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410 CHAPTER 12 Vector Integral Calculus

We are now ready to evaluate F · dR. Parametrize C by x = 3cos(t), y = 3sin(t), z = 3
C
for 0 ≤ t ≤ 2π. The point (3cos(t),3sin(t),3) traverses C in the positive direction as t increases
from 0to2π. Therefore

F · dR = −ydx + xdy − xyz dz
C C
2π

= [−3sin(t)(−3cos(t)) + 3cos(t)(3cos(t))]dt
0
2π
= 9dt = 18π.
0
For the surface integral, compute

∇× F =−xzi + yzj + 2k
and
1
2
2
(∇× F) · n = √ (x − y + 2).
2
Then

(∇× F) · ndσ = (∇× F) · n N dx dy
D
√
1
2
2
= √ (x − y + 2) 2dx dy
2
D

2
2
= (x − y + 2)dx dy
D
2π 3
2
2
2
2
= [r cos (θ) −r sin (θ)]rdrdθ
0 0
2π 3 2π 3
3
= r cos(2θ)dr dθ + 2rdrdθ
0 0 0 0
3
2π
1 1 ! 3
= sin(2θ) r 4 + 2π r 2 = 18π.
2 0 4 0 0
12.9.1 Potential Theory in 3-Space
In Section 12.4, we discussed conservative vector fields and used Green’s theorem to derive a nec-
essary and sufficient condition for a vector field in the plane to have a potential function. Using
Stokes’s theorem we can prove the three-dimensional version of this test, which was suggested
in Section 12.4.
THEOREM 12.10 Test for a Conservative Vector Field
Let D be a simply connected region in 3-space. Let F and ∇× F be continuous on D. Then F is
conservative on D if and only if ∇× F = O in D.
This means that, in simply connected regions, irrotational vector fields (those with curl
(rotation) zero) are exactly the fields with potential functions.

Proof In one direction the proof is simple. If F =∇ϕ, then
∇× F =∇ × (∇ϕ) = O.

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October 14, 2010 14:53 THM/NEIL Page-410 27410_12_ch12_p367-424