1
MATH 332, (Vector Analysis), Spring 2000: Test 2
Instructor: Ivan Avramidi
Vector Analysis
MATH 332
Test 2
Set 1
1. Let ϕ = z 2 x3 y and F = grad ϕ. Find div F and curl F .
2. Let F = x2 i + y 2 j + k . Find: a) the general equation of a flow line, b) the flow line through the point
(1, 1, 2).
A×R
.
R
Find: a) div F and b) curl F . (Hint: Use of tensor notation will simplify your work significantly, especially
when computing the curl F ).
3. Let A = a1 i + a2 j + a3 k be a constant vector field, R = x i + y j + z k , R = | R | and F =
4. Let F = (3x + 4y) i + (2x + 3y 2 ) j + (exyz ) k and C be the counterclockwise oriented circle of I
radius 2 in
2
2
the xy plane with the center at the origin. The equations of the circle are: x + y = 4, z = 0. Find
F ·dR
around C. (Hint: Parametrize the circle by R (t) = 2 cos t i + 2 sin t j , 0 ≤ t ≤ 2π.)
C
5. Let F = [y + z cos(xz)]
i + x j + x cos(xz) k . a) Is F conservative? If yes, b) find the scalar potential. c)
R
Find the line integral C F · d R along the line segment from (0, 0, 0) to (1, 1, π). (Hint: F is conservative!)
2
MATH 332, (Vector Analysis), Spring 2000: Test 2
Set 1: Solutions
1. Let ϕ = z 2 x3 y and F = grad ϕ. Find div F and curl F .
Solution:
F = ∂x ϕ i + ∂y ϕ j + ∂z ϕ k = 3x2 z 2 y i + x3 z 2 j + 2zx3 y k
div F = div grad ϕ = ∆ϕ = (∂x2 + ∂y2 + ∂z2 )ϕ = 3 · 2xz 2 y + 0 + 2x3 y = 6xyz 2 + 2x3 y
curl F = curl grad ϕ = 0
2. Let F = x2 i + y 2 j + k . Find: a) the general equation of a flow line, b) the flow line through the point
(1, 1, 2).
Solution: a) The equation of the flow line is:
dx
dy
dz
= 2 =
x2
y
1
Integrating the first equation, we get
−
1
1
+ C1 = −
x
y
or
y=
x
1 − C1 x
By integrating the second equation, we obtain
−
1
+ C2 = z
x
By choosing a parameter t, we can rewrite the general equation of the flow line in the form
x = t,
y=
t
,
1 − C1 t
1
z = − + C2
t
b) To find the equation of the flow line that passes through (1, 1, 2), we equate
1 = t,
1=
t
,
1 − C1 t
1
2 = − + C2
t
Solving these equations gives
C1 = 0,
C2 = 3.
So, the equation of the flow line that passes through (1, 1, 2) reads
x = t,
y = t,
1
z =− +3
t
A×R
.
R
Find: a) div F and b) curl F . (Hint: Use of tensor notation will simplify your work significantly, especially
when computing the curl F ).
3. Let A = a1 i + a2 j + a3 k be a constant vector field, R = x i + y j + z k , R = | R | and F =
Solution: a) First of all, we use the tensor notation
Fi = εijk aj
xk
R
3
MATH 332, (Vector Analysis), Spring 2000: Test 2
and we note that
p
√
x2 + y 2 + z 2 = xm xm
R=
Next,
x xk k
= εijk aj ∂i
div F = ∂i Fi = ∂i εijk aj
R
R
We compute the derivative
∂i
x k
R
=
(∂i xk )R − xk (∂i R)
R2
Noting that
∂i xk = δik
and
we obtain
2xm δim
xi
2xm (∂i xm )
=
=
2R
2R
R
∂i R =
δik R − xkRxi
δik
xi xk
=
− 3
R
R2
R
R
Now, we can use plug this in the formula for div F to get
xi xk
δik
div F = ∂i Fi = εijk aj
− 3
= 0
R
R
∂i
x k
=
since
εijk δik = 0,
εijk xi xk = 0.
