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For the most recent information go to the end of the page.

CHANGE OF OFFICE HOURS TIME: NEW TIME IS 14:00-16:00

Solutions for test 2a are posted.

Solutions for test 3b are posted.

Solutions for test 4b are posted.

Remember that the final exam is on Apr 21, 19:00, SE2072.

Office hours are the day before the exam, on Apr 20, 14:00-16:00.

Office hours for study week are given by Vijay at SB 4059B, on Tue 13th, 2:30-4:30 and on Tues 20th, 2:30-4:30

Notice that you probably need to refresh this page to get the most recent information.

University of Toronto at Mississauga
Department of Mathematics
MAT368 (Vector Calculus) Spring 2004

INSTRUCTOR: Dmitry Khmelev

OFFICE: SB 4059B dkhmelev (at) math.toronto.edu

OFFICE HOURS: Tuesday 14:00-15:00, Thursday 14:00-15:00 or by appointment.

TEXTBOOK: Vector Calculus, 5th edition, J.Marsden and A. Tromba.

EVALUATION: There will be 4 term tests written during the Tuesday¹ 9 a.m. lecture periods (50 minutes); each will be worth 15% of the total mark. A final examination will account for the remaining 40%.

TIME&PLACE: Tue 9:00-11:00 & Thu 9:00-10:00, SE 2068.

TUTORIALS: Your TAs is Samuel Chun. Tutorial time and place: Fri 15:00-17:00, SE 1104. The first tutorial is on Jan 9, Friday.

MISSED TEST: There is no make-up test. If you miss a test due to illness, you will have to bring a dated doctor's certificate stating that you were too ill to take the tests. In that event we will re-rate your score based on the score you take.

DROP DATE: March 7, 2004

RE-MARKING THE TEST:

Any oversight in the marking of a test paper should be brought to the attention of your tutor immediately after the test was returned.

All matters regarding test marks (e.g. additional errors) must be communicated to me, outside the class, within one week of return date.

No test written in pencil or erasable pen may be submitted for re-marking.

CHANGES: Changes to and clarification of details in this course outline will be announced to the class on at least two occasions to take effect.

SYLLABUS: Chapters 1, 2, 4, 7 and 8.

Recommended problems for test #1, Tuesday, Jan 27, 2003.

Section 1.1. Problems 1-9, 11-19, 23-28².

Section 1.2. Problems 1-18.

Section 1.3. Problems 1-16, 21, 22, 24-32³.

Section 1.4. Problems 1-9.

Review excercises for Chapter 1: 1-9, 11, 12, 21-26, 42-47.

Section 2.1. Problems 1-12.

Section 2.2. Problems 1-4, 15-17.

Section 2.3. Problems 1-3, 5-13, 15-18.

Test 1 with solutions:

t/1a.html, t/1a.ps, t/1a.pdf

t/1b.html, t/1b.ps, t/1b.pdf

Recommended problems for test #2, Tuesday, Feb 24, 2004⁴:

Section 2.4. Problems 1-12, 14-20.

Section 2.5. Problems 1-13.

Section 2.6. Problems 1-21⁵.

Section 4.3. Problems 1-8, 13-16.

Section 4.4. Problems 1-6, 9-26.

Office hours on Thursday Feb 19th 10:00-12:00 will be given by Vijay Patankar in Math Aid Centre Room, 3093H.

Office hours on Monday Feb 23 will be given by Denis Gaydashev, 11:00-13:00, SB 4063 (note the different office number).

Test 2 with solutions:

t/2a.ps, t/2a.pdf

t/2b.ps, t/2b.pdf

SOLUTIONS for 2a: t/2asol.ps, t/2asol.pdf (solutions for test 2b are very similar; do them as an excersize).

Recommended problems for test #3, Tuesday, March 16, 2004:

Section 7.2. Problems 1-3, 4(a), 6, 7, 9-18.

Section 7.3. Problems 1-15.

Review problems on double integrals: section 5.3, 1-10, section 5.4, 1-14.

Section 7.6. Problems 1-7, 9-11, 15-18.

Section 7.4. Problems 1-10.

