Math 232 – Fall 2009
Test 1 Information
1
Logistics
• The exam will be given in SB 120 & 128.
• You may come as early as 7:30 am. I don’t intend to write a long exam, but if you move a bit
slowly in the morning and want to have a more leasiurely time, you may begin early. Exams will
be collected at 8:55 am.
• You will have access to the computers so that you can use sage for numerical and graphical calculations as needed.
• Be sure to indicate how you use sage or your calculator on the exam. No mystry numbers or
expressions, please.
• Use good notation and show all work. Full credit will not be given if your answers are not sufficiently
supported.
Material Covered
Test 1 covers Chapter 13 (Vector Calculus). For more details about what we covered when, consult the
web calendar.
Practice Makes Perfect
If you want more practice for the exam, try these ideas:
• Do the bracketed problems from the homework assignments.
• Do problems from the Concept Check, True-False Quiz, and Exercises on pages 797–799.
• Write down all of the “Fundamental Theorems” on a piece of paper without looking at your notes
or the book, then check to see that you have them all correct (including their names and good
notation).
• Do the same thing for the differential conversion equations (see pages 2 and 3 of this review sheet).
Sage
The number of commands that we have used in sage are relatively few. You should know how to use the
following (or be able to do the equivalent on your calculator):
• plot_vector_field(). (I will not require plot3d_vector_field() since it is not a built-in sage
function.)
• parametric_plot() and parametric_plot3d()
• diff() and integral() (including nesting for multiple integrals)
• numerical_integral() for numerical integrals – only works for single integrals.
• factor() for simplifying expressions; n() for getting a numeric value
c
2009
Randall Pruim ([email protected])
Test 1 Information
Math 232 – Fall 2009
2
Parameterization
You should be able to parameterize curves (with one parameter – often we called it t) and surfaces
(with two parameters – often we called them u and v). If you are required to come up with your own
parameterization, be sure that you record the parameterization as part of your solution. We have several
notatoins for parameterizations, and you may use whichever you prefer. Your parameterization skills
should include at least:
• Commonsly recurring shapes: circles, ellipses, line segments, planes, etc.
• Situations where one variable is defined in terms of the other(s).
For example, if a surface is defined by the equation z = f (x, y), then there is an easy parameterization: Parameterize x and y however you like (even x = u and y = v, if that works but sometimes
other parameterizations work out better), and then let z = f (x(u, v), y(u, v)).
You could get problems that require you to find your own parameterization or problems that provide the
parameterization for you. You should know both how to find your own and how to work with one that
is given.
Notation
Being able to correctly read and write mathematical notation is an important key to success with this
material. In particular, it is essential that you keep straight when you are working with vectors and when
you are working with real numbers. The textbook and this review sheet use bold face to indicate
vector values. On the board, I have been writing a wiggle beneath vectors since bold face is hard to
write. I encourage you to do the same.
It is also important that you correctly use the differential portion of the integrals (see next section).
Don’t drop that dA, dt, ds, dS, dS etc. or use the wrong one, especially if you are going to convert your
integral into a different form.
Finally, note that we have often have multiple ways to denote the same thing. For example, the following
are all equivalent:
Z
2
F · dr
F (x, y, z) = xi + yxj + z k
C
F = x, yx, z
2
Z
F · dr
C
P =x
Q = yz
R=z
2
Z
P dx + Q dy + R dz
C
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2009
Randall Pruim ([email protected])
Test 1 Information
Math 232 – Fall 2009
3
Line Integrals
They are called line integrals, but “curve integral” might be a better name.
Z
• General form:
f ds.
C
◦ Think: value of f times the length of a small piece of the curve.
◦ Evaluate by converting to an integral with respect to dt
r
dr dx
dy
dz
◦ ds = dt =
( )2 + ( )2 + ( )2 dt
dt
dt
dt
dt
Z
Z b
dr ◦
f ds =
f (r(t)) dt
dt
C
a
◦ Example applications
∗ arc length: f = 1
∗ mass: f = density
∗ center of mass: f = (distance from axis/plane) × (density)
Z
Z
• Special case:
F · T ds =
F · dr
C
◦
dr
=
dt
C
dx dy dz
, ,
;
dt dt dt
dr = hdx, dy, dzi
dr = hdx, dy, dzi
dr/dt
r 0 (t)
hx0 (t), y 0 (t), z 0 (t)i
= 0
=p
|dr/dt|
|r (t)|
x0 (t)2 + y 0 (t)2 + z 0 (t)2
◦ f = F · T = tangential component of vector field
◦ T = unit tangent vector =
◦ We get some nice cancellation:
Z
Z
dr/dt
|dr/dt| dt =
F · T ds =
F·
|dr/dt| | {z }
C
C
| {z }
ds
Z
a
b
dr
F·
dt =
dt
T
◦ Example application
∗ work done moving object along path
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2009
Randall Pruim ([email protected])
Z
F · dr
C
Test 1 Information
Math 232 – Fall 2009
4
Surface Integrals
• General form:
x
f dS
S
◦ Think: for each “patch” we multiply area of patch by value of function f .