(because of the antisymmetry of εijk ).
b) Next, we compute the curl F :
x xn n
( curl F )i = εijk ∂j Fk = εijk ∂j εkmn am
= εijk εkmn am ∂j
R
R
By using the formula for ∂j (xn /R) computed above, we obtain
xj xn
δjn
xj xn
δjn
( curl F )i = εijk εkmn am
−
=
(δ
δ
−
δ
δ
)a
−
im
jn
in
jm
m
R
R3
R
R3
δjn
xj xn
1
1
= (ai δjn − δin aj )
−
= (ai δjj − δij aj ) − 3 (ai δjn − δin aj )xj xn
3
R
R
R
R
=
1
1
2ai
1
ai
aj xj xi
(3ai − ai ) − 3 (ai xj xj − aj xj xi ) =
− 3 (ai R2 − aj xj xi ) =
+
R
R
R
R
R
R3
Thus,
curl F =
A
(A · R)R
+
R
R3
4. Let F = (3x + 4y) i + (2x + 3y 2 ) j + (exyz ) k and C be the counterclockwise oriented circle of I
radius 2 in
2
2
the xy plane with the center at the origin. The equations of the circle are: x + y = 4, z = 0. Find
F ·dR
around C. (Hint: Parametrize the circle by R (t) = 2 cos t i + 2 sin t j , 0 ≤ t ≤ 2π.)
Solution: We compute
I
C
=
Z
0
F · dR =
Z
0
2π
F·
dR
dt
dt
2π
(3 · 2 cos t + 4 · 2 sin t)2(− sin t) + (2 · 2 cos t + 3 · 4 sin2 t)2 cos t + 0 dt
C
4
MATH 332, (Vector Analysis), Spring 2000: Test 2
2π
Z
=
−12 cos t sin t + 8 cos2 t − 16 sin2 t + 24 sin2 t cos t dt
0
=
Z
2π
0
1 + cos(2t)
1 − cos(2t)
sin(2t)
2
+8
− 16
+ 24 sin t cos t dt
−12
2
2
2
Z 2π
=
−6 sin(2t) + 12 cos(2t) − 4 + 24 sin2 t cos t dt
0
=
−6
sin(2t)
sin3 t
− cos(2t)
+ 12
− 4t + 24
2
2
3
2π
= −8π
0
5. Let F = [y + z cos(xz)]
i + x j + x cos(xz) k . a) Is F conservative? If yes, b) find the scalar potential. c)
R
Find the line integral C F · d R along the line segment from (0, 0, 0) to (1, 1, π). (Hint: F is conservative!)
Solution: a) F is defined everywhere, so the domain is the
i
∂x
curl F = y + z cos(xz)
whole space, which is simply conected. We check
j
k
∂y
∂z
x x cos(xz) = i (0 − 0) − j [cos(xz) − xz sin(xz) − cos(xz) + xz sin(xz)] + k (1 − 1) = 0
Since the domain of F is simply connected and curl F = 0, the vector field F is conservative , i.e. F = ∇ϕ.
b) To find the calar potential we need to solve the equations
∂x ϕ = y + z cos(xz),
∂y ϕ = x,
∂z ϕ = x cos(xz)
From the first equation we have
ϕ(x, y, z) =
Z
(y + z cos(xz))dx = yx + sin(xz) + f (y, z)
Substituting into the second equation, we get
x + ∂y f = x,
∂y f = 0,
and, therefore,
f (y, z) = h(z),
and
ϕ(x, y, z) = xy + sin(xz) + h(z)
Finally, plugging this into the third equation, we get
x cos(xz) + h0 (z) = x cos(xz),
h0 (z) = 0,
and, therefore,
h(z) = C1 ,
where C1 is a constant (that can be chosen to be zero C1 = 0). Thus, the calar potantial is
ϕ(x, y, ) = xy + sin(xz) + C1 .
c) Since F is conservative, the line integral does not depend on the curve C and is equal to
Z
F · d R = ϕ(1, 1, π) − ϕ(0, 0, 0) = 1 + sin(π) − 0 − sin 0 = 1
C
5
MATH 332, (Vector Analysis), Spring 2000: Test 2
Set 2
1
1. Let ϕ = p
x2
y2
z2
. Find: a) gradϕ, b) the directional derivative
+ +
(− i + 2 j + 2 k ), c) the direction of maximal decrease at (1, 0, 0).
dϕ
at (1, 0, 0) in the direction of
ds
x2
i + y j + k . Find: a) the general equation of a flow line, b) the parametrized equation for the
y
flow line that passes through the point (1, 1, 0). (Hint: Separate variables in the differential equation for the
flow line).
2. Let F =
3. Let A = A1 i + A2 j + A3 k be a constant vector field, R = x i + y j + z k , R = | R | =
using the tensor methods show that for R 6= 0
1
1
+ curl A × grad
= 0.
grad A · grad
R
R
p
x2 + y 2 + z 2 . By
Hints: a) Write this equation in tensor notation, b) Use the tensor identities to transform the product of ε’s into
1
δ’s, c) leave the calculation of the second derivatives of
to the very end, d) show that the first term (coming
R
1
from grad) cancels exactly another term (coming from curl), e) prove the remaining equation ∆ = 0, ∆ being
R
Laplace operator.