Hints for solving problems:

Problem 7.6-7. Notice that the surface integral consists of two pieces, the upper hemisphere U and the unit disc S in xy-plane. Integral over U must be parametrized using spherical coordinates with 0 £ q £ 2p, 0 £ f £ p/2, which define rectangle D in qf-plane. With this parametrization we get normal -sinf(x,y,z), pointing inwards, hence

ó
õ ó
õ

U
F·dS = - ó
õ ó
õ

D
-sin(f) æ
è (x+3y²)x+(y-10xz)z+(z-xy)z ö
ø dqdf

= - ó
õ ó
õ

D
-sin(f)(1+3y⁵x+9xyz)dqdf,

where we must use x²+y²+z²=1. Notice that here all variables (x,y,z)=F(q,f). Next,

- ó
õ ó
õ

D
-sin(f)(1+3y⁵x+9xyz)dqdf = ó
õ ó
õ

D
sinfdqdf+ ó
õ ó
õ

D
3y⁵x sinfdqdf+ ó
õ ó
õ

D
9xyzsinfdqdf,

The first integral in r.h.s. gives 2p, while the others are equal to 0 (which is easily seen after spherical coordinates parametrization is used, but you'll need to integrate òsin⁵qcosqdq = sin⁶q). As an excercise, verify that òò_SF·dS=0 (use parametrization F(u,v)=(cosu,sinu,v), 0 £ u £ 2p, 0 £ v £ 1). Answer. 2p.

Problem 7.6-9. Using spherical coordinates parametrization, (and remembering to reverse orientation), we get

ó
õ ó
õ V·dS = - ó
õ ó
õ

D
-sinf(3x²y²+3x²y²+z⁴) dqdf

= ó
õ ó
õ

D
6cos²qsin²qsin⁵f+cos⁴fsinf    dqdf

= ó
õ ó
õ

D
3
2
sin²(2q)sin⁵f  dqdf+ ó
õ ó
õ

D
cos⁴fsinf  dqdf,

where 0 £ q £ 2p, 0 £ f £ p for (q,f) in D. Resulting iterated integrals my be integrating using change of variable y=cosf, which yields

ó
õ p

0
sin⁵fdf = - ó
õ p

0
sin⁴fd(cosf)= ó
õ 1

-1
(1-y²)²dy= 16
15
,

ó
õ p

0
cos⁴fsinfdf = - ó
õ p

0
cos⁴fd(cosf) = ó
õ 1

-1
y⁴dy= 2
5
.

Similarly you can check that

ó
õ 2p

0
sin²(2q)=p.

Answer. 12p/5.

Problem 7.6-10. Let us choose outward direction of normal to define orientation on surface.

The side surface must be parametrized using cylindrical coordinates (cosu,sinu,v), 0 £ u £ 2p, 0 £ v £ 1 for (u,v) Î D. One can easily compute T_u×T_v=cos(u)i+sin(u)j. This normal looks outward, hence

ó
õ ó
õ

U
F·dS= ó
õ 1

0
dv ó
õ 2p

0
(cosu+sinu)du=0.

Hence this integral is 0.

Answer: 0.

Problem 7.6-10. One can easily modify spherical coordinates to parametrize upper half of ellipsoid U by

x=acosqsinf,y=bsinqsinf,z=ccosf,

where 0 £ q £ 2p, 0 £ f £ p/2.

This yields

T_q×T_f=-bcsin²fcosqi-acsin²fsinqj-absin²fcos²fk.

Therefore (taking orientation into account)

ó
õ ó
õ

U
F·dS = - ó
õ ó
õ

D
a³cos³qsin³f(-bc)sin²fcosqdqdf.

Using that

ó
õ 2p

0
cos⁴(q)dq = 3
4
p,

ó
õ p/2

0
sin⁵(f)= 8
15
p,

we get 2a³bcp/5. Answer: 2a³bcp/5.

Test 3 with solutions:

t/3a.ps, t/3a.pdf

t/3b.ps, t/3b.pdf

SOLUTIONS for 3b: t/3bsol.ps, t/3bsol.pdf (solutions for test 3a are very similar; do them as an excersize).

Recommended problems for test #4, Tuesday, April 6, 2004:

Section 8.1. 1-3, 5-10, 12-15, 30.

Section 8.2. 1-3, 4-7, 9-12, 15-21, 25.

Section 8.3. 1-7, 9-21.

Section 8.4. 1-10.

Test 4 with solutions:

t/4a.ps, t/4a.pdf

t/4b.ps, t/4b.pdf

SOLUTIONS for 4b: t/4bsol.ps, t/4bsol.pdf (solutions for test 4a are very similar).

Recommended problems (additinal to all above) for final exam:

Section 8.6. 1-11.

