292ReviewProblems2Solutions.nb
Review Problems: VC.07VC.11
for Exam 2: Wednesday, 12/09/2009
Mathematica 6.0 Initializations
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292ReviewProblems2Solutions.nb
2
· R.1)
You are faced with a hand calculation of
2
2
Ÿ Ÿ R Ix + y M „ x „ y,
where R is the two-dimensional region consisting of everything inside and
on the circle
x2 + y2 = 9.
Switch paper by using polar coordinates
x = r Cos@tD, and
y = r Sin@tD,
and then calculate the integral.
Switch paper by using polar coordinates
x = r Cos@tD, and
y = r Sin@tD,
with r running from 0 to 3 and t running from 0 to 2Pi.
We calculate the gradients:
D[x[r,t],r]=Cos[t]
D[x[r,t],t] = -rSin[t]
D[y[r,t],r]=Sin[t]
D[y[r,t],t]= rCos[t]
Compute the determinant of the gradient matrix: Axy[r,t]= r(Cos[t])^2 +
r(Sin[t])^2 = r.
2 Pi 3
Ÿ0 Ÿ0 HHr ^ 2L HCos@tDL ^ 2 + Hr ^ 2L HSin@tDL ^ 2 L Axy@r, tD „ r „ t
2 Pi 3
Ÿ0 r ^ 3 „ r „ t
2 Pi
=Ÿ0 H1 ê 4L H3 ^ 4L „ t
=Ÿ0
= (Pi/2)(3^4) = 81Pi/2
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292ReviewProblems2Solutions.nb
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· R.2)
You are faced with a hand calculation of
Ÿ Ÿ R x „ x „ y,
where R is the two-dimensional region consisting of everything bounded
by the lines
x + y = 0, x + y = 1,
x - y = 0, and x - y = 1.
This is just a little nasty, because doing it without going to better paper
involves solving some equations, and then doing two separate integrals.
Switch to paper on which you can evaluate this with one sweet integral and
then calculate its value.
Switch paper by using:
u=x+y, and
v=x-y,
with u running from 0 to 1 and v running from 0 to 1.
Thus x = (1/2)(u+v), y=(1/2)(u-v)
We calculate the gradients:
D[x[u,v],u]=1/2
D[x[u,v],v] = 1/2
D[y[u,v],u]=1/2
D[y[u,v],v]= -1/2
Compute the determinant of the gradient matrix: Axy[u,v]= |-1/2|.
1 1
Ÿ0 Ÿ0 HH1 ê 2L Hu + vL L Axy@u, vD „ u „ v
1 1
=Ÿ0 Ÿ0 H1 ê 4L Hu + vL „ u „ v
1
=Ÿ0 H1 ê 4L HH1 ê 2L + vL „ v
= (1/8)+(1/8) = 1/4
· R.3)
The area conversion factor Axy @u, vD is the absolute value of the
determinant of a matrix composed of what two gradient vectors?
gradx[u,v] and grady[u,v]
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292ReviewProblems2Solutions.nb
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· R.4)
To calculate the net flow of a vector field
Field@x, yD = 8m@x, yD, n@x, yD<
across the boundary C of a region R, you have your choice:
Ø You can go to the labor of parameterizing C and then calculate
ò C -n@x, yD „ x + m@x, yD „ y.
Ø Or, if the field has no singularities inside R, you can put
divField@x, yD = D@m@x, yD, xD + D@n@x, yD, yD,
and calculate the 2D integral
Ÿ Ÿ R divField@x, yD „ x „ y.
Here's a vector field:
Clear@x, y, m, n, FieldD;
8m@x_, y_D, n@x_, y_D< = 9- x2 + Cos@yD, Ex + y2 =
9- x2 + Cos@yD, ‰x + y2 =
Agree that R is everything inside and on the parallelogram you see below:
Plot@8- x, - x + 3, 0.5 x, 0.5 x + 4<, 8x, - 2.67, 2.0<,
PlotStyle Ø [email protected]<<, PlotRange Ø 80, 3.68<, AxesLabel Ø 8"x", "y"<D
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292ReviewProblems2Solutions.nb
5
y
3.5
3.0
2.5
2.0
1.5
1.0
0.5
-2
-1
0
1
2
x
Transform the 2D integral
Ÿ Ÿ R divField@x, yD „ x „ y
to favorable uv-paper to measure the net flow of this vector field across the
parallelogram. Is the net flow of this vector field across the boundary of
this parallelogram from outside to inside, or inside to outside?
divField[x,y]=D[m[x,y],x]+D[n[x,y],y]
= -2x+2y
Switch paper by using:
u=x+y, and
v=y-0.5x,
with u running from 0 to 3 and v running from 0 to 4.
