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18.02 Multivariable Calculus
Fall 2007
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18.02 Practice Exam 4B – Solutions
Problem 1.
� �/2 � 1 � 1
0
0
�
�
r2 r dz dr d�.
�
�
0
Problem 2.
a) sphere: � = 2a cos π.
b) plane: � = a sec π.
� 2� � �/4 � 2a cos �
c)
�2 sin π d� dπ d�.
0
0
a sec �
Problem 3.
�
a) �y
(2xy + z 3 ) = 2x =
�
2
�z(x
+ 2yz) = 2y =
�
�
�
�
�
2
3
2
2
�x (x + 2yz);
�z (2xy + z ) = 3z = �x (y
�
2
2
ρ
�y (y + 3xz − 1); so F is conservative.
+ 3xz 2 − 1);
�
b) Method 1: f (x, y, z) = C1 +C2 +C3 Fρ · dρr;
�(x1 , y1 , z1 )
�
�
�
ρ r = x1 (2xy + z 3 ) dx = x1 0 dx = 0 (y = 0, z = 0)
C1 F · dρ
0
0
C1� C3
�
�
�
�
ρ · dρr = y1 (x2 + 2yz) dy = y1 x2 dy = x2 y1 (x = x1 , z = 0)
F
1
1
�C
C
0
0
2
� 2
�z1 2
�
2 − 1) dz = z1 (y 2 + 3x z 2 − 1) dz = y 2 z + x z 3 − z
ρ · dρr =
F
(y
+
3xz
(x
= x 1 , y = y1 )
1
1
1
1
1
1
1
C3
0
0
So f (x, y, z) = x2 y + y 2 z + xz 3 − z + c.
�f
3
�x = 2xy + z , so f (x, y, z)
�f
�g
�g
2
2
�y = x + �y = x + 2yz, so �y = 2yz.
Therefore g(y, z) = y 2 z + h(z), and
Method 2:
�f
�z
= x2 y + xz 3 + g(y, z).
f (x, y, z) = x2 y + xz 3 + y 2 z + h(z).
= 3xz 2 + y 2 + h� (z) = y 2 + 3xz 2 − 1, so h� (z) = −1.
Therefore h(z) = −z + c, and f (x, y, z) = x2 y + xz 3 + y 2 z − z + c.
Problem 4.
a) S is the graph of z = f (x, y) = 1 − x2 − y 2 , so n̂ dS = �−fx , −fy , 1� dA = �2x, 2y, 1� dA.
��
��
��
��
Therefore S Fρ · n̂ dS = S �x, y, 2(1 − z)� · �2x, 2y, 1� dA = S 2x2 + 2y 2 + 2(1 − z) dA = S 4x2 +
4y 2 dA (since z = 1 − x2 − y 2 ).
Shadow = unit disc x2 + y 2 � 1; switching to polar coordinates, we have
��
� 2� � 1 2
� 2� � 4 �1
ρ
r 0 d� = 2�.
S F · n̂ dS = 0
0 4r r dr d� = 0
ˆ Then
b) Let T =unit disc in the xy-plane, with normal vector pointing down (ˆ
n = −k).
��
��
��
ˆ dS =
Fρ · n
ˆ dS = T�x, y, 2� · (−k)
T −2 dS = −2 Area = −2�. By divergence theorem,
��T
���
��
��
ρ · n̂ dS =
ρ dV = 0, since div Fρ = 1 + 1 − 2 = 0. Therefore
F
div
F
=
−
S+T
D
S
T = +2�.
Problem 5.�
�
�
ˆ� k̂ ��
� ı̂
a) curl Fρ = �� θx θy θz �� = 2yı̂ − 2xˆ�.
� −2xz 0 y 2 �
ˆ so curl Fρ · n
b) On the��unit sphere, n̂ = xˆı + yˆ� + zk,
ˆ = �2y, −2x, 0� · �x, y, z� = 2xy − 2xy = 0;
ρ
therefore R curl F · n̂ dS = 0.
�
��
c) By Stokes, C Fρ · dρr = R curl Fρ · n̂ dS, where R is the region delimited by C on the unit sphere.
�
��
Using the result of b), we get C Fρ · dρr = R curl Fρ · n̂ dS = 0.