9.9
1. Homework
1
9.8 Find the divergence and the curl of the vector function
F(x, y, z ) = (x2 + y 2 + z 2 )3/2 (xi + y j + z k).
Sol. Let
F1 =
x
y
z
, F2 = 2
, F3 = 2
.
(x2 + y 2 + z 2 )3/2
(x + y 2 + z 2 )3/2
(x + y 2 + z 2 )3/2
Notice that
1
3x2
y 2 + z 2 2x2
F1 = 2
=2
.
x
(x + y 2 + z 2 )3/2 (x2 + y 2 + z 2 )5/2
(x + y 2 + z 2 )5/2
Similarly, we have
x2 + z 2 2y 2
F2 = 2
,
y
(x + y 2 + z 2 )5/2
x2 + y 2 2z 2
F3 = 2
.
z
(x + y 2 + z 2 )5/2
Thus
F1 + F2 + F3
x
y
z
2
2
2
y + z 2x
x2 + z 2 2y 2
x2 + y 2 2z 2
=
+
+
(x2 + y 2 + z 2 )5/2 (x2 + y 2 + z 2 )5/2 (x2 + y 2 + z 2 )5/2
= 0.
divF =
Notice that
3zy
3yz
F3 F2 = 2
2
= 0.
y
z
(x + y 2 + z 2 )5/2 (x + y 2 + z 2 )5/2
Similarly, we have
F3 F1 = 0,
x
z
F2 F1 = 0.
x
y
Thus
i
curl F =
j
k
x
y
z
F1 F2 F3
= ( F3 F2 )i ( F3 F1 )j + ( F2 F1 )k
y
z
x
z
x
y
= 0.
Computational Science & Engineering (CSE)
C. K. Ko
Homework
2
2. The velocity vector v(x, y, z ) of an incompressible uid rotating in a cylindrical vessel is of the form v = w r, where w is the constant rotation vector and
r = [x, y, z ]. Show that div v = 0.
Sol.
Let w = [, , ]. Then
, , ] (w r)
x y z
ijk
= [ , , ]
x y z
xyz
= [ , , ] [z y, x z, y x]
x y z
= 0.
div v = [
3. Assuming sucient dierentiability, show that
(a) div (f g ) = f 2 g + f g
(b) div (curlv) = 0
(c) curl(f v) = (gradf ) v + f curlv
(d) div(u v) = v curlu u curlv.
Sol.
(a)
, , ] [ f gx , f g y , f g z ]
x y z
=
(f gx ) + (f gy ) + (f gz )
x
y
z
= (fx gx + fy gy + fz gz ) + f (gxx + gyy + gzz )
=
f g + f 2 g.
div (f g ) = [
(b) Let v = [v1 , v2 , v3 ]. By assumption, (v1 )zy = (v1 )yz , (v2 )zx = (v2 )xz , (v3 )yx =
(v3 )xy . Hence
div(curlv) = [ , , ]
x y z
i
j
k
x
y
z
v1 v2 v3
= [(v3 )yx (v2 )zx ] + [(v1 )zy (v3 )xy ] + [(v2 )xz (v1 )yz ]
= 0.
Computational Science & Engineering (CSE)
C. K. Ko
Homework
3
(c) Let v = [v1 , v2 , v3 ]. Then f v = [f v1 , f v2 , f v3 ] and
i
curl (f v) =
=
=
=
=
j
k
x
y
z
f v1 f v2 f v3
[(f v3 )y (f v2 )z , (f v1 )z (f v3 )x , (f v2 )x (f v1 )y ]
f [(v3 )y (v2 )z , (v1 )z (v3 )x , (v2 )x (v1 )y ]
+[fy v3 fz v2 , fz v1 fx v3 , fx v2 fy v1 ]
ijk
ijk
f x y z + fx fy fz
v1 v2 v3
v1 v2 v3
f curlv + (gradf ) v.
