Answer posted on May 5, 2018

I am very grateful to Mercedes Orús-Lacort for her answer.

The book that she refers to is much better than all the books that I have checked so far concerning the definition and computation of the area of a piece of surface in the ordinary 3-dimensional space.

The book she refers to is:

Calculus in Vector Spaces

Lawrence Corwin and Robert Szczarba

Second edition, 600 Pages, CRC Press, 1994

ISBN-13: 9780824792794

https://www.crcpress.com/Calculus-in-Vector-Spaces-Second-Edition-Revised-Expanded/Corwin-Szczarba/p/book/9780824792794

Thanks to her answer, I realize that I must add conditions to the definition that I proposed in my original question.

The following very simple example shows why additional conditions are necessary:

In the ordinary 3-dimensional space, I consider a half sphere bounded by a circle that lies in a plane P.

I consider a lot of planes parallel to this plane P that intersect the half sphere along circles.

In each of these planes, I choose an equilateral triangle with vertices on the circle of intersection of the plane and the sphere.

Then, these triangles are pairwise disjoint, so the set of these triangles satisfy the IPD condition of my original proposed definition.

By adding more and more planes parallel to the plane P, we can make the sum of the areas of these triangles as large as we want, which contradicts the fact that the area of a half sphere is finite.

So, we need an additional condition.

A 2-dimensional surface is a 2-dimensional manifold.

By definition, a 2-dimensional manifold is locally homeomorphic to a plane.

So, my piece of surface is locally homeomorphic to a plane P through a homeomorphism h.

I propose the following additional condition to my original definition:

I consider the images on the plane P of the vertices on the surface S of the triangles of my original definition through the homeomorphism h.

Then to each of the triangles in my original condition corresponds a triangle in the plane P.

My new condition IPD (Interiors Pairwise Disjoint) is that the interiors of the triangles in the plane P are pairwise disjoint.

With this new definition, my question is still open:

Is there a book that presents a valid definition of the area of a piece of surface and a rigorous proof of the usual formula for computation of such areas by double integral?

See my other projects and questions on the following Web page:

Research Gate: Jean-Claude Evard: Contributions:

https://www.researchgate.net/profile/Jean_Claude_Evard/contributions

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Dear Jean-Claude, you mention triangles. So I think you might be interested in differential geometry, in particular in triangulation. https://en.wikipedia.org/wiki/Differential_geometry_of_surfaces

Answer posted on May 6, 2018

I thank Peter Breuer and Viera Čerňanová for their answers.

I have no problem with the definitions of integration, Riemann integrals, and Lebesgue integrals.

I agree with Peter Breuer that it is important to also check books of measure theory concerning my original question.

I agree with Viera Čerňanová that it is important to look at curvilinear triangulations on surfaces.

So far, we do not know whether the use of these curvilinear triangles to check my IPD condition may be simpler than the use of flat triangles in a piece of plane homeomorphic to my piece of surface.

I still feel that triangles are more “natural” than parallelograms to construct the best possible definition of the area of a bounded piece of a 2-dimensional smooth surface.

I have no doubt that “the” formula with a double integral is correct.

I do not like to take it as a definition.

I feel that a geometric definition is more natural than a double-integral-depending-on-parameterization definition.

On the Web page about “Surface area” in the free online encyclopedia Wikipedia:

https://en.wikipedia.org/wiki/Surface_area

https://en.wikipedia.org/wiki/Surface_area#Definition

They say the following:

“… a rigorous mathematical definition of area requires a great deal of care.”

I strongly agree with this.

With a lot of thanks for your answers and with best regards, Jean-Claude

I feel that a geometric definition is more natural than a double integral depending on parameterization definition.

IPD is not enough. Roghly: we need to make sure that the triangles assume an "increasingly tangent" position as their diameter tend to zero uniformly. If you do not observe this "scruple", the supremum will always be infinite, except for the trivial case of flat surfaces (compare Weierstrass example). A convincing approach, in my opinion, is that of geometric measure theory, in which:

1. The measure of Hausdorff H ^ s is defined (s = 2 corresponds to area measure);

2. The so-called "area formula" is proved, which expresses the Hausdorff H^n measure of the image of a parameterization as the integral with respect to the Lebesgue measure L^n (over the domain of the parametrization) of the Jacobian transformation factor relative to the parameterization.

As a reference, see F. Morgan, "Geometric Measure Theory: A Beginner's Guide" , Academic Press.

