   Chapter 16.4, Problem 27E

Chapter
Section
Textbook Problem

Use the method of Example 5 to calculate ∫C F · dr, where F ( x , y ) = 2 x y   i +   ( y 2 − x 2 )   j ( x 2 + y 2 ) 2 and C is any positively oriented simple closed curve that encloses the origin.

To determine

To find: The line integral of vector field F(x,y)=2xyi+(y2x2)j(x2+y2)2 over curve C CFdr .

Explanation

Given data:

The given vector field is,

F(x,y)=2xyi+(y2x2)j(x2+y2)2

Formula used:

Green’s Theorem:

Consider a positively oriented curve C which is piece-wise smooth, simple closed curve in plane with domain D. Then,

CFdr=D(QxPy)dA (1)

Here,

Py is continuous first-order partial derivative of P,

Qx is continuous first-order partial derivative of Q, and

P and Q have continuous partial derivatives.

Write the expression for circle with center origin.

x2+y2=r2

Here,

Refer to Figure 11 in the textbook of section 16.4 for region D.

Consider a curve C , which is a circle r(t)=acosti+asintj,0t2π with counter-clockwise orientation centered at origin and radius a. The radius a is small and lies inside the circle C. The region between the circles C and C is D as shown in Figure 11.

The curve C is positively oriented, piecewise-smooth, and simply closed curve with domain D and hence Green’s theorem is applicable.

Compare the two expressions CPdx+Qdy and 2xyi+(y2x2)j(x2+y2)2 .

P=2xy(x2+y2)2Q=y2x2(x2+y2)2

Find the value of Py .

Py=y(2xy(x2+y2)2)=(y(2xy))(x2+y2)2(2xy)y((x2+y2)2)((x2+y2)2)2{t(uv)=uvuvv2}=(2x(1))(x2+y2)2(2xy)(2(x2+y2)(0+2y))(x2+y2)4{t(k)=0,t(t)=1,t(tn)=ntn1}=2x(x2+y2)28xy2(x2+y2)(x2+y2)4

Take common term outside in the numerator.

Py=2x(x2+y2)(x2+y24y2)(x2+y2)4=2x(x23y2)(x2+y2)3=2x36xy2(x2+y2)3

Find the value of Qx .

Qx=x(y2x2(x2+y2)2)=(x(y2x2))(x2+y2)2(y2x2)x((x2+y2)2)((x2+y2)2)2{t(uv)=uvuvv2}=(02x)(x2+y2)2(y2x2)(2(x2+y2)(2x+0))(x2+y2)4{t(k)=0,t(t)=1,t(tn)=ntn1}=2x(x2+y2)24x(y2x2)(x2+y2)(x2+y2)4

Qx=2x(x2+y2)24x(y2x2)(x2+y2)(x2+y2)4=2(x2+y2)(x3xy22xy2+2x3)(x2+y2)4=2(x33xy2)(x2+y2)3=2x36xy2(x2+y2)3

Substitute 2x36xy2(x2+y2)3 for Py and 2x36xy2(x2+y2)3 for Qx in equation (1),

CFdr=D(2x36xy2(x2+y2)32x36xy2(x2+y2)3)dA=D0dA=0

Find the value of F(r)

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