Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Property of integrals over surfaces

  1. Apr 28, 2006 #1
    If [itex]f:S \to R[/itex] is a continuous function on a surface S, we define

    \int\limits_{}^{} {\int\limits_S^{} {fdS} } = \int\limits_{}^{} {\int\limits_D^{} {\left( {fo\Phi } \right)} } \left\| {\frac{{\partial \Phi }}{{\partial u}} \times \frac{{\partial \Phi }}{{\partial v}}} \right\|dudv

    where [itex]\Phi :D \subset R^2 \to R^3 [/itex] is a regular, 1-1 parameterisation of [itex]S = \Phi \left( D \right)[/itex].

    The integral

    [tex]\int\limits_{}^{} {\int\limits_S^{} {fdS} } [/tex] does not depend on the choice of parameterisation [itex]\Phi [/tex].

    This follows from the change of variables formula for double integrals.

    The formula for n dimension integrals is:

    [tex]\int\limits_D^{} {f\left( {x_1 ,...,x_n } \right)} dx_1 ...dx_2 = \int\limits_{D^* }^{} {f\left( {h\left( {u_1 ,...,u_n } \right)} \right)\left| {\det Dh\left( {u_1 ,...,u_n } \right)} \right|} du_1 ...du_n [/tex]

    where h(D*) = D.

    It would make sense that the integral of f over S doesn't depend on parameterisation of the surface. But I don't see how that follows from the change of variables formula. (Similarly, an integral of a scalar function over a path doesn't depend on paramterisation of curve and this again follows from change of variables formula). Can someone explain why this is the case to me? Any help would be good thanks.
    Last edited: Apr 28, 2006
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?