P.S. The example is due to Schwarz and not to Weierstrass, I apologize for the oversight :)

Wednesday, May 9, 2018

I am very grateful for the comments of Peter Breuer and for the complete answer from Silvano Delladio to my high school question: “What is the correct definition of the area of a piece of surface?” that goes so far into mathematics that only a minority of mathematicians know Silvano’s complete answer.

To answer the question of Peter Breuer of May 6, the reason why I say “piece of plane homeomorphic to the given piece of surface” is that a surface is a special case of a 2-dimensional topological manifold, and by definition of such manifolds, every point of the manifold has an open neighborhood homeomorphic to an open subset of an ordinary plane.

Concerning Silvano’s complete answer, the best free online description of “Schwarz Lantern” that I have found so far is posted by Jean-Pierre Merx on the following Web page:

http://www.mathcounterexamples.net/the-schwarz-lantern/

The best picture of the Schwarz Lantern that I have found so far is posted by Alexander Bogomolny on the following Web page:

https://www.cut-the-knot.org/Outline/Calculus/SchwarzLantern.shtml

Schwarz Lantern is a counter-example found by Karl Schwarz in 1880.

His illuminating counter-example shows that it is fundamentally impossible to define the area of a piece of surface by means of approximations with triangles with all vertices on the surface.

His counter-example is a piece of surface together with approximations of the area of this piece of surface by triangles with all vertices on the surface.

The piece of surface chosen by Karl is one of the simplest and most well-known pieces of surface. It is so simple that it is taught in all high schools: It is a piece of circular cylinder between two parallel planes, with an axis perpendicular to the two planes, and the intersections of the cylinder with the two planes are circles of the same size with centers on the axis of the cylinder.

What could possibly be simpler, nicer, and gentler than this?

To construct Karl’s approximations of the area of this piece of cylinder with triangles, we first consider planes perpendicular to the axis of the cylinder that lie between the two given planes that bound the piece of cylinder so that the distance between any pair of consecutive planes is constant.

We consider the circles of intersection of these planes with the piece of cylinder.

We consider the regions of the piece of cylinder bounded by two consecutive circles, and we denote by R the number of such regions.

We divide one of these circles into a number A of arcs of circle of the same length.

To divide the other circles in the same way, we draw straight lines parallel to the axis of the cylinder and passing through the end points of the A arcs of circle that we have just constructed on one circle.

The intersections of these straight lines with all the circles give the same division of the circles into A arcs of the same length.

Now, we are going to construct isosceles triangles of the same size.

For every isosceles triangles that is not equilateral, I will call “Top vertex” the vertex at the intersection of the two sides of equal length of the triangle and I will call the third side “Base of the triangle”. (In an equilateral triangle, I will call “Base” any of the three sides of the triangle and will call “Top vertex” the vertex opposite to the base.

Let us start constructing Karl’s isosceles triangles of the same size:

Between every pair of consecutive circles, we draw A isosceles triangles with bases on the A arcs of one circle and top vertex on the other circle.

Then we fill the gaps between the two circles with isosceles triangles with bases on the second circle and top vertex on the first circle: See the picture on the Web page posted by Alexander Bogomolny:

https://www.cut-the-knot.org/Outline/Calculus/SchwarzLantern.shtml

With this construction, every approximation is made of isosceles triangles of the same size, and with all vertices of the triangles on the piece of cylinder, and the set of all these triangles “covers” the piece of cylinder.

On the picture, these flat triangles seem (!!!) so close from the cylinder that it seems (!!!) that we have a tiling of the piece of cylinder.

On the Web page posted by Jean-Pierre Merx:

http://www.mathcounterexamples.net/the-schwarz-lantern/

Jean-Pierre gives a complete computation of the limit of the sums of the areas of the triangles as functions of the number R of regions and the number A of arcs on each circle. He obtains the following results:

Let us denote by L(C) the limit of the sum of the areas of the triangles when the number of triangles goes to infinity under condition C.

Then

L(R/A = constant) gives the correct value of the area of the piece of cylinder.

L(A^2/R = constant = c) is an increasing function of c.

So, if by miracle the limit is correct for one value of the constant c, then it is sure that it is false for all the other values.

L(A^3/R = constant) = infinity.

What is going on?

Consider the extreme case A = 2, where we divide every circle into only two arcs, and consider a very large number of circles, so that consecutive circles are very close from each other.

