   Chapter 16.3, Problem 16E

Chapter
Section
Textbook Problem

(a) Find a function f such that F = ∇ f and (b) use part (a) to evaluate ∫C F · dr along the given curve C.16. F(x, y, z) = (y2z + 2xz2) i + 2xyz j+ (xy2 + 2x2z) k,C: x = t , y = t + 1, z = t2, 0 ⩽ t ⩽ 1

(a)

To determine

To find: The potential function f such that F=f .

Explanation

Given data:

Vector field is F(x,y,z)=(y2z+2xz2)i+2xyzj+(xy2+2x2z)k .

Consider f=fx(x,y,z)i+fy(x,y,z)j+fz(x,y,z)k .

Write the relation between the potential function f and vector field F .

f=F

Substitute fx(x,y,z)i+fy(x,y,z)j+fz(x,y,z)k for f ,

F=fx(x,y,z)i+fy(x,y,z)j+fz(x,y,z)k

Compare the equation F=fx(x,y,z)i+fy(x,y,z)j+fz(x,y,z)k with F(x,y,z)=(y2z+2xz2)i+2xyzj+(xy2+2x2z)k .

fx(x,y,z)=y2z+2xz2 (1)

fy(x,y,z)=2xyz (2)

fz(x,y,z)=xy2+2x2z (3)

Integrate equation (1) with respect to x.

f(x,y,z)=(y2z+2xz2)dx=y2zdx+2z2xdx=y2z(x)+2z2(x22)+g(y,z) {dt=t,tdt=t22}

f(x,y,z)=xy2z+x2z2+g(y,z) (4)

Apply partial differentiation with respect to y on both sides of equation (4).

fy(x,y,z)=y(xy2z+x2z2+g(y,z))=y(xy2z)+y(x2z2)+y(g(y,z))=xzy(y2)+x2z2y(1)+g(y,z)=xz(2y)+0+g(y,z) {t(k)=0,tt2=2t}

fy(x,y,z)=2xyz+g(y,z) (5)

Compare the equations (2) and (5)

(b)

To determine

The value of Cfdr along the curve C.

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