18.02 Practice Exam 3B
Problem 1. a) Draw a picture of the region of integration of
�
1
0
�
2x
dydx
x
b) Exchange the order of integration to express the integral in part (a) in terms of
integration in the order dxdy. Warning: your answer will have two pieces.
Problem 2. a) Find the mass M of the upper half of the annulus,
y
.
1 < x2 + y 2 < 9 (y � 0) with density � = 2
x + y2
b) Express the x-coordinate of the center of mass, x,
¯ as an iterated integral. (Write
explicitly the integrand and limits of integration.) Without evaluating the integral, explain
why x̄ = 0.
Problem 3. a) Show that F = (3x2 − 6y 2 )î + (−12xy + 4y)ĵ is conservative.
b) Find a potential function for F.
3
3
c) Let C be the curve x = 1 + y (1 − y) , 0 � y � 1. Calculate
�
F · dr.
C
Problem 4. a) Express the work done by the force field F = (5x + 3y)î + (1 + cos y)ĵ
on a particle moving counterclockwise once around the unit circle centered at the origin in
� b
the form
f (t)dt. (Do not evaluate the integral; don’t even simplify f (t).)
a
b) Evaluate the line integral using Green’s theorem.
Problem 5. Consider the rectangle R with vertices (0, 0), (1, 0), (1, 4) and (0, 4). The
boundary of R is the curve C, consisting of C1 , the segment from (0, 0) to (1, 0), C2 , the
segment from (1, 0) to (1, 4), C3 the segment from (1, 4) to (0, 4) and C4 the segment from
(0, 4) to (0, 0). Consider the vector field
F = (cos x sin y)î + (xy + sin x cos y)ĵ
a) Find the work of F along the boundary C oriented in a counterclockwise direction.
b) Is the total work along C1 , C2 and C3 , more than, less than or equal to the work along
C?
Problem 6. Find the volume of the region enclosed by the plane z = 4 and the surface
z = (2x − y)2 + (x + y − 1)2 .
Suggestion: change of variables.
18.02 Practice Exam 3B — Solutions
� (1, 2)
��
�
�
��
�
x=1
��
y=2x �
1. a)
��
�
�
(1, 1)
��
�
� ��
y=x
�
�
�
b)
�
1
0
�
y
dxdy +
y/2
�
2
1
�
1
dxdy.
y/2
�
r sin �
rdrd� = sin �drd�.
r2
� �
� � � 3
�
M=
�dA =
sin � drd� =
2. a) �dA =
R
b) x
¯
=
1
M
� �
0
x�dA =
R
1
4
�
1
�
0
�
�
0
�
�
2 sin �d� = − 2 cos � 0 = 4.
3
r cos � sin �drd�
1
The reason why one knows that x
¯
= 0 without computation is that x is odd with respect
to the y-axis whereas the region and the density are symmetric with respect to the y-axis:
(x, y) � (−x, y) preserves the half annulus and �(x, y) = �(−x, y).
3. a) Nx = −12y = My , hence F is conservative.
b) fx = 3x2 − 6y 2 → f = x3 − 6y 2 x + c(y) → fy = −12xy + c� (y) = −12xy + 4y. So
c (y) = 4y, thus c(y) = 2y 2 (+ constant). In conclusion
�
f = x3 − 6xy 2 + 2y 2
(+ constant).
c) The curve C starts at (1, 0) and ends at (1, 1), therefore
�
F · dr = f (1, 1) − f (1, 0) = (1 − 6 + 2) − 1 = −4.
C
4. a) The parametrization of the circle C is x = cos t, y = sin t, for 0 � t < 2�; then
dx = − sin tdt, dy = cos tdt and
� 2�
W =
(5 cos t + 3 sin t)(− sin t)dt + (1 + cos(sin t)) cos tdt.
0
b) Let R be the unit disc inside C;
�
� �
� �
F · dr =
(Nx − My )dA =
(0 − 3)dA = −3 area(R) = −3�.
C
R
R
(0, 4) •�
5. a)
C3
•� (1, 4)
C2
C4
(0, 0) •�
C1
•� (1, 0)
� �
F · dr =
(Nx − My )dxdy
R
C� �
� �
ydxdy
=
(y + cos x cos y − cos x cos y)dxdy =
R
� 4 �R 1
� 4
=
ydxdy =
ydy = [y 2 /2]40 = 8.
�
0
0
0
b) On C4 , F = sin y i, whereas dr is parallel to j. Hence F · dr = 0. Therefore the work
of F along C4 equals 0. Thus
�
�
�
�
F · dr =
F · dr −
F · dr =
F · dr ;
C1 +C2 +C3
C
C4
C
and the total work along C1 + C2 + C3 is equal to the work along C.
�
� u
6. Let u = 2x − y and v = x + y − 1. The Jacobian �� x
vx
1
Hence dudv = 3dxdy and dxdy = dudv, so that
3
V
=
� �
� �
�
uy �� �� 2 −1 ��
=
= 3.
vy � � 1 1 �
4 − (2x − y)2 − (x + y − 1)2 dxdy
(2x−y)2 +(x+y−1)2 <4
1
(4 − u2 − v 2 ) dudv
3
�2
� 2� �
� 2� � 2
2 2
1
1
r − r4
(4 − r2 ) rdrd� =
=
3
3
12
0
0
0
0
� 2�
4
8�
=
d� =
.
3
3
0
=
� �
u2 +v 2 <4