Thus x = (2/3)(u-v), y=(1/3)u+(2/3)v
We calculate the gradients:
D[x[u,v],u]=2/3
D[x[u,v],v] = -2/3
D[y[u,v],u]=1/3
D[y[u,v],v]= 2/3
Compute the determinant of the gradient matrix: Axy[u,v]= 2/3
4 3
Ÿ0 Ÿ0 H-2 x@u, vD + 2 y@u, vDL Axy@u, vD „ u „ v
4 3
=Ÿ0 Ÿ0 H2 ê 3L HH-2 ê 3L u + H8 ê 3L vL „ u „ v
4
=Ÿ0 H-2 + H16 ê 3L vL „ v
= (-8)+(8/3)(4^2) = 104/3
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292ReviewProblems2Solutions.nb
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· R.5)
When you start with the ellipse
2
y 2
I xa M + I b M = r2
on xy-paper and you go to uv-paper using
u@x, yD = xa , and
y
v@x, yD = b ,
then the xy-ellipse plots out as the circle
u2 + v2 = r2 .
On uv-paper, the area enclosed by this circle measures out to
p r2
square units.
Calculate the area conversion factor Auv @x, yD, and use the result to explain
why the xy-area measurement of the region enclosed by the ellipse
2
y 2
I xa M + I b M = r 2
is
p a b r2
square units.
Thus x = a u, y=b v
We calculate the gradients:
D[x[u,v],u]=a
D[x[u,v],v] = 0
D[y[u,v],u]=0
D[y[u,v],v]= b
Compute the determinant of the gradient matrix: Axy[u,v]= ab
Interesting - our area conversion factor is ab.
Thus when we do the integration, we get ab*(area of circle of radius r) =
p a b r 2.
· R.6)
R is the cube with sides of length 2 centered at 81, 0, 2<.
It looks like this:
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292ReviewProblems2Solutions.nb
2.0
3.0
1.5
x 0.5 0.0
1.0
2.5
z2.0
1.5
1.0
-1.0
-0.5
0.0
y
0.5
1.0
Calculate Ÿ Ÿ Ÿ R x „ x „ y „ z.
3
1
2
‡ ‡ ‡ x „x „y „z
1
-1 0
3
1
= ‡ ‡ 2 „y „z
1
-1
3
= ‡ 4 „z
1
= 8
· R.7)
The volume of a certain solid R is measured by calculating
1 2 3
Ÿ Ÿ Ÿ R „ x „ y „ z = Ÿ0 Ÿ0 Ÿ0 „ x „ y „ z.
Describe the solid and measure its volume in cubic units.
1
2
3
‡ ‡ ‡ „x „y „z
0
0
0
1
2
= ‡ ‡ 3 „y „z
0
0
1
= ‡ 6 „z
0
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292ReviewProblems2Solutions.nb
8
= 6
This is the volume of a rectangular box with
dimensions : length = 3, width = 2, height = 1.
· R.8)
When you start with the ellipsoid (egg)
2
y 2
2
I xa M + I b M + I zc M = r2
in xyz-space and you go to uvw-space using
u@x, y, zD = xa ,
y
v@x, y, zD = b , and
w@x, y, zD = zc ,
then the xyz-ellipsoid plots out as the sphere
u2 + v2 + w2 = r 2 .
In uvw-space, the volume enclosed by this sphere measures out to
4
p r3
3
cubic units.
Calculate the volume conversion factor Vuvw @x, y, zD, and use the result to
explain why the xyz-space volume-measurement of the region enclosed by
the ellipsoid
2
y 2
2
I xa M + I b M + I zc M = r 2
is
4
3
p a b c r3
cubic units.