Recent results:

Last 5 dig of SN T1/70 T2/100 T3/50 T4/40

00010 66 93 49

00612 56 97 22 28

01404 63 83 39 32

01745 58 95 46 35

04033 70 85 note 32

04395 61 92 39 36

08752 66 46 24 24

12033 64 65 24 20

12300 59 73 16 29

13255 note note 36 27

13447 66 95 41 20

14380 68 92 48 30

16012 66 76 30 note

16586 39 92 35 26

16617 58 95 34 25

17176 66 97 47 36

18540 89 note note

21544 68 90 49 33

24360 46 note 18

24524 66 81 note 25

24701 42 73 14 note

24774 57 56 34 23

26059 64 97 50 39

26722 68 97 39 note

30590 50 83 48 34

30950 64 93 24 31

31361 note 58 17

33170 35 46 16 21

33750 note 55 note 18

36711 59 48 33 note

40190 66 85 36

41674 64 88 29 29

43036 53 73 31 27

43064 66 99 45 29

45743 65 92 41 30

46615 68 100 35 32

47567 56 58 33 26

48383 note 83 26 note

53103 68 90 41 35

53176 58 85 21 23

54339 60 97 31 31

54954 68 98 48 28

64196 46 97 41 29

66259 68 100 49 37

66585 note 53

67397 51 78 26 25

67754 60 100 50 36

71668 51 81 40 26

73102 44 73 22 note

73609 70 92 40 39

74747 68 86 37 32

77878 54 79 43 28

78350 55 97 29 32

80502 53 90 36 34

81451 59 84 46 27

84593 62 87 48 34

85040 60 88 note 30

89012 56 73 28 note

90020 70 90 27 38

91664 68 88 43 30

91705 66 45 15 32

93619 note 65 25 26

96071 64 90 44 36

96581 64 100 50 36

97261 68 82

99360 38 13 17 13

Footnotes:

¹There was a misprint here. Thanks to students who corrected it.

²21-26 for 4th edition

³1-17, 20-28 for 4th edition

⁴From now on problem numbers will be given for 5th edition of the textbook only

⁵In problem 21 you should find a vector in tangent plane, emanating from point (1,1,z(1,1)), such that its angle A with z-plane has tan(A)=0.03. There are two possible solutions. By the sketch the authors mean the projection of vector at given point and several level curves for the surface. Hint: the level surfaces are ellipses. If you have difficulties with handling positive parameters a, b, and c, try to sketch situation for c=10, a=1, b=4.

File translated from T_EX by T_TH, version 3.00.
On 20 Apr 2004, 16:43.

Last 5 dig of SN	T1/70	T2/100	T3/50	T4/40
00010	66	93	49
00612	56	97	22	28
01404	63	83	39	32
01745	58	95	46	35
04033	70	85	note	32
04395	61	92	39	36
08752	66	46	24	24
12033	64	65	24	20
12300	59	73	16	29
13255	note	note	36	27
13447	66	95	41	20
14380	68	92	48	30
16012	66	76	30	note
16586	39	92	35	26
16617	58	95	34	25
17176	66	97	47	36
18540		89	note	note
21544	68	90	49	33
24360	46	note	18
24524	66	81	note	25
24701	42	73	14	note
24774	57	56	34	23
26059	64	97	50	39
26722	68	97	39	note
30590	50	83	48	34
30950	64	93	24	31
31361	note	58	17
33170	35	46	16	21
33750	note	55	note	18
36711	59	48	33	note
40190	66	85	36
41674	64	88	29	29
43036	53	73	31	27
43064	66	99	45	29
45743	65	92	41	30
46615	68	100	35	32
47567	56	58	33	26
48383	note	83	26	note
53103	68	90	41	35
53176	58	85	21	23
54339	60	97	31	31
54954	68	98	48	28
64196	46	97	41	29
66259	68	100	49	37
66585	note	53
67397	51	78	26	25
67754	60	100	50	36
71668	51	81	40	26
73102	44	73	22	note
73609	70	92	40	39
74747	68	86	37	32
77878	54	79	43	28
78350	55	97	29	32
80502	53	90	36	34
81451	59	84	46	27
84593	62	87	48	34
85040	60	88	note	30
89012	56	73	28	note
90020	70	90	27	38
91664	68	88	43	30
91705	66	45	15	32
93619	note	65	25	26
96071	64	90	44	36
96581	64	100	50	36
97261	68	82
99360	38	13	17	13