Calculus- Early Transcendentals, 2021a

562 Vector Calculus
we might want to measure the flux in the “upwards” direction, or if the surface is closed, like a sphere,
we might want to measure the flux “outwards” across the surface. In the first case we would choose N
to have positive z component, in the second we would make sure that N points away from the origin.
Unfortunately, there are surfaces that are not orientable: they have only one side, so that it is not possible
to choose the normal vectors to point in the “same way” through the surface. The most famous such
surface is the Möbius strip shown in Figure 16.7. It is quite easy to make such a strip with a piece of paper
and some tape. If you have never done this, it is quite instructive; in particular, you should draw a line
down the center of the strip until you return to your starting point. No matter how unit normal vectors are
assigned to the points of the Möbius strip, there will be normal vectors very close to each other pointing
in opposite directions.
Figure 16.7: A Möbius strip.
Assuming that the quantities involved are well behaved, however, the flux of the vector field across the
surface r(u,v) is
∫∫
∫∫
f · NdS = f · ru
∫∫
× r v
D
D |r u × r v | |r u × r v |dA = f · (r u × r v )dA.
D
In practice, we may have to use r v × r u or even something a bit more complicated to make sure that the
normal vector points in the desired direction.
Example 16.22: Flux
Compute the flux of f = 〈x,y,z 2 〉 across the cone z = √ x 2 + y 2 , 0 ≤ z ≤ 2, in the downward direction.
Solution. We write the cone as a vector function: r = 〈vcosu,vsinu,v〉, 0≤ u ≤ 2π and 0 ≤ v ≤ 2. Then
r u = 〈−vsinu,vcosu,0〉 and r v = 〈cosu,sinu,1〉 and r u × r v = 〈vcosu,vsinu,−v〉. The third coordinate
−v is negative, which is exactly what we desire, that is, the normal vector points down through the surface.
Then
∫ 2π ∫ 2
0
0
〈x,y,z 2 〉·〈vcosu,vsinu,−v〉dvdu =
∫ 2π ∫ 2
0
0
xvcosu + yvsinu − z 2 vdvdu