Gauss curvature integral: Why do we have a double integral?

[Mondo]

Hi,

I wonder why in the Gaussian curvature formula we have a double integral? The definition says we calculate "angular excess" of a triangle [imath]T[/imath] by adding up gaussian curvature over interior of [imath]T[/imath]. That Gaussian curvature is defined by [imath]k(p)dA[/imath] hence I would calculate the angular excess by: [math]\sum_{i=1}^{n} k(p)dA[/math] which in the limit gives [imath]\int_{T} k(p)dA[/imath] However for some unknow reason this is defined as [math]\int\int_{T} k(p)dA[/math] So why the double integral?

Thank you.

 

[blamocur]

Where does the formula with a sum come from?

 

[Dr.Peterson]

Hi,

I wonder why in the Gaussian curvature formula we have a double integral? The definition says we calculate "angular excess" of a triangle [imath]T[/imath] by adding up gaussian curvature over interior of [imath]T[/imath]. That Gaussian curvature is defined by [imath]k(p)dA[/imath] hence I would calculate the angular excess by: [math]\sum_{i=1}^{n} k(p)dA[/math] which in the limit gives [imath]\int_{T} k(p)dA[/imath] However for some unknow reason this is defined as [math]\int\int_{T} k(p)dA[/math] So why the double integral?

Thank you.

I'm not sure of the context, but I would say that dA, being two-dimensional, is inherently going to need a double integral. It doesn't surprise me at all.

If you can show the entire context in which you found this, it will help.

 

[Mondo]

Where does the formula with a sum come from?

I came up with this. [imath]k(p)[/imath] is a gaussian curvature at a point [imath]p[/imath] and [imath]dA[/imath] is the area of an infinitesimal triangle of some larger surface we wish to know the angular excess [imath]E()[/imath] of. So naturally [math]E(A) = \sum_{i=1}^{N} k(p)dA[/math] now if we let [imath]N[/imath] tend to infinity while [imath]dA[/imath] tend to [imath]0[/imath] that should become an integral (by a definition), namely [math]E(A) = \int_{T} k(p)dA[/math] A single integral. While in books or on a wiki page here https://en.wikipedia.org/wiki/Gaussian_curvature#Total_curvature I see it is a double integral, why?

@Dr.Peterson did I provide enough context now?

 

[blamocur]

I came up with this. [imath]k(p)[/imath] is a gaussian curvature at a point [imath]p[/imath] and [imath]dA[/imath] is the area of an infinitesimal triangle of some larger surface we wish to know the angular excess [imath]E()[/imath] of. So naturally [math]E(A) = \sum_{i=1}^{N} k(p)dA[/math] now if we let [imath]N[/imath] tend to infinity while [imath]dA[/imath] tend to [imath]0[/imath] that should become an integral (by a definition), namely [math]E(A) = \int_{T} k(p)dA[/math] A single integral. While in books or on a wiki page here https://en.wikipedia.org/wiki/Gaussian_curvature#Total_curvature I see it is a double integral, why?

@Dr.Peterson did I provide enough context now?

You summation looks analogous to 1-dimensional cases, i.e. Rhiemann Integral, although I would not use differentials (i.e., [imath]dA[/imath]) in the summation but would use a limit of the sum.

Your case is 2-dimensional, as @Dr.Peterson has mentioned in an earlier post. You could come up with an expression containing limit of a double summation, but I don't know why you would want to.

 

[Dr.Peterson]

I came up with this. [imath]k(p)[/imath] is a gaussian curvature at a point [imath]p[/imath] and [imath]dA[/imath] is the area of an infinitesimal triangle of some larger surface we wish to know the angular excess [imath]E()[/imath] of. So naturally [math]E(A) = \sum_{i=1}^{N} k(p)dA[/math] now if we let [imath]N[/imath] tend to infinity while [imath]dA[/imath] tend to [imath]0[/imath] that should become an integral (by a definition), namely [math]E(A) = \int_{T} k(p)dA[/math] A single integral. While in books or on a wiki page here https://en.wikipedia.org/wiki/Gaussian_curvature#Total_curvature I see it is a double integral, why?