(d) Let u = [u1 , u2 , u3 ] and v = [v1 , v2 , v3 ]. Then
ijk
div(u v) = [ , , ] u1 u2 u3
x y z
v1 v2 v3
(u2 v3 u3 v2 ) (u1 v3 u3 v1 ) + (u1 v2 u2 v1 )
=
x
y
z
= ((u2 )x v3 (u3 )x v2 ) ((u1 )y v3 (u3 )y v1 ) + ((u1 )z v2 (u2 )z v1 )
+(u2 (v3 )x u3 (v2 )x ) (u1 (v3 )y u3 (v1 )y ) + (u1 (v2 )z u2 (v1 )z )
= v1 ((u3 )y (u2 )z ) v2 ((u3 )x (u1 )z ) + v3 ((u2 )x (u1 )y )
u1 ((v3 )y (v2 )z ) u2 ((v3 )x (v1 )z ) + u3 ((v2 )x (v1 )y )
i
= [ v1 , v 2 , v 3 ]
j
k
i
j
k
x
y
z
x
y
z
[u1 , u2 , u3 ]
u1 u2 u3
= v curl u u curl v.
v1 v2 v3
4. Let u = [y 2 , z 2 , x2 ], v = [yz, zx, xy ] and f = xyz . Find the following expressions.
(a) curl(f u)
(b) curl(u v).
Sol. (a)
curl(f u) = curl[xy 3 z, xyz 3 , x3 yz ]
i
j
k
=
x
y
z
xy 3 z xyz 3 x3 yz
= [x3 z 3xyz 2 , xy 3 3x2 yz, yz 3 3xy 2 z ].
Computational Science & Engineering (CSE)
C. K. Ko
Homework
4
(b) Notice that
uv =
ijk
y 2 z 2 x2
yz zx xy
= [z 2 xy zx3 , x2 yz xy 3 , y 2 zx yz 3 ].
.
i
curl(u v) =
j
k
x
y
z
z 2 xy zx3 x2 yz xy 3 y 2 zx yz 3
= [2xyz z 3 x2 y, 2xyz x3 y 2 z, 2xyz y 3 z 2 x].
Computational Science & Engineering (CSE)
C. K. Ko
Homework
5
10.1
1. Calculate C F(r) dr, where F(x, y, z ) = [z, x, y ] and C : r(t) = [cos t, sin t, t]
from (1, 0, 0) to (1, 0, 4 ).
Sol. Notice that r(0) = [1, 0, 0] and r(4 ) = [1, 0, 4 ].
4
F(r) dr =
C
[t, cos t, sin t] [ sin t, cos t, 1]dt
0
=
4
(t sin t + cos2 t + sin t)dt
0
=
[t cos t]4
0
4
4
cos tdt +
0
0
1
sin 2t
= 4 [sin t]4 + t +
0
2
2
= 4 0 + 2 + 0 = 6.
4
0
1 + cos 2t
dt + [ cos t]4
0
2
+0
2. Find the work done by the force F(x, y, z ) = 9z i + 5xj + 3y k in the displacement
along C the intersection x2 + y 2 = 9 and z = x + 2, counterclockwise.
Sol.
Note that C : r(t) = [3 cos t, 3 sin t, 3 cos t + 2], 0 t 2 .
Work done by the force F
W=
F(r) dr
C
2
=
[9(3 cos t + 2), 5(3 cos t), 3(3 sin t)] [3 sin t, 3 cos t, 3 sin t]dt
0
2
=
0
(81 cos t sin t 54 sin t + 45 cos2 t 27 sin2 t)dt
2
=
(81 cos t sin t 54 sin t +
0
=
45
27
(1 + cos 2t) (1 cos 2t))dt
2
2
2
(81 cos t sin t 54 sin t + 9 + 36 cos 2t)dt
0
81
cos2 t + 54 cos t + 9t + 18 sin 2t]2
0
2
= 18.
