18.02
Final Exam
No books, notes or calculators.
15 problems, 250 points.
Useful formula:
cos2 () = 1 (1 + cos(2)
2
Problem 1. (20 points)
a) (15 pts.) Find the equation in the form Ax + By + Cz = D o
Problems: Simply Connected Regions
1. Which of the regions shown below are simply connected?
(a)
(b)
(c)
Answer: Region (a) is not simply connected the puncture at the center of the disk
would prevent
Problems: Work and Line Integrals
Z
1. Evaluate I =
y dx + (x + 2y) dy where C is the curve shown.
C
y
1
C1
(1, 1)
C2
1
x
Figure 1: Curve C is C1 followed by C2 .
Answer: The curve C is made up of two
7.1IntegrationbyParts
(page287)
CHAPTER 7 TECHNIQUES OF INTEGRATION
7.1
Integration by Parts
(page 287)
Integration by parts aims to exchange a difficult problem for a possibly longer but probably eas
18.02 Problem Set 7, Part II Solutions
1.(a)
(b)
V
=
=
=
Z
Z
Z
4
0
4
0
4
Z
4 x
p
4
xdy dx
0
p
y 4
(4
x
y=4
x
y=0
dx
2
(4
5
x)3/2 dx =
0
4
x)5/2
0
2
64
= 45/2 = .
5
5
2. (a) For simplicity let us assu
Identifying Gradient Fields and Exact Dierentials
1. Determine whether each of the vector elds below is conservative.
a) F = hxex + y, xi
b) F = hxex + y, x + 2i
c) F = hxex + y + x, xi
Answer: We kno
Identifying Potential Functions
1. Show F = h3x2 + 6xy, 3x2 + 6yi is conservative and nd the potential function f such
that F = rf .
Answer: First, My = 6x = Nx . Since F is dened for all (x, y), F is
Identifying Gradient Fields and Exact Dierentials
y xE
,
.
r2 r2
Answer: We know that if F = hM, N i then curlF = Nx My . In this case, M = ry2
and N = rx . Applying the chain rule and dierentiating r
18 02 Practice Final 3hrs.
Problem 1 Given the points P : (1, 1, 1), Q : (1, 2, 0), R : ( 2, 2, 2) nd
b) a plane ax + by + cz = 0 trough P, Q and R
d 1
0
0
1
0 1
1
1 0 c
x
0
Problem 2 Let A = @2 c 1
Problems: Extended Greens Theorem
x dy
exact? If so, nd a potential function.
y2
1
x
Answer: M = and N =
are continuously dierentiable whenever
y
y2
y = 0, i.e. in the two half-planes R1 and R2 both s
Extended Greens Theorem
y
yi + xj
, dened on the punctured plane
r2
D = R2 (0, 0). Its easy to compute (weve done it before) that curlF = 0 in D.
Question:
For the tangential eld F, what do you think
18.01 Exam 3
Problem 1. (20 pts) Evaluate the following integrals
xdx
(1 + x 2 ) 2
a)
2
b)
/2
0
/ 2
sin 6 x cos xdx
Problem 2. (20 pts.) Find the following approximations to
/2
0
cos xdx
(Do not
18.02 Exam 4
Problem 1. (10 points)
Let C be the portion of the cylinder x2 +y 2 1 lying in the rst octant (x 0, y 0, z 0) and
below the plane z = 1. Set up a triple integral in cylindrical coordinate
18.02 Problem Set 9, Part II Solutions
1. (a) If C is a simple closed curve enclosing the region R then
I
Z Z
F dr =
curl F dx dy
C
Z ZR
=
(6x x3 )x (y 3 6y)y dx dy
Z ZR
=
(6 3x2 + 6 3y 2 ) dx dy
Z ZR
18.02 Practice Exam 4A
Problem 1. (15)
a) Show that the vector eld
is conservative.
F = hex yz, ex z + 2yz, ex y + y 2 + 1i
b) By a systematic method, nd a potential for F.
c) Show that the vector eld
18.02 Problem Set 8, Part II Solutions
Problem 1
a) F = (1 x
y)j. (See diagram (a) below.)
y
y
y
(x, 1
C2
x)
C1
F
x
C3
C2
C3
C1
x
x
diagram b(i)
diagram b(ii)
diagram a
b) The two possibilities for C
18.01 Exam 2
Problem 1(20 points) Compute the following derivatives. Show all work, or you will not receive
credit.
(a)(10 points)
d
d
2 tan()
1 (tan()2
Solution to (a) The simplest solution uses the