@Dr.Peterson did I provide enough context now?

You haven't quite shown the part that most interests me: how you came up with your single summation (particularly, the definitions of p and N), and what theorem you used to justify passing to the integral, since this is not a (one-dimensional) Riemann sum. I suspect you are just being a little too informal, and making assumptions about what "should" happen.

Can you show us your starting point? What did you go by in writing your summation?

 

[Mondo]

@Dr.Peterson you are right, sorry I did to many shortcuts. N is number of triangles that we can divide a given area into; p I defined in the previous post, it is just a point at which we calculate the curvature. The whole expression [imath]k(p)dA[/imath] is an angular excess of an infinitesimal triangle. It is a known fact from a differential geometry (I don't have a proof of that but also it is not important here). Since the curvature is adding a bit of confusion maybe we can simplify the integral objective to finding a density of the area? In this case we have say a plate of the area S which we divide into N number of infinitesimal triangles of the area dA whose density at a point p is given a function [imath]\sigma(p)[/imath].
Now we can get the total plate density by a summation [math]\sum_{i=1}^{N} \sigma(p)dA[/math] which in a limit becomes an integral [math]\int_S \sigma(p)dA[/math]
So this is analogous example, we again integrate over two dimensional area and there is only a single integral right?

 

[blamocur]

...integrate over two dimensional area and there is only a single integral right?

The convention is to to use use double integral signs ([imath]\iint[/imath]) when integrating over 2-dimensional areas. You can read on the topic in, among other places, Wikipedia.
P.S. It is also more convention to use [imath]dS[/imath] when integrating over [imath]S[/imath], i.e. [imath]\int_S \sigma(p) dS[/imath]

 

[Dr.Peterson]

@Dr.Peterson you are right, sorry I did to many shortcuts. N is number of triangles that we can divide a given area into; p I defined in the previous post, it is just a point at which we calculate the curvature. The whole expression [imath]k(p)dA[/imath] is an angular excess of an infinitesimal triangle. It is a known fact from a differential geometry (I don't have a proof of that but also it is not important here). Since the curvature is adding a bit of confusion maybe we can simplify the integral objective to finding a density of the area? In this case we have say a plate of the area S which we divide into N number of infinitesimal triangles of the area dA whose density at a point p is given a function [imath]\sigma(p)[/imath].
Now we can get the total plate density by a summation [math]\sum_{i=1}^{N} \sigma(p)dA[/math] which in a limit becomes an integral [math]\int_S \sigma(p)dA[/math]
So this is analogous example, we again integrate over two dimensional area and there is only a single integral right?

I will ask again: What theorem are you using to get from your summation to the integral? You appear to be thinking informally, and just assuming this transition. This is not just a Riemann sum.

In addition, in your summation, shouldn't p at least be indexed? It presumably means a point in the ith "infinitesimal triangle" in a list of N of them. Also, as has been mentioned, you used "dA" prematurely, as the triangles are not really infinitesimal until you pass to the limit. And it is that limit, which you gloss over so lightly, that leads to the double integral you are asking about. All of these observations relate to the informality of your thinking.

We usually write such a summation as a double summation, in two dimensions, and write a surface integral as a double integral, because in practice there will be two variables. Do you have a source that you copied from in using a single summation? On the other hand, it seems reasonable, on the surface [no pun intended], to use a single integral symbol; that just isn't the standard convention. When you invent things yourself, it's a good idea to check on what others do. (And that's why I've been begging for a theorem you might have used to justify your notation, in case someone might actually do that.)

 

[Mondo]

I will ask again: What theorem are you using to get from your summation to the integral? You appear to be thinking informally, and just assuming this transition. This is not just a Riemann sum.

