Please solve the screenshot using the other screenshot! Use the method of this example to calculateF. dr, where= 2xyi + (y2 – x²)j(x2 + y2)²F(x, y) =and C is any positively oriented simple closed curve that encloses the origin.F. dr =/c EXAMPLE If F(x, y) = (-yi+x j)/(x² + y³), show that f. F • dr = 27 for everypositively oriented simple closed path that encloses the origin.SOLUTION Since C is an arbitrary closed path that encloses the origin, it's difficult tocompute 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' liesinside C. (See Figure 1.) Let D be the region bounded by C and C'. Then its positivelyoriented boundary is CU (-C') and so the general version of Green's Theorem givesCõe( P dx + Q dy + L Pds + edy - ( -) daDдуDy² – x²(x² + y²)²y? - x?(x² + y?)?dA = 0FIGURE 1S. P dx + Q dy = P dx + Q dyThereforeF • dr = {_ Fthat is,We now easily compute this last integral using the parametrization given byr(t) = a cos ti + a sin t j, 0 < t < 27. ThusF. dr = F· dr = " F(r(?)) · r'(1) dt(2= (-a sin t)(-a sin t) + (a cos t)(a cos t)a² cos?t + a² sin²t*27dt = " dt = 27

Given F(x,y) = 2xyx2+y22i+y2-x2x2+y22j To Evaluate ∫CF(x,y)dr

Answered: F(x, y) = 2xyi + (y² – x²)j (x2 + y2)2…

F(x, y) = 2xyi + (y² – x²)j (x2 + y2)2 and C is any positively oriented simple closed curve that encloses the origin. F. dr =

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Please solve the screenshot using the other screenshot!

Use the method of this example to calculate
F. dr, where
= 2xyi + (y2 – x²)j
(x2 + y2)²
F(x, y) =
and C is any positively oriented simple closed curve that encloses the origin.
F. dr =
/c

EXAMPLE If F(x, y) = (-yi+x j)/(x² + y³), show that f. F • dr = 27 for every
positively oriented simple closed path that encloses the origin.
SOLUTION Since C is 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 CU (-C') and so the general version of Green's Theorem gives
C
õe
( P dx + Q dy + L Pds + edy - ( -) da
D
ду
D
y² – x²
(x² + y²)²
y? - x?
(x² + y?)?
dA = 0
FIGURE 1
S. P dx + Q dy = P dx + Q dy
Therefore
F • dr = {_ F
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(?)) · r'(1) dt
(2= (-a sin t)(-a sin t) + (a cos t)(a cos t)
a² cos?t + a² sin²t
*27
dt = " dt = 27

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