4. Let F = (y +yz cos(xyz)] i +[x2 +xz cos(xyz)] j +[z +xy cos(xyz)] k and C be the counterclockwise oriented
ellipse in I
the plane z = 1 parallel to the xy plane parametrized by x = 2 cos t, y = 3 sin t, z = 1, 0 ≤ t ≤ 2π.
Evaluate
F · d R around C.
C
5. Let F = [(1 + x)ex+y ] i + [xex+y + 2y] j − 2z k . a) Use the zero curl test to determine whether F is
irrotational.
b) Is F conservative? c) If F is conservative, find the scalar potential, d) Evaluate the line
R
integral C F · d R along the semicircle x = cos t, y = sin t, z = 1, 0 ≤ t ≤ π.
6. (Extra Credit.) With the same notation as in the problem 3, determine whether or not the vector field
A×R
F =
is solenoidal. If so, find the vector potential G.
R2
6
MATH 332, (Vector Analysis), Spring 2000: Test 2
Set 2: Solutions
1
1. Let ϕ = p
x2
y2
z2
. Find: a) gradϕ, b) the directional derivative
+ +
(− i + 2 j + 2 k ), c) the direction of maximal decrease at (1, 0, 0).
Solution.
a) Let R = x i + y j + z k and R = | R | =
∇ϕ = ∇
dϕ
at (1, 0, 0) in the direction of
ds
p
x2 + y 2 + z 2 . Then
1
1
= − 2 ∇R .
R
R
Since
∇R =
R
,
R
we obtain
∇ϕ = −
R
.
R3
b) The unit vector in the direction of (− i + 2 j + 2 k ) is
u=
−i + 2j + 2k
1
√
= (− i + 2 j + 2 k ) .
3
1+4+4
At the point (1, 0, 0) we get
∇ϕ
Therefore,
(1,0,0)
= −i .
1
dϕ
1
= u · ∇ϕ
= (− i + 2 j + 2 k ) · (− i ) =
.
ds
3
3
(1,0,0)
c) The direction of maximal decrease at the point (1, 0, 0) is
−∇ϕ
= i .
(1,0,0)
x2
i + y j + k . Find: a) the general equation of a flow line, b) the parametrized equation for the
y
flow line that passes through the point (1, 1, 0). (Hint: Separate variables in the differential equation for the
flow line).
2. Let F =
Solution.
a) The equations of the flow line are
dx
x2
y
From the equation
=
dy
dz
=
.
y
1
dy
= dz we have
y
ln y = z + C1 ,
y = C2 ez ,
where C2 = eC1 . By separating the variables in the equation
y dx
dy
=
x2
y
7
MATH 332, (Vector Analysis), Spring 2000: Test 2
we obtain
dx
dy
= 2.
2
x
y
By integrating this equation we get
−
1
1
= − + C3 ,
x
y
which can be simplified to
x=
y
.
1 − C3 y
By choosing the parameter z = t we get the parametrized equation of the flow line
x=
C2 et
,
1 + C4 et
y = C2 et
z = t,
where C4 = −C2 C3 is another constant of integration.
b) First we determine the value of the parameter t at the point (1, 1, 0); we find t = 0. By substituting the
initial values we have
C2
1=
,
1 = C2 ,
1 + C4
which means that C2 = 1 and C4 = 0. Thus the flow line through (1, 1, 0) is
x = et ,
y = et ,
z = t.
3. Let A = A1 i + A2 j + A3 k be a constant vector field, R = x i + y j + z k , R = | R | =
using the tensor methods show that for R 6= 0
1
1
+ curl A × grad
= 0.
grad A · grad
R
R
p
x2 + y 2 + z 2 . By
Hints: a) Write this equation in tensor notation, b) Use the tensor identities to transform the product of ε’s into
1
δ’s, c) leave the calculation of the second derivatives of
to the very end, d) show that the first term (coming
R
1
from grad) cancels exactly another term (coming from curl), e) prove the remaining equation ∆ = 0, ∆ being
R
Laplace operator.
