   Chapter 16, Problem 13RE

Chapter
Section
Textbook Problem

Show that F is a conservative and use this fact to evaluate ∫C F · dr along the given curve.13. F(x, y) = (4x3y2 – 2xy3) i + (2x4y – 3x2y2 + 4y3) j, C: r(t) = (t + sin πt) i + (2t + cos πt) j, 0 ⩽ t ⩽ 1

To determine

To show: F is a conservative vector field and evaluate the value of CFdr.

Explanation

Given data:

F(x,y)=(4x3y22xy3)i+(2x4y3x2y2+4y3)j (1)

C:r(t)=(t+sinπt)i+(2t+cosπt)j,0t1 (2)

Formula used:

Consider the standard equation of curl F for F=Pi+Qj+Rk

curlF=|ijkxyzPQR| (3)

If the curl of a vector field F is zero, then F is said to be a conservative vector field.

The relation between the potential function f and vector field F is F=f, where f=fx(x,y)i+fy(x,y)j.

Obtain the value of curlF.

Substitute (4x3y22xy3) for P, (2x4y3x2y2+4y3) for Q and 0 for R in equation (3).

curlF=|ijkxyz(4x3y22xy3)(2x4y3x2y2+4y3)0|={[y(0)z(2x4y3x2y2+4y3)]i[x(0)z(4x3y22xy3)]j+[x(2x4y3x2y2+4y3)y(4x3y22xy3)]k}=(00)i(00)j+(8x3y6xy2+08x3y+6xy2)k=(00)i(00)j+(0)k

On simplification,

curlF=0

Since curlF=0, F is a conservative vector and the domain of F is 2.

Thus, F is a conservative vector field.

Substitute fx(x,y)i+fy(x,y)j for f in F=f.

F=fx(x,y)i+fy(x,y)j (4)

Compare the equation (4) and equation (1).

fx(x,y)=4x3y22xy3 (5)

fy(x,y)=2x4y3x2y2+4y3 (6)

Integrate equation (5) with respect to x.

f(x,y)=(4x3y22xy3)dx=4x44y22x22y3+g(y)

f(x,y)=x4y2x2y3+g(y) (7)

Apply partial differentiation with respect to y on both sides of equation (7)

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