dx dy dz
dx dy dz
◦ dS = |ru × rv | dA;
ru =
; rv =
, ,
, ,
du du du
dv dv dv
◦ Evaluate by expressing everything in terms of parameters (often u and v):
x
x
f dS =
f (r(u, v)) |ru × rv | dA
|
{z
}
S
D
dS
◦ Example applications
∗ surface area: f = 1
∗ mass: f = density
∗ center of mass: f = (distance from axis/plane) × (density)
x
x
• Special case:
F · n dS =
F · dS (flux of F across S)
S
S
◦ Think: for each patch, multiply area of patch by component of F normal to patch.
ru × rv
◦ n=
|ru × rv |
◦ dS = n dS (Note where the boldface is.)
◦ S = surface; D = domain of parameters (often u and v)
◦ dA = du dv = dv du
◦ We get nice cancellation again:
x
x
x
x
ru × rv
F·
|ru × rv | dA =
F · (ru × rv ) dA
F · dS =
F · n dS =
|ru × rv | |
{z
}
| {z }
D
D
S
S
| {z }
dS
dS
n
Fundamental Theorems
We have encountered a number of theorems that all have some similarity to the original Fundamental
Theorem of Calculus you learned in Math 161. These are summarized nicely on page 795.
Each of these theorems presents an equivalence between two expressions. These equations share some
common features:
• The left side involves “one more” integration than the right side.
• The integrand on the left side involves some sort of derivative. (Or you could say that the right
side involves some sort of antiderivative.) This means that you can only use these equations from
left to right if the integrand has a special form.
• Each of the theorems has some additional assumptions (not listed here or on page 795) about
contintuity, partial derivatives, orientation, etc.
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2009
Randall Pruim ([email protected])
Math 232 – Fall 2009
Test 1 Information
5
Review Problems
1 Give a parameterization for each of the following curves.
a) The “upper half” of the circle x2 + y 2 = 4 going counterclockwise from h2, 0i to h−2, 0i.
b) The line segment from h1, 2, 3i to h4, 2, 1i.
x2 y 2
c) The curve that forms the intersection of the (non-circular) cylinder with equation
+
= 1 and
4
9
plane with equation z = x + y.
2 Give a parameterization for each of the following surfaces.
a) The portion of the paraboloid z = 10−x2 −y 2 that lies above the rectangle with corners h±2, ±1, 0i.
b) The plane that includes the points h1, 1, 1i, h1, 2, 3i, and h3, 2, 1i.
c) A sphere centered at h1, 2, 3i with radius 2.
d) An ellipsoid centered at the origin and passing through the points h1, 0, 0i, h0, 2, 0i, and h0, 0, 3i.
Problems 3 – 7 use the following vector fields.
F (x, y) = 4x3 y 2 − 2xy 3 , 2x4 y − 3x2 y 2 + 4y 3
G(x, y, z) = sin y i + x cos y j + sin z k
H = (x − y) i + −yj + k
3 Compute curl G, div G, curl H, and div H. Why can’t we compute the curl and divergence of F ?
4 Make a list of all the places where we have used curl and divergence.
5 Which of F , G, and H are conservative vector fields?
6 Evaluate the following line integrals.
Z
F · dr where C is the curve in 1a.
a)
C
Z
G · dr where C is the curve in 1b.
b)
C
Z
yz ds where C is parameterized by r = ht, 3 cos t, 3 sin ti for t ∈ [0, π/2].
c)
C
7 Evaluate the following surface integrals.
a)
x
b)
x
H · dS where S is the surface in 2d.
S
xy dS where S is the interior of the triangle with corners at h1, 0, 0i, h0, 2, 0i, and h0, 0, 2i.
S
8 Make a list of topics covered on the exam that don’t appear here.
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2009
Randall Pruim ([email protected])
Test 1 Information
Math 232 – Fall 2009
6
Solutions to Problem Set 8
Problem 18.8.7
curl F = hxexy , yexy − y, 2z − zi. Since the curve is oriented counter-clockwise, we use an upward
normal to C, namely n = h0, 0, 1i. (We don’t need to use a cross product to find the normal vector in
this case because our surface is so simple.) Now we plug in and evaluate
Z
x
x
x
x
F · dr =
curl F · dS =
curl F · n dS =
z dS =
5 dS = 5(16π) = 80π.