Thus x = a u, y=b v, z=c w
We calculate the gradients:
D[x[u,v,w],u]=a
D[x[u,v,w],v] = 0
D[x[u,v,w],w]=0
D[y[u,v,w],u]=0
D[y[u,v,w],v]= b
D[y[u,v,w],w]=0
D[z[u,v,w],u]=0
D[z[u,v,w],v]= 0
D[z[u,v,w],w]= c
Compute the determinant of the gradient matrix: Vxyz[u,v,w]= abc
Interesting - our volume conversion factor is abc.
Thus when we do the integration, we get abc*(volume of circle of radius r)
= 43 p a b c r 3 .
· R.9)
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292ReviewProblems2Solutions.nb
9
R.9)
Given a transformation coming from
x = x@u, v, wD, y = y@u, v, wD, and z = z@u, v, wD.
Say how you use
gradx@u, v, wD, grady@u, v, wD, and gradz@u, v, wD
to calculate Vxyz @u, v, wD ¥ 0.
Then discuss the meaning of Vxyz @u, v, wD.
We calculate the gradients:
D[x[u,v,w],u]
D[x[u,v,w],v]
D[x[u,v,w],w]
D[y[u,v,w],u]
D[y[u,v,w],v]
D[y[u,v,w],w]
D[z[u,v,w],u]
D[z[u,v,w],v]
D[z[u,v,w],w]
Compute the Jacobian (absolute value of determinant of the gradient
matrix): Vxyz[u,v,w]
The resulting Vxyz[u,v,w] is our voume conversion factor - it tells us how
much stretching happens in the conversion xyz-paper and uvw-paper.
When we perform the integration in our new coordinate system, we
multiply by the volume conversion factor to account for stretching
involved in converting from our old coordinates to the new ones.
· R.10)
You want to calculate Ÿ ŸR Ix2 + y3 M „ A. Here's a view of the region with a
randomly placed horizontal line.
y0 = Random@D;
hline = [email protected], Brown, LineA99y02 , y0=, 81, y0<=E=E;
pregion = Show@region, hlineD
·
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292ReviewProblems2Solutions.nb
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x = y2
y
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
x
Use what you see to fill in the boxes so that
Ñ Ñ 2
2
3
3
Ÿ ŸR Ix + y M „ A = ŸÑ ŸÑ Ix + y M „ x „ y.
Here it is!
1 1
2
3
2
3
Ÿ ŸR Ix + y M „ A = Ÿ0 Ÿy2 Ix + y M „ x „ y.
· R.11)
You can specify a point in three dimensions as usual by its three
coordinates 8x, y, z<.
Another way to specify a point is to plant the x-, y-, and z-axes at 80, 0, 0<,
and then run a stick from 80, 0, 0< to the point:
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292ReviewProblems2Solutions.nb
11
z
8x,y,z<
y
x
The spherical coordinates 8r, s, t< of the plotted point are specified by the
equations
x = r Sin@sD Cos@tD,
y = r Sin@sD Sin@tD, and
z = r Cos@sD.
Take out a pencil and label the meaning of the measurements r, s, and t on
the plot above.
Here it is!
z
s
r
t
y
x
· R.12a.)
Set numbers r * , slow, shigh, tlow, and thigh so that the xyz-points
8x@s, tD, y@s, tD, z@s, tD< =
8r * Sin@sD Cos@tD, r * Sin@sD Sin@tD, r * Cos@sD<
with slow § s § shigh and tlow § t § thigh describe the part of the sphere
x2 + y2 + z2 = 9
consisting of those points with z ¥ 0.
r*=3, slow=0, shigh=Pi/2, tlow=0, thigh=2Pi.
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· R.12b.)
Set numbers slow, shigh, tlow, and thigh so that the xyz-points
8x@s, tD, y@s, tD, z@s, tD< =
82 Sin@sD Cos@tD, 2 Sin@sD Sin@tD, 2 Cos@sD<
with slow § s § shigh and tlow § t § thigh describe something like this:
For example: slow=Pi/6, shigh=Pi/3, tlow=0, thigh=2Pi.
· R.12c.)
Set numbers s* , rlow, rhigh, tlow, and thigh so that the xyz-points
8x@r, tD, y@r, tD, z@r, tD< =
8r Sin@s* D Cos@tD, r Sin@s* D Sin@tD, r Cos@s* D<
with rlow § r § rhigh and tlow § t § thigh describe something like this:
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292ReviewProblems2Solutions.nb
For example: s*=Pi/6, rlow=1, rhigh=2, tlow=Pi/2, thigh=3Pi/2.
