Select Answers to Section 4.3 Exercises

of Worldwide Multivariable Calculus

1. 0 joules

3.π

2joules

5. 3 joules

7.→∇× F = 0 so f(x, y) = 1

4x4 + 1

6y6 + c for c ∈ R.

9.→∇× F = 0 so f(x, y) = xy2 + x2y3 + c for c ∈ R.

11.→∇× F = −2 sinx

13.→∇× F = 0 so f(x, y) = xy + tan−1(x) + tan−1(y) + c for c ∈ R.

15.

∫C

F · dr = 0 joules

17.

∫C

F · dr = 0 joules

19.

∫C

F · dr = 8 sin(2)− 4 joules

21.

∫C

F · dr = 1 +π

2joules

23. Yes.

25. No.

27.→∇× F = 0, so f(x, y, z) = 1

2x2 + 1

2y2 + 1

2z2 + c for c ∈ R..

29.→∇× F = i + j + k

31.→∇× F = −i− 2j + k

33.→∇× F = 0, so f(x, y, z) = x sin y + y cos z + ez + c for c ∈ R.

35.

∫C

F · dr = 19 joules

1

37.

∫C

F · dr = −11

2joules

39.

∫C

F · dr = 2− e3 − e joules

41.

∫C

F · dr = eπ − π − 1 joules

43.

45. If, for every closed oriented piecewise regular curve C, we have∫C

F · dr =

∫C

G · dr

for the given C1 vector fields F and G, then∫C

(F−G) · dr = 0

and via Theorem 4.3, this means that→∇× (F−G) = 0. But the “curl” operation

is linear on vector fields, so that

→∇× F−

→∇×G = 0

i.e. the curls are equal.

2