- #1
JG89
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When we integrate a scalar map over a manifold M, what exactly are we measuring?
If M is the unit circle in R^2, then regular Riemann integration of the function f = 1 over it will yield the volume of a cylinder of height 1. Okay, no problem.
Now, if we integrate f = 1 over the unit circle in R^2 using the definition of the "integral of a scalar map over a manifold" then I get for my answer 2pi.
If f = 2, then I get 2*2pi = 4pi.
What exactly are we measuring here? I'm very confused.EDIT: Ok, I know that if we integrate f = 1 over the manifold M (which in our example is the unit circle), then we are just measuring the 1-dimensional volume of M, which is its length. So no problem there. But what about f = 2? This gives us an answer of 2*2pi = 4pi, so what is that that we are measuring? The circumference of the circle if we go twice around it?
What about the function f(x,y) = x^2 integrated (using the definition of integration over manifolds) over the unit circle? Whatever the answer may be, and I can certainly calculate it, what will that measure?
If M is the unit circle in R^2, then regular Riemann integration of the function f = 1 over it will yield the volume of a cylinder of height 1. Okay, no problem.
Now, if we integrate f = 1 over the unit circle in R^2 using the definition of the "integral of a scalar map over a manifold" then I get for my answer 2pi.
If f = 2, then I get 2*2pi = 4pi.
What exactly are we measuring here? I'm very confused.EDIT: Ok, I know that if we integrate f = 1 over the manifold M (which in our example is the unit circle), then we are just measuring the 1-dimensional volume of M, which is its length. So no problem there. But what about f = 2? This gives us an answer of 2*2pi = 4pi, so what is that that we are measuring? The circumference of the circle if we go twice around it?
What about the function f(x,y) = x^2 integrated (using the definition of integration over manifolds) over the unit circle? Whatever the answer may be, and I can certainly calculate it, what will that measure?
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