· R.13)
You go to a library reference and learn that the volume of a sphere of
radius r measures out to
4
p r 3 cubic units.
3
You think about this for a second and then you type and execute:
Clear@x, y, z, r, s, tD;
x@r_, s_, t_D = r Sin@sD Cos@tD;
y@r_, s_, t_D = r Sin@sD Sin@tD;
z@r_, s_, t_D = r Cos@sD;
Clear@gradx, grady, gradz, VxyzD;
gradx@r_, s_, t_D = 8D@x@r, s, tD, rD, D@x@r, s, tD, sD, D@x@r, s, tD, tD<;
grady@r_, s_, t_D = 8D@y@r, s, tD, rD, D@y@r, s, tD, sD, D@y@r, s, tD, tD<;
gradz@r_, s_, t_D = 8D@z@r, s, tD, rD, D@z@r, s, tD, sD, D@z@r, s, tD, tD<;
Vxyz@r_, s_, t_D = TrigExpand@Det@8gradx@r, s, tD, grady@r, s, tD, gradz@r, s, tD<DD
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292ReviewProblems2Solutions.nb
r2 Sin@sD
And then you say that you can give your own hand calculation that
explains where the formula
4
p r3
3
comes from.
Do it.
Here goes:
2 Pi Pi rad
Ÿ0 Ÿ0 Ÿ0 Hr ^ 2 Sin@sDL „ r „ s „ t
2 Pi Pi
Ÿ0 HHrad ^ 3L ê 3L Sin@sD „ s „ t
2 Pi
=Ÿ0 HHrad ^ 3L ê 3L H2L „ t
= 43 p I rad 3 M.
=Ÿ0
Done!
· R.14)
Use spherical coordinates to pull off a very quick hand calculation of
2
2
2
Ÿ Ÿ Ÿ R Ix + y + z M „ x „ y „ z
xyz
where Rxyz is everything inside and on the sphere
x2 + y2 + z2 = a2
Here goes:
2 Pi Pi a
Ÿ0 Ÿ0 Ÿ0 Hr ^ 2L Hr ^ 2 Sin@sDL „ r „ s „ t
2 Pi Pi
Ÿ0 HHa ^ 5L ê 5L Sin@sD „ s „ t
2 Pi
=Ÿ0 HHa ^ 5L ê 5L H2L „ t
= 45 p a 5 .
=Ÿ0
Done!
· R.15)
Give the formula for the divergence, divField@x, y, zD, of a threedimensional vector field
Field@x, y, zD = 8m@x, y, zD, n@x, y, zD, p@x, y, zD<.
Say how the sign (positive or negative) of divField@x, y, zD tells you
whether 8x, y, z< is a source or a sink of the three-dimensional flow
represented by F@x, y, zD.
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292ReviewProblems2Solutions.nb
15
Say how the sign (positive or negative) of divField@x, y, zD tells you
whether 8x, y, z< is a source or a sink of the three-dimensional flow
represented by F@x, y, zD.
divField[x, y, z] = D[m[x, y, z], x] + D[n[x, y, z], y] + D[p[x, y, z], z]
If divField[x,y,z]>0, then {x,y,z} is a source.
If divField[x,y,z]<0, then {x,y,z} is a sink.
· R.16)
How do you know that
Field@x, y, zD = 9y z - x, Sin@zD - 2 y, Cos@x yD - 5 z3 =
has sinks at every point 8x, y, z<?
How do you know that the net flow of this vector field across the skin of
any solid surface is from outside to inside?
divField[x, y, z] = -3-15z^2
divField[x,y,z]<0 for all {x,y,z}, so every point is a sink.
Since every point is a sink, the net flow across the skin of any solid surface
is inward.
· R.17)
Determine the sources and sinks of
Field@x, y, zD = 93 z - 3 x2 , 3 z + 2 y, 4 z=.
Eyeball your answer and then respond quickly to the following questions.
If C1 is the sphere of radius 2 centered at 84, 0, 0<, is the net flow of this
vector field across C1 from outside to inside, or is it from inside to outside?
If C2 is the sphere of radius 2 centered at 8-3, 0, 0<, is the net flow of this
vector field across C2 from outside to inside, or is it from inside to outside?
divField[x, y, z] = -6x+6
divField[x,y,z]>0 for all {x,y,z} with x<1, and divField[x,y,z]<0 for
{x,y,z} with x>1.
