The Vector Surface Differential   

We often define the vector surface differential dS = n  dS, so that the flux integral becomes

Flux =   ó õ ó õ S    F· dS

For example, if r( u,v) is the regular parameterization of a surface S, then the unit normal is of the form

n =  ru×rv | | ru×rv| |

and correspondingly, the vector surface differential becomes

dS = n  dS =  ru×rv || ru×rv| |   | | ru×rv| | dudv = ( ru×rv) dudv

That is, dS = ( r_u×r_v)dudv, so that the flux integral becomes

Flux =   ó õ ó õ S  F · dS = ó õ ó õ S  F·( ru×rv) dudv

which eliminates the need to calculate the norm of the cross product.      

EXAMPLE 3    What is the flux of the vector field F( x,y,z) = á y,x,z ñ through the surface

r( u,v) = á cos( u) ,sin(u) ,2v ñ ,    u  in  [ 0,p],  v  in  [ 0,1]

Solution: Since r_u = á -cos( u),sin( u) ,0 ñ and r_v = á0,0,1 ñ , it follows that

ru×rv = á -cos( u),sin( u) ,0 ñ × á 0,0,2 ñ = á 2sin( u) ,2cos( u) ,0 ñ

As a result, dS = á 2sin( u) ,2cos(u) ,0 ñ dudv. Since x = cos( u) , y = sin( u) , and z = 2v, we have

Flux = ó õ ó õ S   á y,x,z ñ ·dS = ó õ p 0  ó õ 1 0  á sin( u) ,cos(u) ,2v ñ · á sin( u) ,cos( u) ,0 ñ dudv = ó õ p 0  ó õ 1 0  sin2( u) +cos2(u) +0  dudv = p    units  of  volume units  of  time

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EXAMPLE 4    What is the flux of F(x,y,z) = á y,-x,z ñ through the surface

r( u,v) = á u+v,u-v,u ñ ,   u  in  [ 0,1] ,  v  in  [ 0,1]

Solution: Since r_u = á 1,1,1 ñ and r_v = á 1,-1,0 ñ , it follows that

ru×rv = á 1,1,-2 ñ

Thus, dS = á 1,1,-2 ñ du dv and the flux is

Flux = ó õ 1 0  ó õ 1 0  á y,-x,z ñ · á 1,1,-2 ñ dudv = ó õ 1 0  ó õ 1 0  ( y-x-2z)   dudv

Since x = u+v, y = u-v, and z = u on the surface, we have

Flux = ó õ 1 0  ó õ 1 0  ( u-v-( u+v) -2u)  dudv = ó õ 1 0  ó õ 1 0  ( -2u-2v) dudv = -2    units  of  volume units  of  time

       

The surface z = g( x,y) can be parametrized by r( x,y) = á x,y,g( x,y) ñ .  Since  r_x = á 1,0,g_x ñ and r_y = á 0,1,g_y ñ, their cross product is

ry×ry = á 1,0,gx ñ× á 0,1,gy ñ = á-gx,-gy,1 ñ

If F = á M,N,P ñ and z = g( x,y) is over a region R in the xy-plane, then

Flux =    ó õ ó õ S  F· dS =  ó õ ó õ R áM,N,P ñ × á -gx,-gy,1 ñ   dA

which reduces to

Flux =  ó õ ó õ S F· dS = ó õ ó õ R (P-Mgx-Ngy) dA

       

EXAMPLE 5    What is the flux of the vector field F( x,y,z) = á x,y,3z ñ through the surface z = x²+y² over the unit circle: Solution: If we identify g( x,y) = x²+y², then g_x = 2x and g_y = 2y, so that

Flux = ó õ ó õ R ( 3z-xgx-ygy) dA = ó õ ó õ R ( 3x2+3y2-x( 2x) -y( 2y)) dA = ó õ ó õ R ( x2+y2) dA

Since R is the unit circle, we transform to polar coordinates to obtain

Flux = ó õ 2p 0  ó õ 1 0  r  rdrdq = ó õ 2p 0  ó õ 1 0  r2drdq = ó õ 2p 0   1 3 dq =  2p 3     units  of  volume units  of  time