Select Answers to Section 4.3 Exercises of Worldwide ... · Select Answers to Section 4.3 Exercises...
Transcript of Select Answers to Section 4.3 Exercises of Worldwide ... · Select Answers to Section 4.3 Exercises...
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.
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