Urgent math problem: integral in 2D space

[cjolley]

One last thought though.
Would the double integral method remove, or at least make unknown, the constants that translate your area away from the x and y axii?
And if so, would that matter?

[VJ]

I don't think it matters: the surface of the area is a value regardless of where the area is positioned.


Jörg

[Fat Tone]

This all reminds of one of Sprial's threads.

[High_Jumbllama]

42

Ahh, one of the answers to the Theta function.

[Fat Tone]

Wasn't that updated to 56 in a later version?

[Wombat]

Wasn't that updated to 56 in a later version?

No, it's 54. People say "42" all the time, but they forget to qualify that as 42, base-13.

[VJ]

In the original BBC tv series, it was said to be 42, without qualifying the base.

Jorg

[VJ]

Just to come back to it.
We have consesus over this notation:

Consider a function f, and A a convex subset of R²

f : R²->R
   (x,y) -> 1  if x in A
   (x,y) -> 0  if x not in A

The surface are of A can be expressed as:

y=+inf  x=+inf
 (          (
  )          )  f(x,y) dx dy
y=-inf  x=-inf

Now, I would like to write this as a shorter form, and the discussion has shifted to this notation:

       (
        )  f(p) dp
p(x,y) in R²

(the main discussion point is the "dp")
This notation could then translate to

       (
        )   dp
p(x,y) in A

(which is what I would like to use)

I have an appointment with THE calculus professor at our university on Mondayafternoon. So I'll get an definate answer from him.


Jörg

[Strahd]

42 still makes more sense to me.

My cat's breath smells like cat food...

[Wombat]

The surface are of A can be expressed as:

y=+inf  x=+inf
 (          (
  )          )  f(x,y) dx dy
y=-inf  x=-inf

Is that strictly true? That would be an integral for volume , but without defining x = g(y), you don't know if you're going to get the volume of the universe minus the object, or the object. Doesn't the inner integral need to be something g(y)?

[VJ]

True, it would be a volume. But as the "height" for the elements of A is 1 (obtained via f(x,y)), the numerical value for the surface as the value for the volume.

g(y) is not needed, f(x,y) is a 3d function with 2 variables.

Jorg

[Wombat]

Without g(y), then how can you guarantee that the outer integral is possible? The inner integral is going to end up with factors such as f(inf,y), and f(-inf,y). Unless you know that the summations are going to be constants with respect to y, can you do this?

It would seem to me that you need bounds on R.

[VJ]

The set A was defined as a convex subset of the 2D space (which implies a finite surface area).

Jorg

[VJ]

Just went to a calculus professor. He says I can denote it in different ways:

(
 )                 d(x,y)
(x,y) in A

and even

(
 )   d(x,y)
A

The use of 'dp' at the end could give rise to an ambiguity, so he would keep it in the above form with d(x,y). So it requires only minimal changes to my notation...

Furthermore, I also consulted him about combining different such integrals, and he agreed with all my steps.
WOOHOO!

He also said that basically, every notation is a matter of agreement. As long as one uses a notation that it unabiguous and clearly explained (not contradicting with other notations), there is no real issue.


Jörg