18.02 Practice Exam 4 A – Solutions
Problem 1.
a)
My = ex z = Nx
MzExam
= ex y4A
=P
18.02 Practice
- x
Solutions
x
NZ = e + 2y = Py
b) We begin with
fx = ex yz
fy = ex z + 2yz
fz = ex y + y 2 + 1
lb) We begin with
Integrating fx we get f = ex yz + g(y, z).
and comparing with the above equations
fzDifferentiating
= e5yz
we get
fy = exz + 2yz�
�
fy = exf,z =
+ e5y
gy
y2 1gy = 2yz
→
x
+ gz + g(y, z).gzDifferentiating
= y2 + 1
Integrating f, we fget
and comparing
z =f e=y e"yz
+ +
with the above equations we get
Integrating gy we get g = y 2 z + h(z). Then gz = y 2 + h� (z) so comparing with the second equation
we get h� (z) = 1. Hence h = z + C. Putting everything together we get
x
+
2
+
= y2z
e yz +
y z Then
+ z +g ,C.= y2 hl(z) so comparing
Integrating g, we get gf =
h(z).
with the second equation above we get hl(z) = 1. Hence h = z+C. Putting
1c) Nz = 0 and Peverything
the field
is not conservative.
y = 1 hence
together
we get
Problem 2.
lc) figure
N, = 0 and P,
a) Consider the
= 1 hence the field is not conservative
2a) Consider the figure
n=
1
2 �x, y, z�
n' = $(x, y, z) hence
hence
�x, y, z�
z2
=
2 get 2
z = 2 cos r+6 and d S = 2' sin r+6 d 4 dB hence we
2
z = 2 cos φ and dS = 2 sin φ dφ dθ hence we get
6"
F · n = �y, −x, z� ·
�
0
2π
4 sin 4 d 4 dB = 1671 �
5cos2
4 sin r$d4 = - 1 6 ~ �
3 � 5π
π
√
6
4 cos2 φ
cos φ 6
4 sin φ dφ dθ = 16π
cos2 φ sin φ dφ = 16π
= 4 3 π
π
π
2
3
π
2b)
6 n' = *(x, y,O) hence 2 . n' = 0. So
6 the flux is 0.
6
�
5π
6
/L /K
2c) diu@ = 1 hence
/k
b) n = ±�x, y, 0� hence F · n = 0. So the flux is 0.
=
1d v =
diu~d=
v
@..ds+//
$.ads = 4 h ~
c) divF =,(R)
1 hence
Cylinder
� � �
� � �
� �
� �
√
Y2 + z2 =
1. = 4 3 π
V ol(R) =3a) C is given
1 dVby=the equations
divFx2
dV+=
F 2· nand
dS z+= 1. So x2 +Fy2· n=dS
Parametrization:
R
x = cost
dx = - s i n t d t
S
y = sin t
dy = cost dt
Cylinder
z =1
dz = 0
Problem 3.
a) C is given by the equations x2 + y 2 + z 2 = 2 and z = 1. So x2 + y 2 = 1.
Parametrization:
x = cost
y = sin t
z=1
dx = − sin t dt
So
dy = cos t dt
dz = 0
2π
�
I=
(− cos t sin t + sin t cos t) dt = 0.
0
b)
�
�
� ı̂
k̂ ��
ĵ
�
�
�
� × F =
� ∂x ∂y ∂z � = ı̂ + xĵ
�
�
�
xz y y �
c) By Stokes’ theorem
�
� �
F · dr =
(� × F) · n dS
C
S
n is the normal pointing upward hence
�
� �
� �
�x, y, z�
x + xy
√
F · dr =
�1, x, 0� · √
dS =
dS
2
2
C
S
S
Problem 4. divF = 0 hence
� �
� � �
F · n dS =
divF dV = 0
S
Problem 5.
a) z = (x2 + y 2 + z 2 )2 ≥ 0
b) z = ρ cos φ and x2 + y 2 + z 2 = ρ2 hence ρ cos φ = ρ4 . Canceling ρ we get cos φ = ρ3 .
c)
� 2π �
π � (cos φ)1/3
2
ρ2 sin φ dρ dφ dθ
0
0
0
Problem 6. The flux is upward so and z = f (x, y) = xy, hence
n dS = +�−fx , −fy , 1� dx dy = �−y, −x, 1� dx dy
So,
� �
� �
� �
F · n dS =
�y, x, z� · �−y, −x, 1� =
x2 +y 2 <1
S
x2 +y 2 <1
where we substituted z = xy. Using polar coordinates we get
� 2π � 1
(−r2 + r2 cos θ sin θ) r dr dθ
0
Inner:
�1
Outer:
� 2π
0
(−r2
0
+
r2 cos θ sin θ) r dr
1
4 (cos θ sin θ
− 1) dθ =
0
= 14 (cos θ sin θ
�2π
�
1 sin2 θ
−
θ
4
2
0
− 1)
= − π2
(−y 2 − x2 + xy) dx dy
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18.02SC Multivariable Calculus
Fall 2010
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