Name______________________________________________________________________________________________________
MAYNARD
MATH 277
TEST 5
VECTOR ANALYSIS
SAMPLE
1)
Find the gradient vector field for the
scalar function. (That is, find the
conservative vector field for the given
potential function.)
f(x, y , z)  z - ye x
2)
Determine whether the vector field F
is conservative. If it is, find a potential
function for the vector field.
F(x, y, z) 
3)
2
1
x
i - 2 j  2z  1 k
y
y
Find the curl of the curl of F ,
that is find   (  F ), for
F(x,y,z) = xyz i + y j  z k
4)
Find the divergence of the curl of F, that is,
find   ( x F) , for
F(x,y,z) = xyz i  y j  z k
5)
Parametrize the given path, C.
C: The counterclockwise traverse
around the triangle having
(0, 0), (1 ,0), and (0, 1) as vertices.
6)
Evaluate
 (x + 4y) ds
C
along the path C given by
C: The counterclockwise traverse
around the triangle having
(0, 0), (1 ,0), and (0, 1) as vertices.
7)
Parametrize the given path, C.
y
(0, 1)
(−1, 0)
8)
C: x2  y2 =1, y0
x
(1, 0)
Find the value of the line integral
 Fdr
, for
C
F(x,y) = y2 i  2xy j, and the path, C, shown in 7).
(Hint: If F is conservative, you may be able to
find an easier path.)
9)
Evaluate the line integral  (y − x) dx + (2x − y) dy
C
for
C: the boundary of the region lying inside the
rectangle bounded by x = −5, x = 5, y = −3, y = 3
and outside the square bounded by
x = −1, x = 1, y = −1, y = 1. Traverse C in the
counterclockwise direction, i.e., keep the region
on the left as C is traversed.
10)
Verify your result from problem 9 using
Green's Theorem by evaluating the double integral
for F(x,y) = (y − x) i  (2x − y) j
 (  F)k dA
R
and
R: region lying inside the rectangle bounded
by x = −5, x = 5, y = −3, y = 3
and outside the square bounded by
x = −1, x = 1, y = −1, y = 1.
11)
Use a line integral to calculate the work done
by the force F(x,y) = xy i + (x + y) j on a
particle that is moving counterclockwise around
the closed path, C, where
C: x2 + y2 = 4
12)
Use Green's Theorem to calculate with a
double integral the work done by the force
F(x,y) = xy i + (x + y) j on a particle
that is moving counterclockwise around the
closed path, C, where
C: x2 + y2 = 4
13)
Find a vector-valued function whose graph
is the specified surface (i.e., parametrize the
surface, S, in terms of two parameters, u and v.
Be sure to state the ranges of u and v.)
S: The part of the plane z = y lying in the 1st octant.
z
y
x
14)
Find a vector-valued function whose graph
is the specified surface (i.e., parametrize the
surface, S, in terms of two parameters, u and v.
Be sure to state the ranges of u and v.)
S: the cylinder x2  y2 = 16.
z
y
x
15)
 f (x, y, z) dS
Evaluate
for
S
f (x, y, z) = x2 + y2 + z2 and
S: z = x2 + y2 ,
x2 + y2  4
z
y
x
16)
Find the flux of F through the surface, S.
Let N in the surface integral  FNdS be the
S
upper unit normal vector to S and
z
F(x, y, z)= x i  y j  z k
S: z = 9 − x2 − y2 , z  0
y
x
17)
Find the flux of F through the closed surface, S.
Let N in the surface integral  FNdS be the
S
outward unit normal vector and
F(x, y, z) = (2x − y) i − (2y − z) j  z k
S: faces of the polyhedral solid bounded by the plane
2x + 4y + 2z = 12 and the z
three coordinate planes.
y
18)
x
Use the Divergence Theorem to verify your answer to
problem 17) by evaluating  FdV
Q
over the solid, Q, where
F(x, y, z) = (2x − y) i − (2y − z) j  z k
z
and
Q: polyhedral solid bounded by
the plane 2x + 4y + 2z = 12
and the three coordinate planes.
y
x
19)
Evaluate the surface integral
 ( F )N dS
for
S
F(x, y, z) = xyz i  y j + z k and
S: the plane 3x + 4y + 2z = 12,
in the 1st octant, oriented by N,
the upward pointing unit normal
vector to S.
z
y
x
20)
Verify your result from problem 19) using
Stoke's Theorem by evaluating the line
integral
with
F(x, y, z) = xyz i  y j + z k
and
C: The boundary of the plane
3x + 4y + 2z = 12,
in the 1st octant, and with C
having an orientation
consistent with that of S in
problem 19).
x
z
y
ANSWERS
12)
2
1)
 2xye x i  e x j  k
2)
Conservative: f(x, y, z) = x/y + z2 − z + K
3)
  (  F ) = zj + yk
C3:
r3 (t) = 0 i + (1 − t) j,
0t1
ds = || r3||dt = 1dt
x4
6)
4)   ( x F) = 0

