   Chapter 16, Problem 15RE

Chapter
Section
Textbook Problem

Verify that Green’s Theorem is true for the line integral ∫C xy2 dx − x2y dy, where C consists of the parabola y = x2 from (-1, 1) to (1, 1) and the line segment from (1, 1) to (-1, 1).

To determine

To Verify: TheGreen’s Theorem is true for the line integral Cxy2dxx2ydy , where C consists of the parabola y=x2 from (1,1) to (1,1) and the line segment from (1,1) to (1,1) .

Explanation

Given data:

Cxy2dxx2ydy where C consists of the parabola y=x2 from (1,1) to (1,1) and the line segment from (1,1) to (1,1) .

Formula used:

Write the expression Green’s Theorem.

CPdx+Qdy=D(QxPy)dA (1)

Find the value of limits as follows.

Draw the region D enclosed by C as shown in Figure 1.

Figure 1, consists of two parts as follows.

Parabolic function represented as folows.

C1:r(t)=ti+t2j, 1t1 (2)

Line function represented as follows.

C2:r(t)=ti+j, 1t1 (3)

Write the expression for r(t) .

r(t)=xi+yj (4)

Consider the expression as follows.

Cxy2dxx2ydy (5)

C is combination of C1 and C2 .

Modify equation (5) as follows.

Cxy2dxx2ydy=C1xy2dxx2ydy+C2xy2dxx2ydy (6)

Compare equations (2) and (4).

x=t (7)

y=t2 (8)

Differentiate equation (7) with respect to t .

dx=dt

Differentiate equation (8) with respect to t .

dy=2tdt

Find the value of C1xy2dxx2ydy .

Substitute t for x , t2 for y , dt for dx and 2tdt for dy ,

C1xy2dxx2ydy=C1(t)(t2)2dt(t)2(t2)2tdt=C1(t52t5)dt=C1t5dt

Integrate and apply limits as follows.

C1xy2dxx2ydy=11t5dt=[t66]11=(1)66((1)66)=16+16

Simplify expression as follows.

C1xy2dxx2ydy=0

Compare equations (3) and (4).

x=t (9)

y=1 (10)

Differentiate equation (9) with respect to t .

dx=dt

Differentiate equation (10) with respect to t .

dy=0

Find the value of C2xy2dxx2ydy .

Substitute t for x , 1 for y , dt for dx and 0 for dy ,

C2xy2dxx2ydy=C2(t)(1)2dt(t)2(1)(0)=C2tdt

Integrate and apply limits as follows.

C2xy2dxx2ydy=11tdt=[t22]11=(1)22((1)22)=12+12

Simplify expression as follows

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