- #1
dspch11
- 2
- 0
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
[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
Last edited by a moderator: