Surface integral to line integral

[dspch11]

I am agonizing about the following integral identity:

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

where ds is the line element. Clearly, using the Heavisite step function, the condition $$g(x,y) \leq t$$ 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 $$g(x,y) \leq t$$ which is growing with t. The increase in the integral can then be evaluated by summing contributions along its circumference with $$\left| \nabla g(x,y) \right|$$ giving the density of isolines.

How can one formally obtain this result?

Thank you for you help,
Daniel

 

[tiny-tim]

Welcome to PF!

Hi Daniel! Welcome to PF!

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

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

 

[dspch11]

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

Is this what you suggest:

$$s = \int \sqrt{ 1 + (\frac{dy}{dx})^2} dx$$

for constant g we have furthermore:

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

and, thus,

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

The jacobian determinant is

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

What about if $$\frac{dy}{dx}$$ goes to infinity?

 

[tiny-tim]

uhh?

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

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