=[
Computational Science & Engineering (CSE)
C. K. Ko
Homework
6
3. Consider the integral C F(r) dr, where F(x, y ) = xy i y 2 j.
(a) Find the value of the integral when r(t) = [cos t, sin t], 0 t /2. Show that
the value remains the same if you set t = p or t = p2 .
(b) Evaluate the integral when C : y = xn , 0 x 1, n = 1, 2, . What is the
limit as n ?
Sol.
(a)
/2
F(r) dr =
C
[cos t sin t, sin2 t] [ sin t, cos t]dt
0
/2
= 2
0
2
/2
cos t sin2 t dt = [ sin3 t]0
3
2
=.
3
If we set t = p then r (p) = r(p) = [cos p, sin p], p : 0 /2.
F(r ) dr
/2
=
C
[ cos p sin p, sin2 p] [ sin p, cos p]dp
0
/2
=
0
2
/2
(2 cos p sin2 p)dp = [ sin3 p]0
3
2
=.
3
If we set t = p2 then r (p) = r(p2 ) = [cos p2 , sin p2 ], p : 0
/2).
F(r ) dr
/2
=
C
0
/2
=
0
/2 or (p : 0
[cos p2 sin p2 , sin2 p2 ] [2p sin p2 , 2p cos p2 ]dp
2 3 2 /2
(4p cos p sin p )dp = [ sin p ]0
3
2
22
2
=,
3
or
F(r ) dr =
0
C
=
0
/2
/2
[cos p2 sin p2 , sin2 p2 ] [2p sin p2 , 2p cos p2 ]dp
2
(4p cos p sin p )dp = [ sin3 p2 ]0
3
2
22
/2
2
=.
3
The value of the integral remains the same.
Computational Science & Engineering (CSE)
C. K. Ko
Homework
7
(b) Fix n. Let C : r(t) = [t, tn ] 0 t 1.
1
F(r) dr =
C
n+1
[t
2n
, t ] [1, nt
n1
1
]dt =
0
(tn+1 nt3n1 )dt
0
1 n+2 1 3n 1
=[
t
t ]0
n+2
3
1
1
=
.
n+2 3
The limit as n is 1 .
3
Note: As n , C converges to r(t) =
ti
if
i + j if
0t<1
.
t=1
10.4
1. Evaluate C F(r) dr counterclockwise around the boundary curve C of the region R, where
(a) F(x, y ) = [ey , ex ], R the triangle with vertices (0, 0), (2, 0), (2, 1).
(b) F(x, y ) = [x2 y 2 , x/y 2 ], R : 1 x2 + y 2 4, x 0, y x.
Sol. (a) Using Greens theorem,
ey dx + ex dy =
F(r) dr =
C
C
2
x
2
=
0
2
ex (ey )dydx
R
x
2
y
e + e dydx =
0
0
x
2
[yex ey ]0 dx
x
xx
x
=
e e 2 + 1dx = [ ex ]2
20
02
e2 1
2
=e
+ 2 e1
2
2
1
e
+ 2e1 + .
=
2
2
2
0
x
ex
dx + [2e 2 + x]2
0
2
(b) Using Greens theorem,
I=
x2 y 2 dx
F(r) dr =
C
C
x
dy =
y2
Computational Science & Engineering (CSE)
R
1
2x2 ydydx.
2
y
C. K. Ko
Homework
8
Using the polar coordinates, we have
/2
2
I=
(
/4
/2
=
/4
/2
=
1
1
+ 2r3 cos2 sin )rdrd
2 sin2
r
2
([ln r]2 csc2 + [ r5 ]2 cos2 sin )d
1
51
(ln 2 csc2 +
/4
62
cos2 sin )d
5
62
/2
/2
= ln 2[cot ]/4 + [cos3 ]/4
15
31 2
= ln 2
.