Solution.
a) In tensor notation the equation has the form
1
1
1
1
∂i Aj ∂j
+ εijk ∂j εkmn Am ∂n
= Aj ∂j ∂i + Am εijk εkmn ∂j ∂n
R
R
R
R
b)-d) We simplify it as follows
= Aj ∂j ∂i
1
1
1
1
1
1
i j
j
+ Am (δm
δn − δni δm
)∂j ∂n = Aj ∂j ∂i + Ai ∂j ∂j − Am ∂m ∂i = Ai ∆
R
R
R
R
R
R
e) Further we compute
∆
1
1
1
1 xi
= ∂i ∂i = ∂i − 2 ∂i R = −∂i
R
R
R
R2 R
hx i
∂i xi
1
i
= −∂i
= − 3 − (−3) 4 xi ∂i R
R3
R
R
8
MATH 332, (Vector Analysis), Spring 2000: Test 2
Since ∂i xi = 3, ∂i R =
xi
, and xi xi = R2 , we obtain finally
R
∆
3
1 xi xi
3
R2
1
=− 3 +3 4
= − 3 + 3 5 = 0,
R
R
R R
R
R
which completes the proof.
4. Let F = (y +yz cos(xyz)] i +[x2 +xz cos(xyz)] j +[z +xy cos(xyz)] k and C be the counterclockwise oriented
ellipse in I
the plane z = 1 parallel to the xy plane parametrized by x = 2 cos t, y = 3 sin t, z = 1, 0 ≤ t ≤ 2π.
Evaluate
F · d R around C.
C
Solution.
First of all, we notice that although F is not conservative, it has a conservative part, i.e.
F = y i + x2 j + z k + ∇ϕ ,
where
ϕ = sin(xyz). The conservative part does not contribute to a circulation about a closed curve, i.e.
I
∇ϕ · d R = 0, so we can just omit it. Therefore, by using the parametrization of the ellipse, we obtain
C
I
F · dR =
C
=
I
2
y i + x j + z k + ∇ϕ · d R =
C
0
2
3 sin t · 2(− sin t) + 4 cos t · 3 cos t dt =
Next, we compute the integrals
Z 2π
Z
2
sin t dt =
0
2π
0
Z
2π
0
3
cos t dt =
1
1 − cos(2t)
dt =
2
2
2
2π
−6 sin2 t + 12 cos3 t dt
sin(2t)
t−
2
(1 − sin t) cos t dt =
Z
0
=
Z
0
2π
Z
ydx + x2 dy + zdz
C
2π
Z
I
1
2π
= · 2π = π ,
2
0
2π
(1 − sin2 t) d(sin t)
0
sin t −
Finally, we obtain
I
1
2π
sin3 t = 0 .
3
0
F · d R = −6π .
C
5. Let F = [(1 + x)ex+y ] i + [xex+y + 2y] j − 2z k . a) Use the zero curl test to determine whether F is
irrotational.
b) Is F conservative? c) If F is conservative, find the scalar potential, d) Evaluate the line
R
integral C F · d R along the semicircle x = cos t, y = sin t, z = 1, 0 ≤ t ≤ π.
Solution.
a) We compute
i
∂x
curl F = (1 + x)ex+y
j
k
∂y
∂z
[xex+y + 2y] −2z
= i (0 − 0) − j (0 − 0) + k ex+y + xex+y − (1 + x)ex+y = 0 .
9
MATH 332, (Vector Analysis), Spring 2000: Test 2
Thus, F is irrotational .
b) Since F is irrotational and is defined in the whole space (which is simply connected), it is conservative , i.e.
F = ∇ϕ
c) To find the scalar potential we solve the differential equations
∂x ϕ = (1 + x)ex+y ,
∂y ϕ = xex+y + 2y,
∂z ϕ = −2z .
From the last equation we obtain
ϕ = −z 2 + f (x, y) .
We substitute this into the second equation to get
∂y f = xex+y + 2y .
Integrating this equation, we get
f (x, y) = xex+y + y 2 + g(x) ,
hence
ϕ = −z 2 + xex+y + y 2 + g(x) .
Finally, substituting this into the first equation we get
ex+y + xex+y + g 0 (x) = (1 + x)ex+y ,
and, therefore, g 0 (x) = 0, so g = C is contant (which can be choosen to be zero anyway). Finally, the scalar
potential is
ϕ = −z 2 + xex+y + y 2 .
d) Since F is conservative the line integral is equal to the contributions of the endpoints of the curve. The
initial point is P = (1, 0, 1) and the terminal point is Q = (−1, 0, 1). Therefore,
Z
F · d R = ϕ(Q) − ϕ(P ) = [−1 + (−1)e−1+0 + 0] − [−1 + e1+0 + 0] = −
C
1
+e .
e
6. (Extra Credit.) With the same notation as in the problem 3, determine whether or not the vector field
A×R
F =
is solenoidal. If so, find the vector potential G.
