Surface integral to line integral

In summary, the integral identity in question involves a Heaviside step function and a Dirac delta-function. By changing variables and using the Jacobian determinant, one can formally obtain the result for a line integral. However, the question of how to handle the case where the derivative goes to infinity remains unclear.
  • #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
 
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  • #2
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


tiny-tim said:
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
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:
 

1. What is the difference between a surface integral and a line integral?

A surface integral is a type of integral that is used to calculate the flux of a vector field across a surface. It involves integrating over a two-dimensional surface. On the other hand, a line integral is used to calculate the work done by a vector field along a one-dimensional curve. It involves integrating over a one-dimensional curve.

2. How are surface integrals and line integrals related?

Surface integrals and line integrals are related through the fundamental theorem of calculus. This theorem states that the surface integral of a vector field over a surface is equal to the line integral of the same vector field along the boundary of that surface. This relationship is known as Stokes' theorem.

3. What is the significance of surface integrals and line integrals in physics?

Surface integrals and line integrals are important tools in physics for calculating physical quantities such as flux and work. They are used in various fields, including electromagnetism, fluid mechanics, and thermodynamics.

4. How is the orientation of the surface or curve taken into account in surface and line integrals?

The orientation of the surface or curve is taken into account by using a unit normal vector. For surface integrals, the orientation is determined by the direction of the normal vector, while for line integrals, the orientation is determined by the direction of the tangent vector.

5. What are some real-world applications of surface and line integrals?

Surface and line integrals have numerous applications in engineering, physics, and mathematics. Some examples include calculating the flow of a fluid through a surface, determining the work done by a force along a curved path, and calculating the electric flux through a closed surface.

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