Property of integrals over surfaces

  • Thread starter Benny
  • Start date
  • #1
584
0
If [itex]f:S \to R[/itex] is a continuous function on a surface S, we define

[tex]
\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
[/tex]

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:

Answers and Replies

Related Threads on Property of integrals over surfaces

  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
14
Views
677
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
5
Views
511
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
1K
Replies
7
Views
2K
Replies
0
Views
775
Replies
2
Views
2K
Replies
2
Views
2K
Top