R2
Solution.
We use the zero divergence test
x εijk Aj xk
δik
k
−3
div F = ∂i
=
ε
A
∂
=
ε
A
+
x
(−2)R
∂
R
ijk j i
k
i
ijk j
R2
R2
R2
δik
xk xi
= εijk Aj
− 2 4 = 0.
R2
R
This is zero since εijk δik = 0 and εijk xi xk = 0. So F is solenoidal , i.e.
F = ∇ × G,
or
Fi =
εijk Aj xk
= εijk ∂j Gk .
R2
(1)
Since there are only two different vectors A and R , and the vector potential must be linear in A , we can
choose the vector potential G in the form
G = A f (R) + R ( A · R )g(R) + R ψ(R)
10
MATH 332, (Vector Analysis), Spring 2000: Test 2
or
Gi = Ai f (R) + Ak xi xk g(R) + xi ψ(R) ,
where f , g and ψ are some unknown scalar functions of R. On the other hand, we note that R ψ(R) is an
irrotational field, i.e.
h
xk i
{ curl [ R ψ(R)]}i = εijk [δjk ψ + xj ψ 0 ∂k R] = εijk δjk ψ + xj ψ 0
= 0.
R
We see that ψ does not contribute to F , so it is arbitrary, therefore can be set to zero, ψ = 0. Therefore, this
part can be omitted so that the vector potential is
G = A f (R) + R ( A · R )g(R)
or
Gk = Ak f (R) + Ap xp xk g(R) .
By noting that ∂j xk = δjk and ∂j R = xj /R we compute curl G
εijk ∂j Gk = εijk [Ak f 0 ∂j R + Ap δjp xk g + Ap xp δjk g + Ap xp xk g 0 ∂j R]
h
xj
xj i
= εijk Ak f 0
+ Aj xk g + Ap xp δjk g + Ap xp xk g 0
R
R
By noting that εijk δjk = 0 and εijk xj xk = 0, we get finally
0
h
i
xj
f
εijk ∂j Gk = εijk Ak f 0
+ Aj xk g = εijk Aj xk − + g ,
R
R
or, in vector notation
curl G = A × R
f0
− +g
R
.
Comparing this with F = ∇ × G from eq. (1) we obtain
−
f0
1
+g = 2 .
R
R
Therefore
g=
1
f0
+
,
R2
R
and the vector potential is
G = A f (R) + R ( A · R )
1
1 df (R)
+
2
R
R dR
.
Here f (R) remains an arbitrary function of R that can be choosen arbitrarily. The simplest choice is f = 0
resulting in the vector potential
(A · R)R
G =
.
R2
Another choice is g = 0, which gives an equation
f0 = −
1
.
R
This equation can be easily integrated leading to
f = − ln R ,
(the integration constant has been set to zero; it does not affect F .) In this case the vector potential is
G = − A ln R .
11
MATH 332, (Vector Analysis), Spring 2000: Test 2
Set 3
1. Let ϕ = xey+z . Find: a) gradϕ, b) div F and c) curl F .
2. Let F = y 2 i + x2 j + k . Find the general equation of a flow lines. (Hint: Separate variables in the
differential equation for the flow line).
3. Let A = A1 i + A2 j + A3 k be a constant vector field, R = x i + y j + z k , R = | R | =
1
a) Find F = grad ,
R
1
1
b) Show that div F = div grad = ∆ = 0,
R
R
c) Find div ( A × F ).
p
x2 + y 2 + z 2 .
4. Let F = y i +x2 j +z k and G = yz cos(xyz) i +xz cos(xyz) j +xy cos(xyz) k and C be the counterclockwise
oriented ellipse inIthe plane z = 1 parallel to the xy plane parametrized by x = 2 cos t, y = 3 sin
I t, z = 1, 0 ≤
t ≤ 2π. Evaluate
G ·d R = 0).
( F + G )·d R . (Hint: Show that G is a conservative field and, therefore,
C
C
5. Let F = [(1 + x)ex+y ] i + [xex+y + 2y] j − 2z k . a) Use the zero curl test to determine whether F is
irrotational.
b) Is F conservative? c) If F is conservative, find the scalar potential, d) Evaluate the line
R
integral C F · d R along the semicircle x = cos t, y = sin t, z = 1, 0 ≤ t ≤ π. (Hint: F is conservative!)