   # (a) Find a function f such that F = ∇ f and (b) use part (a) to evaluate ∫ C F ⋅ d r along the given curve C . F ( x , y , z ) = ( y 2 z + 2 x z 2 ) i + 2 x y z j + ( x y 2 + 2 x 2 z ) k , C : x = t , y = t + 1 , z = t 2 , 0 ≤ t ≤ 1 ### Calculus (MindTap Course List)

8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285740621

#### Solutions

Chapter
Section ### Calculus (MindTap Course List)

8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285740621
Chapter 16.3, Problem 16E
Textbook Problem
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## (a) Find a function f such that F = ∇ f and (b) use part (a) to evaluate ∫ C F ⋅ d r along the given curve C. F ( x , y , z ) = ( y 2 z + 2 x z 2 )   i + 2 x y z   j + ( x y 2 + 2 x 2 z ) k , C : x = t ,   y = t + 1 , z = t 2 ,     0 ≤ t ≤ 1

To determine

(a)

To find:

f such that F=f

Solution:

fx,y, z=xy2z+x2z2

Explanation:

1) Concept:

If F=f then Fx,y=fxi+fyj+fzk

2) Given:

F=f,Fx,y=(y2z+2xz2)i+2xyzj+xy2+2x2zk

C:x=t, y=t+1, z=t2,  0t1

3) Calculation:

If F=f then fx=y2z+2xz2                                  1

fy=2xyz                                             2

fz=xy2+2x2z                                 (3)

Integrate 1 with respect to x

fx,y, z=xy2z+x2z2+gy, z                    4

Notice that the constant of integration is a constant with respect to x, that is, a functionof y, so consider it as gy, z.

Now differentiate above equation with respect to  y

fyx,y, z=2xyz+g'y,z(5)

Comparing (2), and (4)

g'y,z= 0

Integrating with respect to y

gyy,z=hz, where h(z) is a constant.

Hence (4) becomes

fx,y, z=xy2z+x2z2+h z                         (6)

Differentiate with respect to z,

fzx,y, z=xy2+2x2z+h'(z)

Comparing this equation with (3)

h'z=0

Integrate with respect to z

hz=K

Therefore, (6) becomes,

fx,y, z=xy2z+x2z2+K

fx,y, z=xy2z+x2z2     taking K=0

Conclusion:

fx,y, z=xy2z+x2z2 is a function such that F=f

(b)

To evaluate:

CF· dr along the given curve C

Solution:

5

Explanation:

1) Concept:

Fundamental theorem of line integral:

Let C be a smooth curve given by the vector function rt, atb. Let f be a differential function of two or three variables whose gradient vector f is continuous on C. Then

Cf ·dr=f(rb)-f(ra)

2) Given:

F=f,Fx,y=(y2z+2xz2)i+2xyzj+xy2+2x2zk

C:x=t, y=t+1, z=t2,      0t1

3) Calculation:

C is a smooth curve with initial point  r0 and terminal point 1

So, by using concept,

Cf ·dr=fr1-fr0=f0, 1, 0-f(1, 2, 1)

Since x=t, y=t+1, z=t2 ,

rt=t,t+1,t2

r0=(0, 1, 0)       and      r1=(1, 2, 1)

From the answer of part (a), fx,y=xy2z+x2z2Cf ·dr becomes

Cf ·dr=141+1212-0 =5-0=5

Therefore,

Cf ·dr=5

Conclusion:

Cf ·dr=5

### Explanation of Solution

1) Concept:

If F=f then Fx,y=fxi+fyj+fzk

2) Given:

F=f,Fx,y=(y2z+2xz2)i+2xyzj+xy2+2x2zk

C:x=t, y=t+1, z=t2,  0t1

3) Calculation:

If F=f then fx=y2z+2xz2                                  1

fy=2xyz                                             2

fz=xy2+2x2z                                 (3)

Integrate 1 with respect to x

fx,y, z=xy2z+x2z2+gy, z                    4

Notice that the constant of integration is a constant with respect to x, that is, a functionof y, so consider it as gy, z.

Now differentiate above equation with respect to  y

fyx,y, z=2xyz+g'y,z(5)

Comparing (2), and (4)

g'y,z= 0

Integrating with respect to y

gyy,z=hz, where h(z) is a constant.

Hence (4) becomes

fx,y, z=xy2z+x2z2+h z                         (6)

Differentiate with respect to z,

fzx,y, z=xy2+2x2z+h'(z)

Comparing this equation with (3)

h'z=0

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