Some highlights before the exam
WARNING: In the final exam you are responsible from all the
topics that were explained in the announcements. What is given
below is NOT a complete list of items that may come up in the
exam, but they are merely some basic points which you may find
useful.
•When converting a double or triple integral over a region
into an iterated one, remember that the bounds of the outer-
most integral must be constants, but inner integrals may have
bounds depending on the integration variables of the outer
integrals. For example, iterated integrals like
Zx
2Z4
1
xy dxdy, or Z2
1Zx
0Z1
0
z dxdydz
are incorrect, but
Z4
2Zy
1
xy dxdy, and Z2
1Zz
0Zy+z
z
z dxdydz
are valid expressions.
•The area of a 2-dimensional region Dis RRAdA and the volume
of a 3-dimensional region Eis RRREdV .
•A region Din the xy-plane could be of both Type I or Type
II. In this case, double integrals on Dcan be converted to it-
erated integrals in two different ways. Sometimes, integrating
a function f(x, y)on Dis very hard with one of the iterated
integrals and considerably easier with the other.
•The substitution rule for changing from Cartesian to polar
coordinates is given by
x=rcos θ, y =rsin θ, dA =r drdθ (or r dθdr).
•Make sure that you are able to recognize “polar rectangles”,
that is, sets that are of the form
(r, θ)∈[a, b]×[α, β]
in polar coordinates. These are disks or annuli centered at the
origin, or their intersections with angular sectors.
•Circles with equations of the form x2+y2=ax or x2+y2=ay
(where a= 0 is a constant) also have simple representations
in polar coordinates, given by r=acos θand r=asin θ,
respectively.
•The cylindrical coordinate system is usually useful when the
integration domain has rotational symmetry about the z-axis
and/or the function to be integrated has a simple representa-
tion in terms of the variable r=px2+y2. The substitution
rule for converting a Cartesian integral into cylindrical coor-
dinates is:
x=rcos θ, y =rsin θ, z =z, dV =r drdθdz.
One can also think of variants where the polar coordinate
transformation is done on the pair (x, z)or (y, z ), instead of
(x, y). For example, in the first case it would be:
x=rcos θ, z =rsin θ, y =y, dV =r drdθdy.
•The spherical coordinate system is usually useful when the
integration domain has spherical symmetry about the origin
and/or the function to be integrated has a simple representa-
tion in terms of the variable ρ=px2+y2+z2. The substi-
tution rule for converting a Cartesian integral into spherical
coordinates is:
x=ρsin ϕcos θ, y =ρsin ϕsin θ,
z=ρcos ϕ, dV =ρ2sin ϕ dρdϕdθ.
Recall that the range of values for ϕ(latitude) is [0, π]and for
θ(longitude) it is [0,2π]. The set described by the equation
ϕ=cis the positive z-axis when c= 0, it is the negative
z-axis when c=π, and is the xy-plane when c=π/2. In all
other cases it is a cone. For example, the cone z=px2+y2
is given by ϕ=π/4in spherical coordinates.
•Arclength integrals (integrals of the form RCfds where f
is a function defined at the points of C) are orientation-
independent. That is, when evaluating these integrals you
can use any parametrization of Cyou like, without worrying
about its orientation.
•The length of a curve Cis given by the integral RCds.
•Once the curve Cis parametrized as (x(t), y(t)),t∈[a, b](or
as (x(t), y(t), z(t)) in 3 dimensions), one can evaluate RCf ds
by making the following substitutions and integrating from
t=ato t=b:
x=x(t), y =y(t), ds =p(x′(t))2+ (y′(t))2dt
for curves in 2 dimensions, or
x=x(t), y =y(t), z =z(t),
ds =p(x′(t))2+ (y′(t))2+ (z′(t))2dt
in 3 dimensions.
•The integrals RC(P dx+Qdy)(or RC(P dx+Qdy +Rdz)in 3D)
where P, Q and Rare functions, are orientation-dependent:
To evaluate these integrals correctly, you must parametrize
the curve Cusing the orientation indicated in the question
text. If you use a parametrization in the opposite direction,
the result it will give will have the opposite sign of the correct
result.
•Once the curve Cis parametrized correctly as (x(t), y(t)),t∈
[a, b], the substitutions to compute RC(P dx +Qdy)are
x=x(t), dx =x′(t)dt, y =y(t), dy =y′(t)dt
(in 3D one also has z=z(t)and dz =z′(t)dt). The resulting
expression should be integrated from t=ato t=b.
•If F= (P, Q)is a vector field (or F= (P, Q, R)in 3D, where
P,Qand Rare functions) and Cis a curve with a give orien-
tation, then
ZC
F·Tds and ZC
F·dr
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