18.02 Exam 4 – Solutions
Problem 1.
� π/2 � 1 � 1
0
0
�
�
r2 r dz dr dθ.
0
Problem 2.
a) sphere: ρ = 2a cos φ.
b) plane: ρ = a sec φ.
� 2π � π/4 � 2a cos φ
c)
ρ2 sin φ dρ dφ dθ.
0
0
a sec φ
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
�
∂y (y + 3xz − 1); so F is conservative.
+ 3xz 2 − 1);
�
b) Method 1: f (x, y, z) = C1 +C2 +C3 F� · d�r;
�(x1 , y1 , z1 )
�
�
�
� r = x1 (2xy + z 3 ) dx = x1 0 dx = 0 (y = 0, z = 0)
C1 F · d�
0
0
C1� C3
�
�
�
�
� · d�r = 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
� · d�r =
F
(y
+
3xz
(x
= x1 , 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 F� · 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
��
� 2π � 1 2
� 2π � 4 �1
�
r 0 dθ = 2π.
S F · n̂ dS = 0
0 4r r dr dθ = 0
ˆ Then
b) Let T = unit disc in the xy-plane, with normal vector pointing down (ˆ
n = −k).
��
��
��
ˆ dS =
F� · n
ˆ dS = T �x, y, 2� · (−k)
T −2 dS = −2 Area = −2π. By divergence theorem,
���
��
��
��T
� ·n
� dV = 0, since div F� = 1 + 1 − 2 = 0. Therefore
F
ˆ
dS
=
div
F
=
−
S+T
D
S
T = +2π.
Problem 5.
�
�
�
ĵ k̂ ��
� ı̂
a) curl F� = �� ∂x ∂y ∂z �� = 2yı̂ − 2xĵ.
� −2xz 0 y 2 �
ˆ so curl F� · n
b) On the��unit sphere, n̂ = xˆı + yˆj + zk,
ˆ = �2y, −2x, 0� · �x, y, z� = 2xy − 2xy = 0;
�
therefore R curl F ·n̂ dS = 0.
�
��
c) By Stokes, C F� · d�r = R curl F� · n̂ dS, where R is the region delimited by C on the unit sphere.
�
��
Using the result of b), we get C F� ·d�r = R curl F� · n̂ dS = 0.
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18.02SC Multivariable Calculus
Fall 2010
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