Surface Integrals (click here for discussion of surface integrals with oriented surfaces)

In a Nut Shell: Two questions are relevant. What is a surface integral and how do

you evaluate a surface integral?

Recall the line integral, I, provides the value of a function, f(x,y,z), evaluated along

a curve, C, in space. Here the line integral is I = ∫ f(x, y, z) ds

where ds is the arc length along the curve, C C

The surface integral, I_s, is analogous to the line integral in that it provides the value of a

function, f(x,y,z), evaluated over a “smooth” surface, S, in space. Here the surface integral

is I_s = ∫ ∫ f(x, y, z) dS where dS is the element of surface area on the spatial surface

How do you evaluate a surface integral?

Usually the surface, S, in space is somewhat complicated. So one strategy is to

transform the element of surface area, dS, from the x-y-z space into a parallelogram

dA = du dv in the u-v plane.

where r(u,v) is the parametric representation of the surface, S.

r(u,v) = < x(u,v), y(u,v), z(u,v) > is the position vector to point on surface, S

Click here to continue with discussion of surface integrals.

Return to Calculus 3 Notes

Copyright © 2017 Richard C. Coddington

All rights reserved.