Calculus- Early Transcendentals, 2021a

16.8. Stokes’ Theorem 565
are difficult, but you may be able to find a second surface E so that
∫∫
(∇ × f) · NdS
E
has the same value but is easier to compute.
In the previous example, the line integral was easy to compute. But we might also notice that another
surface E with the same boundary is the flat disk y 2 + z 2 ≤ 1.
Example 16.25
Let f = 〈y 2 z,x 2 z,xy 2 〉 and the surface E be y 2 + z 2 ≤ 1. Compute the surface integral.
Solution. The unit normal N for this surface is simply i = 〈1,0,0〉. We compute the curl:
Since x = 0 everywhere on the surface,
so the surface integral is
∇ × f = 〈2xy − x 2 ,0,2xz − 2yz〉.
(∇ × f) · N = 〈0,0,−2yz〉·〈1,0,0〉 = 0,
∫∫
E
0dS = 0,
as before. In this case, of course, it is still somewhat easier to compute the line integral, avoiding ∇ × f
entirely.
♣
Now let’s look at the proof of Stokes’ Theorem.
Proof. We can prove here a special case of Stokes’ Theorem, which perhaps not too surprisingly uses
Green’s Theorem.
Suppose the surface D of interest can be expressed in the form z = g(x,y),andletf = 〈 f 1 , f 2 , f 3 〉.Using
the vector function r = 〈x,y,g(x,y)〉 for the surface we get the surface integral
∂D
∫∫
D
∫∫
∇ × f · dS =
∫∫
=
a
E
E
〈 ∂ f 3
∂y − ∂ f 2
∂z , ∂ f 1
∂z − ∂ f 3
∂x , ∂ f 2
∂x − ∂ f 1
∂y 〉·〈−g x,−g y ,1〉dA
− ∂ f 3
∂x g x + ∂ f 2
∂z g x − ∂ f 1
∂z g y + ∂ f 3
∂x g y + ∂ f 2
∂x − ∂ f 1
∂y dA.
Here E is the region in the xy-plane directly below the surface D.
For the line integral, we need a vector function for ∂D. If〈x(t),y(t)〉 is a vector function for ∂E then
we may use r(t)=〈x(t),y(t),g(x(t),y(t))〉 to represent ∂D. Then
∫ ∫ b
∫
dx
f · dr = f 1
dt + f dy
2
dt + f dz b
(
3
dt dt = dx
f 1
dt + f dy ∂z
2
dt + f dx
2
∂x dt + ∂z )
dy
dt.
∂y dt
using the chain rule for dz/dt. Now we continue to manipulate this:
∫ b
(
dx
f 1
dt + f dy ∂z
2
dt + f dx
3
∂x dt + ∂z )
dy
dt
∂y dt
a
a