We usually write such a summation as a double summation, in two dimensions, and write a surface integral as a double integral, because in practice there will be two variables. Do you have a source that you copied from in using a single summation? On the other hand, it seems reasonable, on the surface [no pun intended], to use a single integral symbol; that just isn't the standard convention. When you invent things yourself, it's a good idea to check on what others do. (And that's why I've been begging for a theorem you might have used to justify your notation, in case someone might actually do that.)

@Dr.Peterson, Yes I was using Riemann sum, why do you say this is not it? I did a summation which in a limit, where number of summation elements go to infinity while each element size tends to 0 by definition should be an integral right?

In addition, in your summation, shouldn't p at least be indexed? It presumably means a point in the ith "infinitesimal triangle" in a list of N of them. Also, as has been mentioned, you used "dA" prematurely, as the triangles are not really infinitesimal until you pass to the limit. And it is that limit, which you gloss over so lightly, that leads to the double integral you are asking about. All of these observations relate to the informality of your thinking.

Yes you are right, p should be indexed. Thanks for catching it. Let's refer to my post #7, where is dA used prematurely? In the summation it's just means the area of a particular element. Also, I made a mistake in the description of this example, this summation/integral is to find a total mass and not a density (which is know and given by [imath]\sigma()[/imath] function. So mass of a single, ultimately infinitesimal element is given by [math]\sigma_i dA_i[/math] hence correcting my summation in post #7 we have [math]\sum_{i=1}^{N} \sigma_i dA_i[/math] Now in the limit it becomes an integral (by definition of integral, to answer your question what theory I am based on) [math]\int_A \sigma(x,y) dA[/math]Now, to restate my original question, why in the Gauss theorem they use double integral, is it just as @blamocur suggested a "convention" to emphasize 2D space or it is needed to calculate the integral correctly?

To answer my last question I am working on a real example now. I would appreciate any hints in the meantime.

Thank you

 

[blamocur]

Now, to restate my original question, why in the Gauss theorem they use double integral, is it just as @blamocur suggested a "convention" to emphasize 2D space or it is needed to calculate the integral correctly?

You are right, it is often needed to calculate the integral correctly: a typical way to compute surface integrals is to use surface parametrization which reduces this integration over the surface to integration over some subset of [imath]\mathbb R^2[/imath].

 

[Dr.Peterson]

Yes I was using Riemann sum, why do you say this is not it?

A Riemann sum, as I've always seen it, is taken over a single variable. And there is more detail in its description than you have shown. The Riemann integral is defined to integrate a function over an interval of the real numbers. What you are talking about is a generalization.

where is dA used prematurely?

As @blamocur mentioned in post #5, we use deltas, not differentials, in the summation, and we write a limit of the summation. Again, you are just being informal about all this.

To see what I'm talking about on both points, see any textbook explanation of the Riemann integral, such as this:

5.2: The Definite Integral

If f(x) is a function defined on an interval [a,b], the definite integral of f from a to b is given by \[∫^b_af(x)dx=\lim_{n→∞} \sum_{i=1}^nf(x^∗_i)Δx,\] provided the …

math.libretexts.org

why in the Gauss theorem they use double integral, is it just as @blamocur suggested a "convention" to emphasize 2D space or it is needed to calculate the integral correctly?

Yes, the double integral represents how it is actually evaluated. Look at any textbook explanation of surface integrals, such as this:

https://math.libretexts.org/Bookshe.../16:_Vector_Calculus/16.06:_Surface_Integrals

(in particular, the section "Surface Integral of a Scalar-Valued Function", which shows a double summation and a double integral)

Again, I can't say that no one uses a single summation and a single integral to represent these ideas; there may well be a theorem you could use. But you asked why the sources you've found show a double integral, and that is the reason.

 

[Mondo]

@Dr.Peterson , @blamocur thanks for the help. I need to make some more study, especially in the area of surface parametrization and double integral evaluation. I will be back with an update, maybe in a separate, dedicated thread.