For C1, all points have x coordinates at least 2, thus all points are sinks.
This means the net flow is inward (outside to inside).
For C2, all points have x coordinates at most -1, thus all points are sources.
This means the net flow is outward (inside to outside).
· R.18)
Here's a vector field:
In[6869]:=
Clear@x, y, z, m, n, p, FieldD;
8m@x_, y_, z_D, n@x_, y_, z_D, p@x_, y_, z_D< = 9x2 + E-y , y2 , x - 2 y z2 =;
Field@x_, y_, z_D = 8m@x, y, zD, n@x, y, zD, p@x, y, zD<
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Out[6871]=
9‰-y + x2 , y2 , x - 2 y z2 =
Use Gauss's formula to help come up with a measurement of the flow of
this vector field across the surface of the three dimensional box consisting
of all points 8x, y, z< with
0 § x § 2, -1 § y § 3, and 0 § z § 3.
Is the net flow of this vector field across this skin from inside to outside,
from outside to inside, or 0?
divField[x, y, z] = 2x+2y-4yz
So:
3 3 2
Ÿ0 Ÿ-1 Ÿ0 divField@x, y, zD „ x „ y „ z
3 3
=Ÿ0 Ÿ-1 H4 + 4 y - 8 yzL „ y „ z
3
= Ÿ0 H32 - 32 zL „ z
=32(3)-16(9) = -48
The integral of the divergence is negative, thus the flow is inward (outside
to inside)
· R.19a.)
Suppose C is the skin of a solid region in three dimensions.
Suppose Field@x, y, zD is a given vector field with divField@x, y, zD = 0 at
all points 8x, y, z<, and suppose this vector field has no singularities inside
C. Explain how you know that the net flow of this vector field across C is
0.
The divergence is 0 (and there are no singularities) - so by Gauss again, the
flow across the skin is just 0. No integrals necessary.
We can sort of reason it out, because we know there are no sinks or
sources, so the flow must be 0.
· R.19b.)
Suppose C is the skin of a solid region in three dimensions.
Suppose Field@x, y, zD is a given vector field with divField@x, y, zD > 0 at
all points 8x, y, z< inside and on C, and suppose this vector field has no
singularities inside C. Explain how you know that the net flow of this
vector field across C is from inside to outside.
The divergence is >0 (and there are no singularities) - so by Gauss again,
the flow across the skin outward. No integrals necessary.
We can sort of reason it out, because we know there are only sources
inside C, so the flow must Printed
be outward.
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292ReviewProblems2Solutions.nb
17
The divergence is >0 (and there are no singularities) - so by Gauss again,
the flow across the skin outward. No integrals necessary.
We can sort of reason it out, because we know there are only sources
inside C, so the flow must be outward.
· R.20)
Suppose you are going with a function f@x, y, zD, and you know that its
Laplacian
Ûf@x, y, zD =
∂2 f@x,y,zD
∂x2
+
∂2 f@x,y,zD
∂y2
+
∂2 f@x,y,zD
∂z2
=0
no matter what x, y, and z you go with.
Suppose also that f@x, y, zD has no singularities.
Explain why it is impossible to find a point 8x0 , y0 , z0 < with the property
that
f@x0 , y0 , z0 D > f@x, y, zD
for all points 8x, y, z< nearby 8x0 , y0 , z0 <, but not the same as 8x0 , y0 , z0 <.
Think of it this way:
If there is a point 8x0 , y0 , z0 < with the property that
f@x0 , y0 , z0 D > f@x, y, zD
for all points 8x, y, z< near 8x0 , y0 , z0 < but not the same as 8x0 , y0 , z0 <, then
what is the net flow of gradf@x, y, zD across small spheres centered at
8x0 , y0 , z0 <?
Since the laplacian is identically 0, we know there can be no sources or
sinks in the gradient field. The flow across a closed surface (like a sphere)
must be 0. If there is some point 8x0 , y0 , z0 < with the property that
f@x0 , y0 , z0 D > f@x, y, zD for all points 8x, y, z< near 8x0 , y0 , z0 < but not the
same as 8x0 , y0 , z0 <, then the flow across a small sphere centered at
8x0 , y0 , z0 < would be from outside to inside - a contradiction!
