SM221 Final Exam – May 7, 2012 – Answer Key 1.

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SM221 Final Exam – May 7, 2012 – Answer Key
Part I
1.
c
6.
c
11.
d
16.
a
2.
a
7.
b
12.
a
17.
d
3.
b
8.
c
13.
b
18.
d
4.
e
9.
c
14.
b
19.
b
5.
b
10.
a
15.
a
20.
b
SM221 Final Exam – May 7, 2012 – Answer Key
Part II
21. See Theorem 15, Section 14.6, p. 939 and Theorem 3, Section 16.5, p. 1092 of text.
Part III
22. 𝑓 (π‘₯, 𝑦) = π‘₯ 3 − 3π‘₯𝑦 + 𝑦 3. 𝑓π‘₯ = 3π‘₯ 2 − 3𝑦, 𝑓𝑦 = −3π‘₯ + 3𝑦 2
Setting 𝑓π‘₯ = 0 and 𝑓𝑦 = 0 gives two critical points, (0,0) and (1,1).
2
The discriminant is 𝐷 = 𝑓π‘₯π‘₯ 𝑓𝑦𝑦 − 𝑓π‘₯𝑦
= (6π‘₯)(6𝑦) − (−3)2 .
At (0,0), we have 𝐷 = −9 < 0 so (0,0) is a saddle point of 𝑓.
At (1,1), we have 𝐷 = 27 > 0 and 𝑓π‘₯π‘₯ = 6 > 0 so 𝑓 has a local minimum at (1,1).
23.
οΏ½
2πœ‹
0
2
οΏ½ οΏ½
0
5
1+π‘Ÿ 2
π‘Ÿ 𝑑𝑧 π‘‘π‘Ÿ π‘‘πœƒ = 8πœ‹
24. a. 𝑓 (π‘₯, 𝑦) = 3π‘₯𝑦 + 2𝑦 2 .
b. 𝑓 (0,2) − 𝑓(1,0) = 8 − 0 = 8
c. π‘₯ = 1 − 𝑑, 𝑦 = 2𝑑, 0 ≤ 𝑑 ≤ 1, 𝑑π‘₯ = −𝑑𝑑, 𝑑𝑦 = 2𝑑𝑑
οΏ½ 𝑭 βˆ™ 𝑑𝒓 = οΏ½ 3𝑦 𝑑π‘₯ + (3π‘₯ + 4𝑦)𝑑𝑦
𝐢
1
𝐢
1
= ∫0 3(2𝑑)(−1) + οΏ½3(1 − 𝑑) + 4(2𝑑)οΏ½2 𝑑𝑑 = ∫0 6 + 4𝑑 𝑑𝑑 = 8
25.
𝑇(π‘₯, 𝑦, 𝑧) =
30
1+π‘₯ 2 +𝑦 2 +𝑧 2
, 𝑇π‘₯ =
−60π‘₯
(1+π‘₯ 2 +𝑦 2 +𝑧2 )2
, with similar formulas for 𝑇𝑦 and 𝑇𝑧 .
The rate of change of temperature at the point (1,1,1) in the direction of ⟨2, −2,1⟩ is
−60
1
5
1
⟨1,1,1⟩ βˆ™ ⟨2, −2,1⟩ = −
∇𝑇(1,1,1) βˆ™ ⟨2, −2,1⟩ =
16
3
4
3
i.e., the temperature is decreasing at 1.25o C per meter.
26. a.
b. ∫𝐢 2𝑦 𝑑π‘₯ − 2π‘₯ 𝑑𝑦
2πœ‹
5
=∬interior of 𝐢 −4 𝑑𝐴 = ∫0 ∫0 −4π‘Ÿ π‘‘π‘Ÿ π‘‘πœƒ =
−4(Area of circle of radius 5)= -100π.
c. No, 𝑭 is not conservative because the
integral of 𝑭 around the closed curve 𝐢 is not
zero.
27. The mass of the wire in grams is
πœ‹
πœ‹
π‘š = ∫0 0.1√1 + 𝑑 2 οΏ½(−2 sin(𝑑))2 + (2 cos(𝑑))2 + 12 𝑑𝑑 = ∫0 0.1√1 + 𝑑 2 √5 𝑑𝑑 ≅ 1.366.
28. The mass of the solid in grams is
2πœ‹
πœ‹
3
π‘š = ∫0 ∫02 ∫0 2𝜌 𝜌2 sin(πœ‘) π‘‘πœŒ π‘‘πœ‘ π‘‘πœƒ = 81πœ‹.
29. Let 𝑆 be the given surface. A parametrization of 𝑆 is 𝒓(π‘₯, 𝑦) = ⟨π‘₯, 𝑦, 3 − π‘₯ − 𝑦⟩, for (π‘₯, 𝑦) in
the region 𝐷 given by 0 ≤ π‘₯ ≤ 3 and 0 ≤ 𝑦 ≤ 3 − π‘₯.
𝒓π‘₯ = ⟨1,0, −1⟩, 𝒓𝑦 = ⟨0,1, −1⟩, 𝒓π‘₯ × π’“π‘¦ = ⟨1,1,1⟩ (which is upward).
3
3−π‘₯
βˆ¬π‘† 𝑭 βˆ™ 𝑑𝑺 = ∫0 ∫0
3
3−π‘₯
⟨𝑦, 2π‘₯, 3 − π‘₯ − 𝑦⟩ βˆ™ ⟨1,1,1⟩ 𝑑𝑦 𝑑π‘₯ = ∫0 ∫0
3 + π‘₯ 𝑑𝑦 𝑑π‘₯ = 18.
30. Let 𝐷 be the region enclosed by 𝑆. By the Divergence Theorem, the charge 𝑄 is
2
2
2
πœ€0 ∭𝐷 div 𝑬 𝑑𝑉 = πœ€0 ∫0 ∫0 ∫0 2 + 1 + 3 𝑑𝑧 𝑑𝑦 𝑑π‘₯ = 48 πœ€0 .
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