Surface integral to line integral

  • Thread starter dspch11
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  • #1
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I am agonizing about the following integral identity:

[tex]
\frac{d}{dt} \int \int_{g(x,y) \leq t} f(x,y) dx dy = \int_{g(x,y)=t} f(x,y) \frac{1}{\left| \nabla g(x,y) \right|} ds,
[/tex]

where ds is the line element. Clearly, using the Heavisite step function, the condition [tex]g(x,y) \leq t[/tex] is transferred into the integrand. Differentiation with respect to t yields a Dirac delta-function. However, how can I eventually arrive at the line integral. The picture is quite clear, if one imagines a domain defined by [tex]g(x,y) \leq t[/tex] which is growing with t. The increase in the integral can then be evaluated by summing contributions along its circumference with [tex]\left| \nabla g(x,y) \right|[/tex] giving the density of isolines.

How can one formally obtain this result?

Thank you for you help,
Daniel
 
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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi Daniel! Welcome to PF! :smile:

(have a del: ∇ and an integral: ∫ and a ≤ :wink:)

Have you tried changing variables from x and y to s and g ?
 
  • #3
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Have you tried changing variables from x and y to s and g ?

Is this what you suggest:

[tex]
s = \int \sqrt{ 1 + (\frac{dy}{dx})^2} dx
[/tex]

for constant g we have furthermore:

[tex]
g(x,y) = t = const \Rightarrow
g_x + g_y \frac{dy}{dx} = 0
[/tex]

and, thus,

[tex]
s = \int \sqrt{ 1 + (g_x/g_y)^2} dx
[/tex]

The jacobian determinant is

[tex]
J = 1/(\left| g_x s_y - g_y s_x \right|) = 1/(\left| 0 - g_y \sqrt{ 1 + (g_x/g_y)^2} \right|) = 1/(\left|\nabla g \right|)
[/tex]

What about if [tex] \frac{dy}{dx} [/tex] goes to infinity?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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uhh? :confused:

dxdy is obviously dsdw, for some w perpendicular to g = constant …

what is that w, in terms of x y and g ? :smile:
 

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