· R.21)
C1 is the top half of a distorted ellipsoid centered at the origin
parameterized as indicated:
Clear@x1, y1, z1, s, tD;
x1@s_, t_D = 3 Sin@sD Cos@tD;
y1@s_, t_D = 2 Sin@sD Sin@tD;
z1@s_, t_D = I2 - SinA6 s2 EM Cos@sD;
C1plot = ParametricPlot3DB8x1@s, tD, y1@s, tD, z1@s, tD<, :s, 0,
p
2
>, 8t, 0, 2 p<,
PlotPoints Ø 830, 30<, AxesLabel Ø 8"x", "y", "z"<, ViewPoint Ø CMView, Boxed Ø FalseF
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18
C2 is just a plain elliptical disk in the xy-plane parameterized as indicated:
Clear@x2, y2, z2, r, tD;
x2@r_, t_D = 3 r Cos@tD;
y2@r_, t_D = 2 r Sin@tD;
z2@r_, t_D = 0;
C2plot = ParametricPlot3D@8x2@r, tD, y2@r, tD, z2@r, tD<, 8r, 0, 1<, 8t, 0, 2 p<,
PlotPoints Ø 82, 30<, AxesLabel Ø 8"x", "y", "z"<, ViewPoint Ø CMView, Boxed Ø FalseD
Note that C1 and C2 have the same boundary curve:
Show@C1plot, C2plot, ViewPoint Ø 84, 2, - 4<D
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19
Explain how you know that when you go with a vector field Field@x, y, zD
with divField@x, y, zD = 0 throughout the solid region whose top skin is C1
and whose bottom skin is C2 , then
Ÿ Ÿ C Field.upperunitnormal „ A
1
= Ÿ Ÿ C Field.upperunitnormal „ A;
2
so that the flow of this vector field across both surfaces is the same.
What calculational nightmare does this help you avoid?
Note that since there are no
singularities and since the divergence is 0,
the flow across this closed surface is 0. If
we break the flow integral into 2 pieces,
we see that the flow through the bottom
has to be the flow through the top. Hence
‡‡
Field.topunitnormal „ A =
C1
‡‡
Field.topunitnormal „ A.
C2
So let ' s calculate the second
one because that is easier.
HNote that topunitnormal for the
bottom piece is just 80, 0, 1<.
· R.22)
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292ReviewProblems2Solutions.nb
20
R.22)
A 3D vector field
Field@x, y, zD = 8m@x, y, zD, n@x, y, zD, p@x, y, zD<
comes into your hands. You calculate curlField@1, 0, 0D and learn
curlField@1, 0, 0D = 8-2, 0, 1<.
You stick the tail of the unit vector
V=:
1
2
,
1
2
, 0>
at the point 81, 0, 0< and push your finger onto V so that V spikes through
the center of your finger, and the tip of your finger is at 81, 0, 0<. Does the
tip of your finger feel a net clockwise or counterclockwise swirl?
What happens when you take
1
2
V = :0,
,
1
2
>?
What unit vector V do you take if you want to feel the biggest possible net
counterclockwise swirl at the point 81, 0, 0<?
What unit vector V do you take if you want to feel the biggest possible net
clockwise swirl at the point 81, 0, 0<?
Calculate curlField[1,0,0].V={-2,0,1}.:
1
2
1
2
,
, 0> =
-2
2
. Thus we feel a
net clockwise swirl.
Try for the other vector: {-2,0,1}.:0,
1
2
,
1
2
>=
1
2
. Thus we feel a net
counterclockwise swirl.
To get the largest possible counterclockwise swirl, I'd take the unit vector
in the direction of curlField[1,0,0], which is { -2 ,0, 1 }. To get the
5
5
largest possible clockwise swirl, I'd move in the opposite direction.
