Multivariate Calculus - Bruce E. Shapiro

LECTURE 25. STOKES’ THEOREM 203
Since C, and hence S, both lie in the xy-plane we have (n) = (0, 0, 1). Hence
(∇ × F) · n = (P y − N z , M z − P x , N x − M y ) · (0, 0, 1)
= N x − M y
= ∂
∂x (xz2 + x) − ∂ ∂y (yz2 − y)
= z 2 + 1 − z 2 + 1 = 2
Hence
∮
C
F · T ds = 2
S
dS = 2π(3 2 ) = 18π
Example 25.2 Find ∮ C
F · dr for
where C is a cirlce defined by
using Stoke’s theorem.
F = (z − 2y, 3x − 4y, z + 3y
x 2 + y 2 = 4, z = 1
Solution We are again going to use the formula
∮
F · T ds = (∇ × F) · n dS
C
We again have n = (0, 0, 1), and C is a circle of radius 2. Hence
S
(∇ × F) · n = (P y − N z , M z − P x , N x − M y ) · (0, 0, 1)
= N x − M y
= ∂
∂x (3x − 4y) − ∂ (z − 2y)
∂y
3 + 2 = 5
Hence
∮
C
F · T ds = 5
S
dS = 5π(1 2 ) = 20π
Math 250, Fall 2006 Revised December 6, 2006.