C2
 tdt   1  t   4
1
0
y
x
14)
t

0
C3

19 1 
2dt   4 1  t dt 
6
0
1
2
z
r(u, v) = 4cos u i + 4sin u j + vk ,
C: r (t) = −cos t i + sin t j
8)
F is conservative. So use the straight line path
from (−1, 0) to (0, 1). Then  Fdr = 0.
0  u < 2π, ∞ < v < ∞

, 0tπ
7)
C
y
x
15)
z S: r(u,v) = (v cos u) i + (v sin u) j + v k 0  u  2π , 0  v  2 ;
S
y
C2
9)
C3
C1
x
C5
16)
C10
C12
C11
C9
C7
C8
C6
N
x
z
17)
y
3
N3 = − k
18)
z
S
6
 FdV = 18
C
Q
 Fdr +  Fdr +  Fdr +  Fdr +  Fdr +  Fdr +
C
1
C
2
C
7
C
8
C
3
C
4
C
5
C
10
C
11
C
12
= 25.5 − 30 + 60 − 30 + 34.5 + 12 − 2.5 + 2 − 4
+ 2 − 1.5 −12 = 56 (The hard way. Really ugly!)
R
10)
−2
−2
x 6
19)
z
6
N = (3i + 4j + 2k)/29
3
R
=Area large rectangle
− Area small square
= 60 − 4 = 56
(The easy way. Beautiful!)
2
3
 (  F)k dA = 1dA
R
11)
y
C
6
 Fdr +  Fdr +  Fdr +  Fdr +  Fdr +  Fdr
C
9
S1: r = 0i  uj  vk ;
0  u  3, 0  v  6  2u
S2: r = ui  0j  vk ;
0  u  6, 0  v  6  u
S3: r = vi  uj  0k ;
0  u  3, 0  v  6  2u
S4: r = vi  uj  (6  v  2u)k ;
0  u  3, 0  v  6  2u
 FNdS = 18
N1 = −i
N4 = (i + 2j + k)/6
N2 = − j
For F(x, y) = (y − x) i + (2x − y) j we can express
(y − x)dx + (2x − y)dy as  Fdr and then we have
C
NdS = (ru x rv )dudv
NdS = [(2v2 cos u) i + (2v2 sin u) j  vk]dudv
F∙ NdS = (v3  9v)dudv ;
 FNdS = 243π/2
S
6
x 6
(Other parameterizations are possible.)
S
r(u, v) = (vcos u)i + (vsin u)j + (9  v2)k ,
0  u  2π, 0  v  3
ru x rv = (2v2 cos u) i + (2v2 sin u) j  vk ;
y
C2: r(t) = 5(1 − 2t)i + 3j ;
C4: r(t) = 5(2t − 1)i − 3j ;
C6: r(t) = (5 − 4t)i ;
C8: r(t) = (1 − 2t)i − j ;
C10: r(t) = (2t − 1)i + j ;
C12: r(t) = (1 + 4t)i.
ru = (−v sin u) i + (v cos u) j + 0 k
rv = (cos u) i + (sin u) j + k
ru x rv = (v cos u) i + (v sin u) j −v k
dS = || ru x rv || du dv = v 2 du dv
f(x(u,v), y(u,v), z(u,v)) = v 2
 f (x, y, z) dS = 2v2 dudv = 32π/3
S
z
C4
For 0  t  1,
C1: r(t) = 5i + 3tj ;
C3: r(t) = −5i + 3(1 − 2t)j ;
C5: r(t) = 5i + 3(t − 1)j ;
C7: r(t) = i − tj ;
C9: r(t) = −i + (2t − 1)j ;
C11: r(t) = i + (1 − t)j ;
0  u < ∞, 0  v < ∞
r(u, v) = u i + v j + v k ,
yds   x  4 yds   x  4 yds
C1
1
z
13)
C1:
r1 (t) = t i + 0 j,
0t1
ds = || r1||dt = 1dt
x4
yds 
C
R
R
= (1  rcos )rddr = (r  r2cos )ddr = 4 π .
R
R
C2:
r2 (t) = (1 − t) i + t j,
0t1
ds = || r2||dt = 2dt
1
1
5)
Work =  (  F)k dA =  (1  x)dA
2
C: r(t) = 2cos t i + 2sin t j , 0  t < 2π
x(t) = 2cos t, y(t) = 2sin t
dr = r'∙dt = [(2sin t)i  (2cos t)j]dt
2 F(x(t), y(t)) = (4cos t sin t) i  (2cos t  2sin t) j
F  dr = [8cos t sin2t  4cos2t  4sin t cos t]dt
Work =  Fdr = 4π
C
S: r = ui  vj  [6  (3/2)u  2v]k ;
0  u  4, 0  v  3  (3/4)u
 ( F )N dS = 0
y
S
4
x
20)
z
6
C2 : r(t) = 3(1 − t) j + 6t k
 Fdr = 0
C
y
x
4
3
C1 : r(t) = 4(1 − t) i + 3t j
C2 : r(t) = 4t j + 6(1 − t) k