V={ 2 ,0, -1 }
5
5
· R.23)
Here is a calculation of
[email protected], 0.9, 0.2D.V
for a given 3D vector field and a given unit vector V:
V = 80.205092, 0.489097, 0.847774<;
Clear@Field, x, y, z, m, n, p, curlFieldD;
·
Printed by Mathematica for Students
292ReviewProblems2Solutions.nb
Field@x_, y_, z_D =
8x, - y, z<
x2 + y2 + z2
21
;
8m@x_, y_, z_D, n@x_, y_, z_D, p@x_, y_, z_D< = Field@x, y, zD;
curlField@x_, y_, z_D = 8D@p@x, y, zD, yD - D@n@x, y, zD, zD,
D@m@x, y, zD, zD - D@p@x, y, zD, xD, D@n@x, y, zD, xD - D@m@x, y, zD, yD<;
[email protected], 0.9, 0.2D.V
0.869092
When you push your finger onto V so that V spikes through the center of
your finger, and the tip of your finger is at 80.3, 0.9, 0.2<, does the tip of
your finger feel a net clockwise or net counterclockwise swirl?
curlField[0.3,0.9,0.2].V is positive, so I'd feel a net counterclockwise swirl.
· R.24)
Here's a surface shown with a selection of normal vectors with tails planted
at the points at which the normals are calculated:
Here's the same surface shown with a selection of curlField@x, y, zD vectors
for a certain 3D vector field, Field@x, y, zD, with tails planted at the same
place the tails of the normal vectors are planted above:
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292ReviewProblems2Solutions.nb
22
Here they are together:
On the basis of these two plots, which of the following plots do you expect
to indicate the direction of the net flow of the given 3D vector field along
the boundary curve of the surface?
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292ReviewProblems2Solutions.nb
23
curlField.normal appears to be positive over this surface. This means that
as you look down from the top of each normal, you see Field@x, y, zD
swirling around the tail of each normal vector in the counterclockwise
way. Thus, I'd go with the picture on the left. Counterclockwise.
· R.25a)
You are given a certain 3D vector field
Field@x, y, zD = 8m@x, y, zD, n@x, y, zD, p@x, y, zD<,
and learn that no matter what point 8x, y, z< you go with, curlField@x, y, zD
points in the same direction as 80, 0, 1<.
You drop some paddle wheels into the flow represented by this vector field
so that the axis of each paddle wheel is parallel to the vector 80, 0, 1< like
this:
Printed by Mathematica for Students
292ReviewProblems2Solutions.nb
0.4
x
0.2
24
0.0
0.6
0.6
0.4
z
0.2
0.0
0.0
0.2
y
0.4
0.6
How do you know in advance that the flow represented by the given vector
field will make these paddle wheels rotate? Which way will they rotate?
How about in the case that curlField@x, y, zD points in the same direction as
80, 0, -1< at all points 8x, y, z<?
I can tell that they will rotate, because their axis is aligned with the axis of
the swirl of the field at each point. If you look from the top, then the
wheels are rotating counterclockwise.
In the second case they are rotating clockwise, when viewed from above
(in positive z direction).
· R.25b.)
You are given a 3D vector field
Field@x, y, zD = 8m@x, y, zD, n@x, y, zD, p@x, y, zD<,
and learn that no matter what point 8x, y, z< you go with, curlField@x, y, zD
points in the same direction as 80, 0, 1<.
You drop some small paddle wheels into the flow represented by this
vector field so that the axis of each paddle wheel is parallel to the vector
80, 1, 0< like this:
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292ReviewProblems2Solutions.nb
0.6
0.6
0.4
x
0.2
25
0.0
0.4
z
0.2
0.0
0.0
0.2
y
0.4
0.6
How do you know in advance that the flow represented by the given vector
field will not make these paddle wheels rotate?
I can tell that they will not rotate, because their axis is aligned
perpendicularly to the axis of the swirl of the field at each point.
· R.26a.)
What is the gradient test for a 3D vector field?
We have a gradient field if the curlField[x,y,z]={0,0,0}
· R.26b.)
Is there any way to stick your finger into the flow described by a 3D
gradient field so that the tip of your finger feels a net clockwise or net
counterclockwise swirl? How do you know? (Assume no singularities.)
No. curlField[x,y,z]={0,0,0}, thus there is no rotation (if there are also no
singularities, of course).
· R.27a.)
Write down Stokes's formula and say how the formula is to be interpreted.
Stokes ' s formula is the 3 D analogue
of this formula. Stokes ' s formula says
that if you go with a 3 D vector field
Field@x, y, zD =
8m@x, y, zD, n@x, y, zD, p@x, y, zD<
and R is a surface in three dimensions
with boundary curve C, then
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292ReviewProblems2Solutions.nb
26
with boundary curve C, then
‡ ‡ curlField.topunitnormal „ A
R
= ® Field.unittan „ s
C
= ® m@x, y, zD „ x + n@x, y, zD „ y + p@x, y, zD „ z
C
= net flow of Field@x, y, zD along C.