30
2. (a) Show that for a solution w(x, y ) of Laplaces equation
R with boundary curve C and outer unit normal vector n,
R
w
x
2
+
w
y
2
dxdy =
w
C
2
w = 0 in a region
w
ds.
n
(b) Show that w(x, y ) = 2ex cos y satises Laplaces equation 2 w = 0 and, using
(a), integrate w w counterclockwise around the boundary curve C of the square
n
0 x 2, 0 y 2.
Sol. (a) Let C : r(s) = [x(s), y (s)], a s b, a unit tangent vector r = [ dx , dy ].
ds ds
dy
dx
Then the outer unit normal vector n of is C n = [ ds , ds ] and
w
=
n
wn=[
w w
dy dx
w dy w dx
,
] [ , ] =
.
x y
ds ds
x ds
y ds
Using Greens theorem, we have
w
C
w
ds =
n
=
Since
2
w
w
dy w dx
x
y
w 2
w 2
2w 2w
+
+ w[ 2 + 2 dydx.
x
y
x
y
R
w
C
w = 0 in the region R, we obtain
w
C
w
ds =
n
R
w
x
2
+
w
y
2
dydx.
(b) w(x, y ) = 2ex cos y satises Laplaces equation
2
w = wxx + wyy = 2ex cos y 2ex cos y = 0.
Computational Science & Engineering (CSE)
C. K. Ko
Homework
9
Using (a),
w
C
w
ds =
n
w
x
R
2
+
2ex cos y
=
R
2
=
2
w
y
2
2
+ 2ex sin y
2
2x
4e dydx =
0
4
dydx
0
2
dydx
8e2x dx
0
= 4(e 1).
3. Let R and C be as in Greens theorem, r a unit tangent vector, and n the outer
unit normal vector of C .
Show that
R
F2 F1
dxdy =
x
y
(F1 dx + F2 dy )
C
may be written
div Fdxdy =
R
F nds
C
or
(curl F) k dxdy =
R
F r ds
C
where k is a unit vector perpendicular to the xy plane.
Sol. Let C : r(s) = [x(s), y (s)], a s b, a unit tangent vector r = [ dx , dy ]. Then
ds ds
the outer unit normal vector n of C is
n=[
dy dx
, ].
ds ds
Computational Science & Engineering (CSE)
C. K. Ko
Homework
Let F = F2 i F1 j. Then
F2 F1
dxdy =
y
R x
div Fdxdy,
R
F nds =
C
10
[F2 , F1 ] [
C
=
dy dx
, ]ds
ds ds
(F1 dx + F2 dy ),
C
and so
F2 F1
dxdy =
x
y
R
(F1 dx + F2 dy )
C
may be written
div Fdxdy =
R
F nds.
C
Now, let F = F1 i + F2 j. Notice that
i
(curlF) k =
j
k
x
y
z
k=
F1 F2 0
F2 F1
,
x
y
F1 dx + F2 dy = F r ds.
Thus
F2 F1
dxdy =
x
y
R
(F1 dx + F2 dy )
C
may be written
(curl F) k dxdy =
F r ds.
R
C
10.6
1. Evaluate the surface integral
S F ndA, where
(a) F(x, y, z ) = [2x, 5y, 0], S : r(u, v ) = [u, v, 4u + 3v ], 0 u 1, 8 v 8
(b) F(x, y, z ) = [y 2 , x2 , z 4 ], S : z = 4 x2 + y 2 , 0 z 8, y 0.
ijk
Sol. (a) N = ru rv = 1 0 4 = [4, 3, 1].
013
8
1
F ndA =
S
8
1
F(r(u, v )) N(u, v )dudv =
8
8
0
8
0
1
=
0
8
(8u 15v )dudv =
= 64.
[2u, 5v, 0] [4, 3, 1]dudv
8
Computational Science & Engineering (CSE)
(4 15v )dv
8
C. K. Ko
Homework
11
(b) Let g (x, y, z ) = z 4 x2 + y 2 , x2 + y 2 4, 0 y.
