Calculus- Early Transcendentals, 2021a

564 Vector Calculus
where e is an electric field and ε 0 (the permittivity of free space) is a known constant; N is oriented
outward. Use Gauss’s Law to find the charge contained in the cube with vertices (±1,±1,±1) if the
electric field is e = 〈x,y,z〉.
16.8 Stokes’ Theorem
Recall that one version of Green’s Theorem (see Equation 16.2)is
∫ ∫∫
f · dr = (∇ × f) · kdA.
∂D
D
Here D is a region in the xy-plane and k is a unit normal to D at every point. If D is instead an orientable
surface in space, there is an obvious way to alter this equation, and it turns out still to be true:
Theorem 16.23: Stokes’ Theorem
Provided that the quantities involved are sufficiently nice, and in particular if D is orientable,
∫ ∫∫
f · dr = (∇ × f) · NdS,
∂D
if ∂D is oriented counter-clockwise relative to N.
D
The proof of Stokes’ Theorem will follow a discussion and several examples of the Theorem in use.
Note how little has changed: k becomes N, a unit normal to the surface, and dA becomes dS, since this
is now a general surface integral. The phrase “counter-clockwise relative to N” meansthatifwetakethe
direction of N to be “up”, then we go around the boundary counter-clockwise when viewed from “above”.
Example 16.24
Let f = 〈y 2 z,x 2 z,xy 2 〉 and the surface D be x = √ 1 − y 2 − z 2 , oriented in the positive x direction. It
quickly becomes apparent that the surface integral in Stokes’ Theorem is intractable, so compute
the line integral.
Solution. The boundary of D is the unit circle in the yz-plane, r = 〈0,cosu,sinu〉,0≤ u ≤ 2π. The integral
is
because x = 0.
∫ 2π
0
〈y 2 z,x 2 z,xy 2 〉·〈0,−sinu,cosu〉du =
∫ 2π
0
0du = 0,
An interesting consequence ofStokes’TheoremisthatifD and E are two orientable surfaces with the
same boundary, then
∫∫
∫ ∫ ∫∫
(∇ × f) · NdS = f · dr = f · dr = (∇ × f) · NdS.
Sometimes both of the integrals
∫∫
D
D
∂D
∂E
(∇ × f) · NdS and
∫
∂D
E
f · dr
♣