Math 166 - Study Guide: Exam #1
1. Total Area:
• One Curve: Use sign table to find where region is above and below the x-axis.
• Between Two Curves: Find the intersection points and then look at a slice.
2. Solids of Revolution:
• Disks: ∆V = π(radius)2 ∆w
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2 • Washers: ∆V = π routside − rinside
∆w
• Shells: ∆V = 2π(radius)(height)∆w
• Note 1: Disk and washer methods use slices perpendicular to axis of revolution
• Note 2: Shell method uses a slice parallel to axis of revolution
• Note 3: The radius is measured with respect to the axis. Thus the radius is not necessarily to the function itself. Hence, it would behoove you to make sure that you understand
how to account for shifting the axis of revolution.
3. Arclength:
• y = f (x): This has the parameterization x(t) = t and y(t) = f (t) with arclength
equation:
s
2
Z b
dy
dt
1+
L=
dt
a
• x = g(y): This has the parameterization y(t) = t and x(t) = g(t) with arclength
equation:
s
Z b 2
dx
L=
+ 1dt
dt
a
• x = f (t) and y = g(t): The parameterization is already given with arclength equation:
s
Z b 2 2
dy
dx
L=
+
dt
dt
dt
a
• Note 1: The first two equations are just special cases of the third equation based on the
parameterization of the problem.
• Note 2: When we are trying to find the surface area for the revolution of a curve then,
by the first fundamental theorem of calculus, ds is one the above equations without the
integral based on its parameterization.
4. Spring Problems:
• Hooke’s Law: F (x) = kx where k is the spring constant.
Z b
• Work: W =
F (x)dx
a
1
• Note 1: We need to determine the spring constant k before we can solve the problem
exactly.
• Note 2: The amount a spring is either compressed or stretched is always measured as in
terms of the springs natural length.
5. Center of Mass and Centroids:
• Discrete in one dimension:
Pn
xi mi
M
x̄ =
= Pi=1
n
m
i=1 mi
• Continuous in one dimension:
Rb
xδ(x)dx
M
x̄ =
= Ra b
m
a δ(x)dx
• Discrete in two dimensions:
Pn
xi mi
My
,
x̄ =
= Pi=1
n
m
i=1 mi
• Continuous in two dimensions:
Z b
δ
x [f (x) − g(x)] dx
My
x̄ =
,
= Za b
m
δ
[f (x) − g(x)] dx
a
Pn
yi mi
Mx
ȳ =
= Pi=1
n
m
i=1 mi
ȳ =
Mx
=
m
δ
2
Z
b
f (x)2 − g(x)2 dx
Za
b
[f (x) − g(x)] dx
δ
a
• Note 1: For the one dimensional continuous case, the density is allowed to be a functions
of condition, but for the two dimensional continuous case we only know how to deal
with uniform density. This is why δ(x) needs to be integrated for the one dimensional
continuous case, but the uniform density cancels out of the two dimensional continuous
case.
• Note 2: The moments are with respect to the x and y axes respectively. Thus My yields
how far (left or right) the moment is from the y-axis, and Mx yields how far (up or
down) the moment is from the x-axis. While the labeling my seem counter intuitive,
this is why they appear in their respective center of mass equations for the centroid.
6. Other Notes:
• I will expect that you know the integrals of the basic (polynomial, rational, logarithmic,
exponential, roots, trig, and inverse trig) functions.
• You will be expected to perform u-substitutions (but not manipulate any integrals) in
order to evaluate a few of the questions on the non-calculator portion of the exam.
• You will have a pumping water problem. That said, if I think the integrals are going to
be complicated you will only be asked to setup the integral. Thus it is very important
that you READ THE DIRECTIONS for every question.
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