18.02: Practice Exam 3A
1. Let (¯
x, y)
¯ be the center of mass of the triangle, with vertices at (−2, 0),
(0, 1), (2, 0) and uniform density � = 1.
a) Write an integral formula for ȳ. Do not evaluate the integral(s), but write
explicitly the integrand and limits of integration.
b) Find x̄.
2. Find the polar moment of inertia of the unit disk with density equal to
the distance from the y-axis.
3. Let F = (ax2 y + y 3 + 1)i + (2x3 + bxy 2 + 2)j be a vector field, where a
and b are constants.
a) Find the values of a and b for which F is conservative.
b) For these values of a and b, find f (x, y) such that F = �f .
�
c) Still using the values of a and b from part (a), compute
F · dr along
C
the curve C such that x = et cos t, y = et sin t, 0 � t � �.
�
4. For F = yx3 i + y 2 j find C F · dr on the portion of the curve y = x2 from
(0, 0) to (1, 1).
5. Consider the region R in the first quadrant bounded by the curves y = x 2 ,
y = x2 /5, xy = 2, and xy = 4.
a) Compute dxdy in terms of dudv if u = x 2 /y and v = xy.
b) Find a double integral for the area of R in uv coordinates and evaluate
it.
6. a) Let C be a simple closed curve
going counterclockwise around a region
�
R. Let M = M (x, y). Express
M dx as a double integral over R.
C
�
b) Find M so that
M dx is the mass of R with density �(x, y) = (x + y) 2 .
C
7. Consider the region R enclosed by the x-axis, x = 1 and y = x 3 .
Travelling in a counterclockwise direction along the boundary C or R, call
C1 the portion of C that goes from (0, 0) to (0, 1), C 2 the portion that goes
from (1, 0) to (1, 1) and C3 the portion that goes from (1, 1) to (0, 0).
a) Find the total work of F = (1 + y 2 )i around the boundary C of R, in a
counterclockwise direction.
b) Calculate the work of F along C1 and C2 .
c) Use parts (a) and (b) to find the work along the third side C 3 .
1
2
18.02 Practice Exam 3A Solutions
1. a) Area of triangle is base times height = 2, so ȳ =
1 � 2−2y
1
2
�
�
�/2
0
y dxdy
2y−2
b) By symmetry x̄ = 0
2. � = |x| = r| cos �|. I0 =
�
2�
0
�
� �
1
2
r |r cos �|rdrd� = 4
0
�
r 2 �rdrd� =
D
�/2
0
�
1
4
r cos �drd� = 4
0
0
1
4
cos �d� =
5
5
3. a) Nx = 6x2 + by 2 , My = ax2 + 3y 2
. Nx = My provided a = 6 and b = 3.
b) fx = 6x2 y + y 3
+ 1 =� f = 2x3 y + xy 3 + x + c(y). Therefore,
fy = 2x3 +3xy 2 +c� (y). Setting this equal to N , we have 2x 3 +3xy 2 +c� (y) =
2x3 + 3xy 2 + 2 so c� (y) = 2 and c = 2y (+constant). In all,
f = 2x3 y + xy 3
+ x + 2y
c) C starts at (1, 0) and ends at (−e� , 0), so
(+constant)
�
F · dr = f (−e� , 0) − f (1, 0) =
C
−e� − 1.
�
� 1
� 1
4.
yx3 dx + y 2 dy =
x2 x3 dx + (x2 )2 (2xdx) =
3x5 dx = 1/2
C
0
0
� �
�
�
� ux uy � � 2x/y −x2 /y 2 �
�=�
� = 3x2 /y. Therefore,
5. a) ��
�
vx vy � � y
x
dudv = (3x2 /y)dxdy = 3u dxdy
=�
dxdy =
1
dudv
3u
� 4
1
1
2
b)
dudv =
ln 5 dv = ln 5
3
2
1 3u
2 3
�
� �
6. a)
M dx =
−My dA
�
4� 5
C
R
1
b) We want M such that −My = (x + y)2 . Use M = − (x + y)3
3
��
7. a) For F, My = 2y and Nx = 0, hence C F · dr =
R −2y dA =
� 1
6
� 1 � x3
1
0 0 −2y dydx = 0 −x dx = − 7 .
b) For the work through C1 , we have F · i = 1 + y 2 = 1 + 0 = 1. The length
of C1 is 1, so the total work through C1 is 1.
The work through C2 is zero because F · j = 0.
�
�
�
�
1
8
c)
F · dr =
F · dr −
F · dr −
F · dr = − − 1 − 0 = −
7
7
C2
C3
C
C1