Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro 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
LECTURE 22. LINE INTEGRALS 189 The function h(y, z) is the constant of integration: we integrated over x, so the constant may still depend on y or z. Differentiate with respect to y: By equation 22.37 f y = −2xy + h y (y, z) (22.40) −2xy + h y (y, z) = −2xy h y (y, z) = 0 Therefore h does not depend on y, only on z, and we can write Differentiating with respect to z, f(x, y, z) = 1 3 x3 − y 2 x + h(z) (22.41) f z = dh dz (22.42) By equation 22.38 Hence From equation 22.41 ∫ h(z) = z dh dz = f z = z (22.43) ∫ h ′ (z)dz = z zdz = 1 2 z2 + C (22.44) f(x, y, z) = 1 3 x3 − y 2 x + 1 2 z2 + C (22.45) where C is any arbitrary constant. Math 250, Fall 2006 Revised December 6, 2006.
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LECTURE 22. LINE INTEGRALS 189<br />
The function h(y, z) is the constant of integration: we integrated over x, so the<br />
constant may still depend on y or z. Differentiate with respect to y:<br />
By equation 22.37<br />
f y = −2xy + h y (y, z) (22.40)<br />
−2xy + h y (y, z) = −2xy<br />
h y (y, z) = 0<br />
Therefore h does not depend on y, only on z, and we can write<br />
Differentiating with respect to z,<br />
f(x, y, z) = 1 3 x3 − y 2 x + h(z) (22.41)<br />
f z = dh<br />
dz<br />
(22.42)<br />
By equation 22.38<br />
Hence<br />
From equation 22.41<br />
∫<br />
h(z) =<br />
z<br />
dh<br />
dz = f z = z (22.43)<br />
∫<br />
h ′ (z)dz =<br />
z<br />
zdz = 1 2 z2 + C (22.44)<br />
f(x, y, z) = 1 3 x3 − y 2 x + 1 2 z2 + C (22.45)<br />
where C is any arbitrary constant.<br />
<br />
Math 250, Fall 2006 Revised December 6, 2006.
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