Then N = g = [ 42x 2 , 42y 2 , 1] and
x +y
x +y
F(x, y, 4 x2 + y 2 ) [
F ndA =
S
R
[y 2 , x2 , 16(x2 + y 2 )] [
=
R
=
(
4xy 2 + 4x2 y
x2
R
2
=
0
=
+
y2
4x
x2 + y 2
4x
x2 + y 2
,
,
4y
x2 + y 2
4y
x2 + y 2
, 1]dydx
, 1]dydx
+ 16(x2 + y 2 ))dydx
[4r2 (cos sin2 + cos2 sin ) + 16r2 ]rdrd
0
[16(cos sin2 + cos2 sin ) + 64]d
0
sin3 cos3
= [16(
)]0 + 64
3
3
32
= + 64.
3
tan1 (y/x)dA, where S : z = x2 + y 2 , 1 z
2. Evaluate the surface integral
S
9, x 0, y 0.
Sol. Let R = {(x, y )|1 x2 + y 2 9, x 0, y 0}.
tan1 (y/x)dA =
tan1 (y/x)
S
R
/2
3
=
0
1 + 4x2 + 4y 2 dydx
1 + 4r2 rdrd
1
/2
(1 + 4r2 )3/2 ]3 d
1
12
0
/2
(37 37 5 5)d
=
12
0
(37 37 5 5) 2
=
.
96
=
[
Computational Science & Engineering (CSE)
C. K. Ko
Homework
12
10.7
1
1. Find the total mass of a mass distribution of density (x, y, z ) = 2 (x2 + y 2 )2 in
a region T the cylinder x2 + y 2 4, |z | 2.
Sol. Let R = {(x, y )|x2 + y 2 4}.
total mass =
(x, y, z )dV
T
12
(x + y 2 )2 dV
T2
2
12
(x + y 2 )2 dz dydx
2 2
=
=
R
2(x2 + y 2 )2 dydx
=
R
2
2
=
0
128
=
.
3
2r4 rdrd
0
2. Evaluate the surface integral
S F ndA by the divergence theorem.
(a) F(x, y, z ) = [4x, 3z, 5y ] and S the surface of the cone x2 + y 2 z 2 , 0 z 2.
(b) F(x, y, z ) = [x3 y 3 , y 3 z 3 , z 3 x3 ] and S the surface of x2 + y 2 + z 2 25, z 0.
Sol. (a) Let T = {(x, y, z )|x2 + y 2 z 2 , 0 z 2}. By the divergence theorem,
F ndA =
div FdV =
S
2
T
2
2
=4
rdzdrd
0
2
0
=4
16
3
r
2
r(2 r)drd
0
=
4dV
T
2
0
d =
0
32
.
3
(b) Let T = {(x, y, z )|x2 + y 2 + z 2 25, z 0}. By the divergence theorem,
F ndA =
S
div FdV
T
3(x2 + y 2 + z 2 )dV.
=
T
Computational Science & Engineering (CSE)
C. K. Ko
Homework
13
Using the spherical coordinates,
x = sin cos , y = sin sin , z = cos ,
0 5, 0 2, 0 /2,
we have
2
2
2
2
/2
5
3(x + y + z )dV = 3
T
0
0
2
2 2 sin ddd
0
/2
= 1875
sin dd
0
0
2
= 1875
d
0
= 3750.
10.8
1. Show that a region T with boundary surface S has the volume
V=
xdydz =
1
=
3
ydzdx =
S
S
zdxdy
S
(xdydz + ydzdx + zdxdy ).
S
Sol. When F = [F1 , F2 , F3 ] and n = [cos , cos , cos ] the outward unit normal
vector,
div F dV =
F ndA
T
S
can be written as
(
T
F1 F2 F3
+
+
)dxdydz =
x
y
z
(F1 cos + F2 cos + F3 cos )dA
S
=
F1 dydz + F2 dzdx + F3 dxdy.
