Surface Integrals

The concept of a line integral can be extended to the concept of an integral over a  surface. Moreover, as we will see in the next two sections, this extension allows us to extend Green's theorem to higher dimensional settings.

To begin with, if r( u,v) is a chart for a surface in 3 dimensional space, then the parallelogram spanned by r_udu and r_vdv is practically a section of the surface when du and dv are sufficiently small. The area of the small parallelogram is the magnitude of the crossproduct || r_udu×r_vdv|| , which simplifies to

dS = || ru×rv|| dudv

We say that dS is the surface area differential for the surface.

If r( u,v) maps a region S in the uv-plane to a surface S, then a partition of S in the uv-plane leads to a partition of S in which each partition is practically the same as the parallelogram spanned by vectors r_uDu_j and r_vDv_k. Since the area of the parallelogram spanned by r_uDu_j and r_vDv_k is DS_jk = || r_u×r_v|| Du_jDv_k, the approximate area of the surface is

Surface  Area » n å j = 1  m å k = 1  DSjk = n å j = 1  m å k = 1  || ru×rv||   DujDvk

The limit as the partition becomes finer and finer results in the surface area formula

Surface  Area = ó õ ó õ S  dS = ó õ ó õ S  || ru×rv||   dudv

             

EXAMPLE 1    Find dS for the unit sphere in spherical coordinates. What is the area of the spherical cap with q in [0,2p] and f in [ 0,p/4] ?

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Solution: The parameterization of the unit sphere in spherical coordinates is given by

r( f,q) = á sin( f)cos( q) ,sin( f) sin( q) ,cos( f) ñ

(i.e., r = 1). Thus, r_f = á cos( f) cos( q) ,cos( f) sin(q) ,-sin( f) ñ and r_q = á -sin( f) sin( q),sin( f) cos( q) ,0 ñ . Since r_f and r_q are orthogonal and since r_f is a unit vector, we have

|| rf×rq|| = || rf||   || rq||   sin æ è  p 2 ö ø = 1· sin2( f) sin2( q)+sin2( f) cos2( q) = sin2( f)

Since f is between 0 and p, we thus have || r_f×r_q|| = sin( f) and dS = sin( f) dfdq. Thus, the area of the spherical cap is

Area =    ó õ ó õ   spherical  cap   dS =   ó õ 2p 0  ó õ p/4 0  sin(f) dfdq = p( 2-Ö2)

               

More generally, suppose that f( x,y,z) is defined on a surface S. Given h > 0, let's divide S into small patches with area DS_jk such that DS_jk < h for each jk patch. Also, let's choose a point P_jk = ( x_jk,y_jk,z_kj) in each patch, The surface integral of f( x,y,z) over S is defined to be

ó õ ó õ S   f( x,y,z) dS =   lim h®0  n å j = 1  m å k = 1  f( xjk,yjk,zkj) DSjk

It follows that if r( u,v) = á x(u,v) ,y( u,v) ,z( u,v) ñ  maps a region S in the uv-plane to the surface S, then

ó õ ó õ S  f( x,y,z) dS =   ó õ ó õ S  f( x( u,v),y( u,v) ,z( u,v) ) || ru×rv||   dudv