If F is the vector field of this example, show that F. dr = 0 for every simple closed path that does not pass through or enclose the origin.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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EXAMPLE
If F(x, y) = (-yi+ x j)/(x² + y²), show that f. F· dr = 2n for every
positively oriented simple closed path that encloses the origin.
SOLUTION Since Cis an arbitrary closed path that encloses the origin, it's difficult to
compute the given integral directly. So let's consider a counterclockwise-oriented circle C'
with center the origin and radius a, where a is chosen to be small enough that C' lies
inside C. (See Figure 1.) Let D be the region bounded by C and C'. Then its positively
oriented boundary is C U (-C') and so the general version of Green's Theorem gives
y
C'
P dx + Q dy + P dx + Q dy = ||
õe
ӘР
dA
ду
D
-C'
y? – x?
(x² + y²)²
y? – x?
(x² + y²)².
dA = 0
FIGURE 1
Therefore
Рӑх + Q dy 3D | Pӑх + Qdy
S. F • dr - F• dr
that is,
We now easily compute this last integral using the parametrization given by
r(t) = a cos ti + a sin t j, 0 < t < 27. Thus
F· dr = F· dr = " F(r(t)) · r(t) dt
Jc'
(27 (-a sin t)(-a sin t) + (a cos t)(a cos t)
a cos?t + a²sin?t
dt =
" = 27
Transcribed Image Text:EXAMPLE If F(x, y) = (-yi+ x j)/(x² + y²), show that f. F· dr = 2n for every positively oriented simple closed path that encloses the origin. SOLUTION Since Cis an arbitrary closed path that encloses the origin, it's difficult to compute the given integral directly. So let's consider a counterclockwise-oriented circle C' with center the origin and radius a, where a is chosen to be small enough that C' lies inside C. (See Figure 1.) Let D be the region bounded by C and C'. Then its positively oriented boundary is C U (-C') and so the general version of Green's Theorem gives y C' P dx + Q dy + P dx + Q dy = || õe ӘР dA ду D -C' y? – x? (x² + y²)² y? – x? (x² + y²)². dA = 0 FIGURE 1 Therefore Рӑх + Q dy 3D | Pӑх + Qdy S. F • dr - F• dr that is, We now easily compute this last integral using the parametrization given by r(t) = a cos ti + a sin t j, 0 < t < 27. Thus F· dr = F· dr = " F(r(t)) · r(t) dt Jc' (27 (-a sin t)(-a sin t) + (a cos t)(a cos t) a cos?t + a²sin?t dt = " = 27
If F is the vector field of this example, show that
F. dr = 0 for every simple closed path that does not pass through or enclose the origin.
Transcribed Image Text:If F is the vector field of this example, show that F. dr = 0 for every simple closed path that does not pass through or enclose the origin.
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