the correct area, to the first
order in dx, and this is what we must integrate, after expressing ds in
terms of dx.

I suppose you might also say that ds is not *exactly* Sqrt( 1 + f'(x)^2 )
dx? I.e. not just a simple multiple of dx, because the graph of f is
typically curved at (x,f(x))? The answer to this is exactly the same as
above - we need ds to be "a multiple of dx" + "terms of higher order in dx,
like multiples of (dx)^2 or (dx)^3 etc.". This is what all my "pretty
much" brackets above were meaning - looking only at terms of first order in
dx.

Similarly if you say the radius of the band is not exactly f(x) because f is
changing, the answer is that the changes will be of the same order as dx,
and so negligible in comparison with f(x) as dx becomes smaller and smaller.
(f(x) is just a value which doesn't change as dx shrinks.)

I'm hoping you can see now that the expression within the Integral is the
right factor to give us the area of an infinitesimal band corresponding to
the slice of area between x and x+dx, TO FIRST ORDER IN dx.

------

So, the quick answer to your question "why dx for volume and ds for area?"
is:

Considering an infinitesimal element being "summed" in each integral:
To first order in dx, the element volume is (x-section area) * dx, but
To first order in dx, the element area is (x-section circumference) * ds
= (x-section circumference) * Sqrt( 1 + f'(x)^2 ) * dx

When people talk about integrals, they generally don't keep saying "to first
order in dx", because they are already considering dx to have shrunk to an
arbitrarily small value, so compared to dx, all the higher order terms
(dx)^2 etc. are just taken to have become negligible - it's just taken as
read: understood as part of the meaning of dx, ds, etc.. Really this is
just an informal/intuitive way of viewing the integral, and to properly
understand this you would need to study a precise definition of e.g.
"Riemann integral" to see how the summing and limits really work out.


------

OK having written all the above, it now occurs to me that maybe your
conception of "integral" is not the problem, i.e. you have no problem with
seeing that you have to only consider terms of order dx and ignore higher
order terms [which are negligible in comparison to dx as we take smaller and
smaller elements to sum].

Perhaps it's simply that you don't see that the width of an "infinitesimal
band element" is ds and not dx !! If that's the case, the problem is easy
to explain :) Just think of a ruler laying flat on the table. Think of the
ruler as representing a straightened and flattened out band, and the x axis
is measured along the table in the direction of the width of the ruler. s
measures distance across the width of the ruler.

If the ruler is 1 inch wide, then while the ruler is flat on the table, we
can say x and s measure the same for the ruler: 1 inch. This corresponds to
the graph of f(x) being parallel to the x-axis.

But if you swivel the ruler on one edge so that it makes an angle of 60
degrees with the table, then we (obviously) still have s = 1 inch (the width
of the ruler hasn't changed!) but x is just 0.5 inch now. I.e. we have a K
factor of 2 relating s and x like in my concrete example above: s=2x. And
the area of the ruler of course is still s * Length, not x * Length. (This
corresponds to a graph where f'(x) has a value of Sqrt(3)/2 at some point
(tangent making 60 degree angle with x-axis)


In a similar vein, Suppose we want to calculate the length of a curve for a
graph of f(x). For an infinitesimal element of the curve the length will be
ds, not dx. ds measures the infinitesimal change in distance measured along
the curve of the graph, while dx just measures the infinitesimal change
measured along the x axis. Clearly ds>dx as with the ruler example, and we
must calculate Integral (Sqrt( 1 + f'(x)^2 )) dx. (I.e. relating ds to dx
as above.) If we were to follow the suggestion of "using dx and not ds" we
would just be calculating Integral 1*dx which would just give us the length
of x-axis in the range, not the length of the graph curve in the range. I
think this example is perhaps more obvious to understand as there is no
complication of "revolution" to confuse things - but it's exactly the same
issue again.


Hope this helps.
Mike.