Summing a Function Value Over an Interval

[mwspice]

Essentially, I have a function of two variables f(x, y), and I want to find the total value (sum) of that function for all values of (x, y) over an interval. How would I do that? The closest thing I can think of is the mean value theorem for integrals, but I want the total value, not the average value. Can anybody help?

Thanks

 

[fresh_42]

What exactly do you mean by total value or sum? The ordinary integral or some exotic measure? Is your function real valued? Is it continuous? Is your 'interval' a rectangle?

 

[mwspice]

What exactly do you mean by total value or sum? The ordinary integral or some exotic measure? Is your function real valued? Is it continuous? Is your 'interval' a rectangle?

It is both real valued and continuous, and yes, my interval is a rectangle.
I basically have a varying force, which depends on the position (x, y), and I want to find the sum of the forces at each point to find a total force. I don't think an integral would work because I don't want to multiply by the area. I don't want Fdxdy, I just want F.

 

[fresh_42]

If you have the sum of all $F(x,y)$ over intervals for $x$ and $y$, i.e. at each point a positive value, it is the integral.
Imagine a vector at each point, the force $F$. This defines you a function. Since the length of such a vector is the total amount of the force at this point the volume defined by the integral is the sum. If all vectors point in the same direction it's a function to ℝ, if not then to some $ℝ^n$.

 

[mwspice]

If you have the sum of all $F(x,y)$ over intervals for $x$ and $y$, i.e. at each point a positive value, it is the integral.
Imagine a vector at each point, the force $F$. This defines you a function. Since the length of such a vector is the total amount of the force at this point the volume defined by the integral is the sum. If all vectors point in the same direction it's a function to ℝ, if not then to some $ℝ^n$.

I think I understand what you are saying, but I'm not sure if it makes sense when I think of the physical system. For example, if we simplify the problem to one dimension and assign a constant value to the force, $F=1$, we could think of a line with length $0.5$ units, then the integral would return a value of $0.5$, which is smaller than the force at each point.

 

[fresh_42]

You must not forget the unities. Integration gives you $Nm$ or $Nm^2$ as in your original description. Considering a single point again means to divide the result by $m$ or $m^2$. You've said that the a force acts on every single point of the area. If you have only one vector of force acting on the whole area you get pressure which is something else. Imagine an object moving through your vector field. Then it continuously accelerates because at every point there is an additional force.

Edit: If you want to add up all forces along a path through your vector field, you have to define the path first and then integrate along this path.

Edit 2: If I remember correctly there is a continuous and surjective mapping (onto) $ƒ: [0,1] → [a,b]$ x $[c,d]$ which would give you $\int_0^1 F(ƒ_1(t), ƒ_2(t)) dt$ as result but it's not an easy path and of course not injective, i.e. you double cross a lot of points.

 

[mwspice]

Edit 2: If I remember correctly there is a continuous and surjective mapping (onto) $ƒ: [0,1] → [a,b]$ x $[c,d]$ which would give you $\int_0^1 F(ƒ_1(t), ƒ_2(t)) dt$ as result but it's not an easy path and of course not injective, i.e. you double cross a lot of points.

Do you know of any references I could find to help with this?

 

[fresh_42]

Sorry. I once read it in a popular written book about set theory and cardinal numbers. The function itself has been drawn but not explicitly defined.
It was about showing that you can paint out a square by a single line. All I remember is: it is not a path you'd want to integrate! Plus, as I mentioned before, you would get several points, i.e. forces multiple times counted. May I ask why you want to do this?

 

[mwspice]

Partially curiosity. I got the idea from something else I was working on. I was thinking about if there was a varying frictional force between surfaces, and there was a function giving the force at any point of contact, if the force could be modeled as a sum of the forces at each point of contact.

 

[Svein]

Edit 2: If I remember correctly there is a continuous and surjective mapping (onto) ƒ:[0,1]→[a,b]ƒ:[0,1]→[a,b]ƒ: [0,1] → [a,b] x [c,d][c,d] [c,d] which would give you ∫10F(ƒ1(t),ƒ2(t))dt∫01F(ƒ1(t),ƒ2(t))dt\int_0^1 F(ƒ_1(t), ƒ_2(t)) dt as result but it's not an easy path and of course not injective, i.e. you double cross a lot of points.

There are several such curves, two of them are the Sierpinski curve:

and the Hilbert curve:

 

[WWGD]

You must not forget the unities. Integration gives you $Nm$ or $Nm^2$ as in your original description. Considering a single point again means to divide the result by $m$ or $m^2$. You've said that the a force acts on every single point of the area. If you have only one vector of force acting on the whole area you get pressure which is something else. Imagine an object moving through your vector field. Then it continuously accelerates because at every point there is an additional force.

Edit: If you want to add up all forces along a path through your vector field, you have to define the path first and then integrate along this path.

Edit 2: If I remember correctly there is a continuous and surjective mapping (onto) $ƒ: [0,1] → [a,b]$ x $[c,d]$ which would give you $\int_0^1 F(ƒ_1(t), ƒ_2(t)) dt$ as result but it's not an easy path and of course not injective, i.e. you double cross a lot of points.

A Math Pop comment. If the curve was injective (implying bijective), it would be a homeomorphism, as a continuous bijection between compact and Hausdorff, meaning the interval [0,1] is homeomorphic to [0,1]x[0,1] , which contradicts previous results.