Then on every circle, each of the two arcs is a half-circle, and the base of the corresponding triangles is a diameter of a circle. Because the circles are very close to each other, the top of triangles is almost on the same circle as the base, and since the base is a diameter, the top vertex is almost (not on the same circle, but very close) at a quarter of circle of each of the end points of the diameter.

Since the base is a diameter and the top vertex is almost on the intersection of the circle with a radius perpendicular to the diameter, the isosceles triangle is a triangle with a large area.

Since its 3 vertices are almost on the same circle, the triangle is almost in a plane perpendicular to the axis of the cylinder, so that the corresponding curvilinear triangle on the cylinder is almost a half-circle with area zero!!!

That is, the triangle has a large area corresponding to almost zero on the cylinder.

Therefore, as Karl observed in 1880, it is fundamentally impossible to define the area of a piece of surface by means of triangles with vertices on the surface.

Since the definition of the area of a piece of surface is so dangerous, it is really a miracle that the definition with double integral and parallelograms with just one vertex on the surface works 100% for differentiable pieces of surface. It is a miracle that these hanging parallelograms kind of cover the piece of surface at the limit, without gaps, without overlap. We have to pray that it will work, and actually, this shaky thing works. I am extremely surprised that these horrible parallelograms work and my nice triangles do not work.

Therefore, I agree 100% with Silvano that the best and most natural definition of the area of a piece of surface is given by the Hausdorff measure.

The mathematicians who have written the page on Hausdorff measure in the free online encyclopedia Wikipedia have done a wonderful work: Every mathematician, including beginners, can fully understand this wonderful definition in a few minutes, thanks to the excellent work of the writers of the article posted on Wikipedia, on the following Web page:

https://en.wikipedia.org/wiki/Hausdorff_measure

Hausdorff measure is 100% geometric (no integral), very natural, it is defined even for manifolds that are just continuous, not differentiable, even more than this, it is defined for any subset of R^n, and even more than this, it is defined for any subset of a metric space.

The best introductory book to Hausdorff measure mentioned by Silvano is the following:

Geometric Measure Theory

A Beginner's Guide

Frank Morgan

ISBN-13: 9780128044896

5th edition, 272 pages, Academic Press, 2016

https://www.elsevier.com/books/geometric-measure-theory/morgan/978-0-12-804489-6

With a lot of thanks to Peter and Silvano for all the wonderful information and with best regards, Jean-Claude

Please visit also:

https://conan777.wordpress.com/2011/11/28/the-schwartz-lantern/

it is well-written, simple and provides nice pictures, it can be useful for your purposes.

Thanks and best regards, Silvano

I guess one possibility for a definition of surface area is the so called "Lebesgue area". A concise definition can be found in the Dictionary of Analysis, Calculus, and Differential Equations - Page 154 : https://tinyurl.com/y7w4llc2

There are several books on the "Lebesgue Area"

For example "Surface Area" by Lamberto Cesari. Other books can be found by searching "Lebesgue area" in google books.

If you want a coordinate free develop of integration on a surface - or smooth 2 dimensional manifold that is quite readable see Harley Flander's wonderful little book "Differential Forms with Applications to the Physical Sciences"

https://www.amazon.com/Differential-Applications-Physical-Sciences-Mathematics/dp/0486661695

The correct way to develop the theory is with chains and cycles chains and boundaries. On a manifold one has to have a couple things to describe area. First is a metric tensor. That defines the element of area. Then integration follows from the standard Lebesgue integration.

Comments posted on Friday, May 11, 2018

I thank Silvano Delladio, Manuele Santoprete, and Truman Prevatt a lot for all the major pieces of information that they have added to this search for the best definition of the concept of area of a piece of surface.

By “Best” definition, I mean the most geometrical, the most meaningful, the most “natural”, the most real world, like for a piece of right circular cylinder, we start from a rectangular sheet of paper, then everyone agrees how to compute the area of a rectangle, and then we fold this sheet into the wanted piece of cylinder; then everyone agrees that we did not change the area by folding the sheet.

The following link given by Silvano on May 9, 2018 is wonderful:

The Schwartz Lantern

Conan Wu (alias Conan777)

November 28, 2011

https://conan777.wordpress.com/2011/11/28/the-schwartz-lantern/

Conan’s pictures of the Schwarz Lantern are of top quality and extremely helpful.