C
S
S
S
S
Problem 18.8.9
If we let our surface S be the “flat” interior C, we can parameterize S in one of two ways:
• r = hx, y, 1 − x − yi for x2 + y 2 ≤ 9
(or r = hu, v, 1 − u − vi for u2 + v 2 ≤ 9, if you prefer.)
• r = hv cos u, v sin u, 1 − v cos u − v sin ui for u ∈ [0, 2π], v ∈ [0, 3].
Using the first parameterization, ru × rv = rx × ry = h1, 1, 1i, so using Stoke’s Theorem, we get
Z
x
x
F · dr =
curl F · dS =
curl F · (ru × rv ) dA
C
S
=
D
x
D
h0, x2 , y 2 i · h1, 1, 1i dA
x
=
(x2 + y 2 ) dx dy
x2 +y 2 ≤9
Z 2π Z 3
=
0
r2 r dr dθ = 81π/2
0
Problem 18.9.5
div F = ex sin y − ex sin y + 2yz = 2yz. So
Z
x
y
F · dS =
div F dV =
S
0
E
1Z 1Z 2
2yz dz dy dx = · · · = 2
0
0
Here’s the sage code for the integral:
var(’x y z’)
integral(integral(integral(2*y*z, z,0,2),y,0,1),x,0,1)
Problem 18.9.10
div F = 2xy + 2xy + 2xy = 6xy. So
Z
x
y
F · dS =
div F dV =
S
0
E
2 Z 2−x Z 1− x2 − z2
0
6xy dx dz dy = · · · = 2/5
0
This integral can be done is sage as follows:
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2009
Randall Pruim ([email protected])
Test 1 Information
Math 232 – Fall 2009
var(’x y z’)
integral(integral(integral( 6*x*y, y,0,(2-x-z)/2), z,0,2-x), x, 0, 2)
Solutions to Review Problems
1
a) r = h2 cos t, 2 sin ti for t ∈ [0, π]
b) r = h1, 2, 3i + th3, 0, −2i = h1 + 3t, 2, 3 − 2ti for t ∈ [0, 1]
c) r = h2 cos t, 3 sin t, 2 cos t + 3 sin ti for t ∈ [0, 2π]
2
a) r = hx, y, 10 − x2 − y 2 i for x ∈ [−2, 2], y ∈ [−1, 1].
b) r = h1, 1, 1i + uh0, 1, 2i + vh2, 1, 0i = h1 + 2v, 1 + u + v, 1 + 2ui
c) r = h1, 2, 3i + h2 cos u cos v, 2 sin u cos v, sin vi = h1 + 2 cos u cos v, 2 + 2 sin u cos v, 3 + sin vi
d) r = hcos u cos v, 2 sin u cos v, 3 sin vi
3 curl G = 0
div G = 0 − x sin y + cos z
curl H = h0, 0, −1i = −k
div H = 1 − 1 + 0 = 0.
F is not a 3-dimensional vector field, so we can’t compute curl F .
5 F is conservative since
∂P
∂y
= 8x3 y − 6xy 2 =
∂Q
∂x .
G is conservative since curl G = 0.
H is not conservative since curl G 6= 0.
6
a) F is conservative. A potential function is f = x4 y 2 − x2 y 3 + y 4 so
Z
F · dr = f (−2, 0) − f (2, 0) = 0 − 0 = 0
C
b) G is conservative. A potential function if g = x sin y + cos z, so
Z
G · dr = g(4, 2, 1) − g(1, 2, 3) = 4 sin(2) + cos(1) − sin(2) − cos(3)
C
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2009
Randall Pruim ([email protected])
7
Test 1 Information
Math 232 – Fall 2009
8
c)
Z
Z
π
yz ds =
√ Z
√
9 cos t sin t 1 + 9 dt = 9 10
0
C
0
1
√
9 10
u du =
2
7
a) div H = 0, so using the Divergence Theorem, we get
x
y
H · dS =
0 dV = 0
S
E
b) The triangle lies in the plane with the equation 2x + y + z = 2 (use cross-products and dot-products
to get this result, for example), so r = hx, y, 2 − 2x − yi (for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2 − 2x) is a
parameterization of S. Using this ru × rv = rx × ry = h2, 1, 1i, so
x
S
xy dS
x
xy
√
√ Z
6 dA = 6
D
1 Z 2−2x
0
xy dy dx =
0
Here is sage code for the double integral:
integral( integral( x*y, y, 0, 2-2*x),x , 0 ,1)
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2009
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√
6/6