S
Applying F = xi to the relation, we have
V=
dxdydz =
T
xdydz.
S
Applying F = y j to the relation, we have
V=
dxdydz =
T
Computational Science & Engineering (CSE)
ydzdx.
S
C. K. Ko
Homework
14
Applying F = z k to the relation, we have
V=
dxdydz =
T
and
1
3
V=
zdxdy,
S
(xdydz + ydzdx + zdxdy ).
S
2. Let f and g be functions that are harmonic in some domain D containing a region
T with boundary surface S such that T satises the assumptions in the divergence
f
theorem. Show that if n = 0 on S , then f is constant in T , and show that if
f
g
n = n on S , then f = g + c is constant in T , where c is a constant.
Sol. Notice that Greens rst formula
2
(f
g+
f
g )dV =
f
T
S
g
dA.
n
If g = f , then
(f
2
f |2 dV =
f +|
T
Since
2
f = 0 on T and
f
n
f
S
f
dA.
n
= 0 on S , we have
|
f |2 dV = 0.
T
Since | f | is continuous in T and nonnegative, f is zero everywhere in T . Hence
f is constant in T .
f
g
Let n = n on S , that is, (f g) = 0 on S . By the same argument, f g = c, f = g +c
n
in T , where c is a constant.
10.9
1.(a) Evaluate the surface integral
S (curl F) ndA directly for F(x, y, z ) =
[y 3 , x3 , 0] and S : x2 + y 2 1, z = 0.
(b) In (a), using Stokess theorem, calculate the surface integral.
i
Sol. (a) curlF n =
j
k
x
3
y
z
y
k = 3(x2 + y 2 ).
x3 0
Computational Science & Engineering (CSE)
C. K. Ko
Homework
15
Let R = {(x, y )|x2 + y 2 1}.
3(x2 + y 2 )dxdy
(curl F) ndA =
S
R
2
1
=
0
3
= .
2
3r2 rdrd
0
(b) Let C : r(t) = cos ti + sin tj, (0 t 2 ). Using Stokess theorem, we have
(curl F) ndA =
S
F r ds
C
2
=
[sin3 t, cos3 t, 0] [ sin t, cos t, 0]dt
0
2
=
0
2
=
0
sin4 t + cos4 tdt
1 2 sin2 t cos2 tdt
2
=
12
sin (2t))dt
2
31
+ cos(4t)dt
44
(1
0
2
=
0
3
= .
2
2. Calculate the line integral C F r ds by Stokess theorem, clockwise as seen by
a person standing at the origin, for F(x, y, z ) = [x2 , y 2 , z 2 ] and C the intersection
of x2 + y 2 + z 2 = 4 and z = y 2 .
Sol.
Let S be a smooth surface with the boundary C . Since
i
curlF =
j
k
x
2
y
2
z
2
x
y
= 0,
z
by Stokess theorem,
F r ds =
C
(curl F) ndA = 0.
S
Computational Science & Engineering (CSE)
C. K. Ko
Homework
16
3. Evaluate C F r ds, where F(x, y ) = (x2 + y 2 )1 [y, x] and C : x2 + y 2 = 1, z = 0,
oriented clockwise. Why can Stokess theorem not be applied?
Sol. Let C : r(t) = cos ti + sin tj, (t : 0 2 ).
2
F r ds =
[ sin t, cos t] [ sin t, cos t]dt
0
C
2
=
dt = 2.
0
i
Note: curlF =
j
k
x
y
x2 +y 2
y
x
x2 +y 2
z
= 0 and if Stokess theorem can be applied,
0
F r ds =
C
(curl F) ndA = 0.
S
If we regard F as a vector in the plane, then the surface S with boundary C is the
disk z = 0(x2 + y 2 1). Since F is not continuous at the origin, Stokess theorem
cannot be applied.
Computational Science & Engineering (CSE)
C. K. Ko