Multivariate Calculus - Bruce E. Shapiro

188 LECTURE 22. LINE INTEGRALS
5. The line integral around a closed curve is zero, i.e.,
∮
F · dr = 0
for all closed curves C.
Example 22.5 Determine if
is a gradient field.
C
F = ( x 2 − y 2 , −2xy, z )
Solution. We calculate the curl F,
⎛
⎞ ⎛
0 −D z D y x 2 − y 2 ⎞ ⎛
⎞
−D z (−2xy) + D y (z)
∇F = ⎝ D z 0 −D x
⎠ ⎝ −2xy ⎠ = ⎝ D z (x 2 − y 2 ) − D x (z) ⎠
−D y D x 0 z −D y (x 2 − y 2 ) + D x (−2xy)
= (0, 0, 2y − 2y) = 0
Hence the field is a gradient field.
Example 22.6 Determine if F = (−y, z, x) is a gradient field.
Solution.
⎛
0 −D z D y
∇F = ⎝ D z 0 −D x
−D y D x 0
= (−1, −1, 1) ≠ 0
⎞ ⎛ ⎞ ⎛
⎞
−y −D z (z) + D y (x)
⎠ ⎝ z ⎠ = ⎝ D z (−y) − D x (x) ⎠
x −D y (−y) + D x (z)
Hence the field is not a gradient field.
Example 22.7 Find the potential function f(x, y, z) for the gradient field in example
22.5.
Solution. The vector field is
F = (x 2 − y 2 , −2xy, z) = (M, N, P )
where M = x 2 − y 2 , N = −2xy, and P = z. Since F is a gradient field, then for
some scalar function f,
f x = M = x 2 − y 2 (22.36)
f y = N = −2xy (22.37)
f z = P = z (22.38)
Integrating the first equation with respect to x,
∫ ∫
f(x, y, z) = f x dx = (x 2 − y 2 )dx = 1 3 x3 − y 2 x + h(y, z) (22.39)
Revised December 6, 2006. Math 250, Fall 2006