Urgent math problem: integral in 2D space

[VJ]

Hello,

I have an urgent problem that has surfaced during one of the final proofreadings of my phd text. It is just a matter of notation though, but needs to be resolved soon.

Suppose one is working in the two-dimensional real space (R square). In this space I have a conxex subset A.
How can I express the surface area of this convex subset as an integral over the points in A?

(the discussion is about single or double integral, and about what is the expression to be integrated)

Thanks!


Jorg

[MultimediaMan]

42

[cjolley]

Wouldn't it be an integral over the points defining the boundarys of A not the points in A?

An area is given by a single integral.
Or the sum of single integrals if both lower and upper bound of the area you want are not on the x axis. (if I recall correctly)

Not sure how anyone but you would know what expression is to be integrated though.
(is the expression to be integrated not the expression of the boundaries of A?)

Maybe I don't understand the question.


PS Nice quickie on wikipedia

[VJ]

mmm:
I wasn't looking for the answer to life, the universe and everything.

cjolley:
Nope, you understand perfectly.
The problem is that the points that make up the outline of A are not known (no mathematical equation that binds them).

The origination for the double integral came from my promotor, but at present it is unclear to me why (but it may have given an idea).
Thanks,

Jorg

[Wombat]

You can set up a double integral when your area is defined as a function of two variables, as long as you can somehow find a relation between the two variables, even if that relation is different on each border.

http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/255doub/255doub.html

[cjolley]

Do the points define a line of some sort or are the just scattered around?
After all, the area under a bunch of unrelated points would be 0 because the area under each point would be 0.
The area has to be under a line or between two lines.

I mean you could just do the rectangles:

(p2x - p1x)((p1y + p2y)/2) +
(p3x - p2x)((p2y + p3y)/2) +
(p4x - p3x)((p3y + p4y)/2) +
...



That would define sort of the "average" area.
Maybe there is a general form for that.
Of course it uses algebra and not calculus.
Though, I suppose if you add summation/limit symbols it's really just the definition of an integral.

If your promoter wants you to do some sort of curve fitting thing and take the integral of that, well that's way beyond me.

@Wombat but it's supposed to be in a plane, I don't see how a volume equation can help.
Or why his promoter would bring up the idea (which is the main reason I worry I don't really understand the problem)

[VJ]

How about this:

Numerically, the surface area of A is equal to the volume below a constant function which is 1 at all points p in A (basically a volume with the same convex set as surface and with height 1).

So I'm now thinking that if I were to consider the double integral over "all points p in A" of the function "1", with "dp"...
Wouldn't this be a formula that represents the numerical value I'm looking for?

Jorg

[cjolley]

How about this:

Numerically, the surface area of A is equal to the volume of a 3D volume with the same convex set as surface and with height 1.

So I'm now thinking that if I were to consider the double integral over "all points p in A" of the function "1", with "dp"...
Wouldn't this be a formula that represents the numerical value I'm looking for?

Jorg

Umm. would that make the original integral (the area) the derivative of fx = 1 and therefor = 0?

PS Come to think of it, how can you know it is convex if you don't have an equation describing the points?

[VJ]

The set A is defined elsewhere, and is defined to be convex (and without holes). As the points are not isolated, randomly distruted, the surface area is greater than 0.

Jorg

[cjolley]

The set A is defined elsewhere, and is defined to be convex (and without holes). As the points are not isolated, randomly distruted, the surface area is greater than 0.

Jorg

I think I see what you mean, but I can't figure out how to get it input.

The double integral across your area (region) with f(x,y) = 1?

[VJ]

Yes, that is what I'm currently thinking about.

The result appears to yield what I want, and I think that notation can be used. (the required correcgions to my work are also minimal :-))


Jorg

[Umfriend]

42 still makes more sense to me.

[VJ]

I just came across this:

http://mathworld.wolfram.com/SurfaceIntegral.html

where the first integral in formula 1 basically also expresses the surface area (this resembles my initial notation). So hopefully my promotor agrees with this notation (in which case I need to change nothing).


Jörg

[cjolley]

I think you should use the double derivative on the triple integral over one second with a height of 1 across a region defined by your points.

But I hope she accepts your proposed notation.

[VJ]

I think the double integral as proposed this morning stands a better chance of becoming accepted...

But thanks for the input!
(it helped me organize things in my mind )

Jörg