· R.27b.)
Take a surface R with boundary curve C. If you are given a 3D vector
field Field@x, y, zD with the extra property that
curlField@x, y, zD
is tangent to the surface R at all points 8x, y, z< on the surface R, then how
do you use Stokes's formula to tell you once and for all the net flow of
Field@x, y, zD along C is 0?
Stoke's Theorem says that the flow along is the double integral of
CurlField[x,y,z] . unitnormal.
If CurlField[x,y,z] is tangent to the surface everywhere, it is perpendicular
to the unitnormal everywhere.
So the flow is the integral of 0, and hence the net flow is 0.
· R.28a)
If C is a closed curve in three dimensions, and
Field@x, y, zD = 8m@x, y, zD, n@x, y, zD, p@x, y, zD<
with
curlField@x, y, zD = 80, 0, 0<
at all points, then what is the flow along C measurement
ò C m@x, y, zD „ x + n@x, y, zD „ y + p@x, y, zD „ z?
How does Stokes's formula help you to back up your answer?
If curlField[x,y,z]= 0 at all points, that means that the flow of the field
along any closed curve is 0. Consider, the curlField.topunitnormal
integration, for example. Clearly we would get 0 from this computation.
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292ReviewProblems2Solutions.nb
27
· R.28b)
A 3D vector field
Field@x, y, zD = 8m@x, y, zD, n@x, y, zD, p@x, y, zD<
comes into your hands. You are given two curves C1 and C2 , both
parameterized to start at a given point 8as , bs , cs < and end at a given point
8ae , be , ce < You calculate curlField@x, y, zD and are pleased to find that
curlField@x, y, zD = 80, 0, 0<
at all points 8x, y, z<.
What is the relationship between the flow along the curve measurements
Ÿ C1 m@x, y, zD „ x + n@x, y, zD „ y + p@x, y, zD „ z
and
Ÿ C2 m@x, y, zD „ x + n@x, y, zD „ y + p@x, y, zD „ z?
Do you expect this relationship to hold up in the case that curlField@x, y, zD
is not 80, 0, 0< at all points 8x, y, z<?
If curlField[x,y,z]= 0 at all points, that means that the flow of the field
along any closed curve is 0. Since the two above curves together make a
closed curve, we know that the flow along each curve is the same.
I would certainly not expect this to hold if curlField[x,y,z] was not {0,0,0},
we'd have to calculate the flow along each curve separately.
· R.29)
Give a hand calculation of curlField@x, y, zD by calculating
“ ä Field@x, y, zD
for Field@x, y, zD = 8x, y, z<.
Del cross field[x,y,z]
= {-Dz n[x, y, z] + Dy p[x, y, z], Dz m[x, y, z] - Dx p[x, y, z], -Dy m[x, y,
z] + Dx n[x, y, z]}
= {0+0, 0+0, 0+0}
· R.30)
Calculate
“ .Field@x, y, zD
for Field@x, y, zD = 8x, y, z< by hand, and interpret the result in terms of
divField@x, y, zD.
Del.field[x,y,z]=
= Dx m[x, y, z] + Dy n[x, y, z] + Dz p[x, y, z]
= 1+1+1=3.
This means that all points Printed
are sources.
by Mathematica for Students
292ReviewProblems2Solutions.nb
Del.field[x,y,z]=
= Dx m[x, y, z] + Dy n[x, y, z] + Dz p[x, y, z]
= 1+1+1=3.
This means that all points are sources.
· R.31)
Calculate the Laplacian
Ûf@x, y, zD = “ .“ f@x, y, zD
for
f@x, y, zD = ex Cos@yD + z2
by hand, and interpret the result in terms of
∂2 f@x,y,zD
∂x2
+
∂2 f@x,y,zD
∂y2
+
∂2 f@x,y,zD
,
∂z2
and the divergence of the gradient field of f@x, y, zD.
We know:
Ûf@x, y, zD =
“ .“ f@x, y, zD = Dx2 f@x, y, zD + Dy2 f@x, y, zD + Dz2 f@x, y, zD
= ex Cos@yD - ex Cos@yD + 2
= 2.
Thus, every point in the gradient field is a source.
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28