Conan’s explanations of what can go wrong if we try to define surface areas by means of triangles with all vertices on the surface are illuminating.

In addition, in Conan’s comments, the following one is dramatically important:

“Let’s note that although the triangles are getting uniformly smaller, they do become ‘thinner and thinner’ in the example.

In fact, this is the only way it can go wrong, i.e. it can be shown that if we further require the triangles to have bounded eccentricity then the area does converge.”

In the above comment, the dramatic piece of information is that:

… “it can be shown that if we further require the triangles to have bounded eccentricity then the area does converge.”

So, someone, let us call this person “Doctor Who”, has showed that we can save the method of triangles.

This means that I was wrong in my comments posted on May 9, 2018 to say that it is fundamentally impossible to construct a definition of area by means of triangles. Karl Schwarz did not kill the method with his famous counterexample. There were no international funerals for the method. The method is still alive. Or perhaps, Karl did kill the method of triangles with his lantern, but Doctor Who knows how to extract the cadaver of the method from its grave and how to resuscitate it.

I think that everyone would be interested to know who Doctor Who is, where Doctor Who’s method of resuscitation is published, what definition of eccentricity of a triangle is used in this resuscitation method.

Manuele’s suggestion posted on May 10, 2018 to use “Lebesgue Area” for the definition of area, it is also wonderful.

Also the following book suggested by Manuele is wonderful:

Surface Area

Lamberto Cesari

Annals of Mathematics Studies 35

ISBN-13: 9780691095851

595 pages, Princeton University Press, 1956

https://press.princeton.edu/titles/3958.html

Free “DjVu” copy on the following Web site:

BookSC

http://booksc.org/

http://b-ok.xyz/book/1300261/fdf4e8

Download a free “DjVu” reader (not browser) to be able to open the DjVu copy of the book.

“DjVu” stands for "Déjà Vu".

Search on the Internet, I got lost, I lost the link, and I am not able to find the same again.

Try to get the most recent free DjVu reader on the Internet.

Both the free DjVu copy of the book and the DjVu-reader that I downloaded landed in the folder “Download” on my PC (with Windows 10).

I clicked on the reader to unpack and install it.

I had to find where the unpacked and installed form of the reader went in the folders of my PC by looking at the dates in the “Programs File’ folder to find a folder with today’s date.

On my PC, DjVu-reader went to a new folder named Cuminas.

Then I right clicked on the DjVu copy of the book, I selected “Open With”, and in a small Microsoft window, I navigated to the folder Cuminas, and I selected DjVu-reader, to have this reader opening the DjVu copy of the book.

The book suggested by Manuele is absolutely wonderful.

I recommend to the readers to wear sunscreen and sunglasses before opening the book, because there is so much light in this book that we get a sunburn on our face.

On page 3 of Lamberto’s book, there is a history around the year 1900 of the struggle to construct a valid definition of the concept of area.

After this, there is the geometric definition of “Lebesgue Area”, and then, Lamberto very carefully deduces the well-known double integral formula with parallelograms from the definition of Lebesgue Area. The bad news is that these horrible parallelograms work while my wonderful triangles do not work. The reason is what Silvano told me on May 8, 2018, we need some tangent position.

The book suggested by Truman on May 11, 2018 is the following:

Differential Forms with Applications to the Physical Sciences

Harley Flander

240 pages, Dover Publications, revised edition, with improvements by the author, 1989

ISBN-13: 978-0486661698

http://store.doverpublications.com/0486661695.html

I just had a first look at this book with the marvelous “Look inside the book” on the following Web page of Amazon given by Truman:

https://www.amazon.com/Differential-Applications-Physical-Sciences-Mathematics/dp/0486661695

In a foreword by George Hawkins on page vii, George says that there are problems that are difficult to solve by using only tensors, and that are much easier to solve by using differential forms.

So, now we have 4 competing geometric definitions of areas in 3-dimensional spaces:

1. Hausdorff measure, suggested by Silvano.

2. Lebesgue Area, suggested by Manuele.

3. Differential forms, suggested by Truman.

4. Triangles with bounded eccentricity, suggested by Doctor Who.

A new problem is to check which of these best methods is the most best.

More comments on this important topic will be helpful to most mathematicians.

With a lot of thanks to Mercedes, Peter, Viera, Silvano, Manuele, and Truman for all the wonderful pieces of information that they have posted about the quest for best definition of area and with best regards, Jean-Claude