Final Exam Topics
Chapters 16 and 17
• In a very broad sense, the two major topics of this exam will be line and surface integrals.
Both of these have versions for scalar functions and vector fields, and both have various
versions of the fundamental theorem of calculus that apply to them.
• Parameterizations: Setting up a line or surface integral requires us to parameterize the
curve or surface we are integrating over. Here are a few parameterization techniques to be
familiar with.
– Curves: We always parameterize a curve with a single variable, for example r(t) =
hx(t), y(t)i.
– If our curve is the graph of a function y = f (x) then we can use x as a parameter. The
parameterization would be r(x) = hx, f (x)i.
Ex: If the curve C is the graph of y = x4 with 2 ≤ x ≤ 4, we can parameterize this as
r(t) = hx, x4 i with 2 ≤ x ≤ 4.
– Lines are also the graph of a function, but sometimes it is easier to parameterize them by
finding a direction vector v and a point P on the line. In this case the parameterization
is r(t) = vt + P .
Ex: To parameterize the line between P = (1, 2, 5) and Q = (3, 5, 9) we find a vector
between these points v = h3 − 1, 5 − 2, 9 − 5i = h2, 3, 4i. If we just want the segment
of the line between these points, the parameterization r(t) = h2, 3, 4it + h1, 2, 5i =
h2t + 1, 3t + 2, 4t + 5i with 0 ≤ t ≤ 1 will start at P and end at Q.
– Circles can be parameterized with trig. functions. The segment of the circle of radius
R between the angles θ1 and θ2 is parameterized by r(t) = hR cos(t), R sin(t)i with
θ1 ≤ t ≤ θ2 . If we are in a 3D setting, just add a z-coordinate to this parameterization.
Ex: The circle of radius 3 at height z = 4, with x ≥ 0 and y ≥ 0 could be parameterized
by r(t) = h3 cos(t), 3 sin(t), 4i with 0 ≤ t ≤ π2 .
– Surfaces: We parameterize surfaces with two variables, for example G(u, v) = hx(u, v), y(u, v), z(u, v)i.
We also need to specify what domain (2D region) u and v come from.
– If the surface S is the graph of a function of two variables z = f (x, y), we can parameterize by setting x = x, y = y and z(x, y) = f (x, y). Whatever domain in the xy-plane
the surface sits over is the domain of parameterization.
Ex: We can parameterize the part ofp
the ellipsoid 2x2 + 3y 2 + z 2 = 1 sitting over the
xy plane by solvingpfor z = f (x, y) = 1 − 2x2 − 3y 2 and letting our parameterization
be G(x, y) = hu, v, 1 − 2x2 − 3y 2 i. In the xy plane z = 0, so 2x2 + 3y 2 = 1. We want
the region inside this ellipse, so a domain for x and y would be 2x2 + 3y 2 ≤ 1.
– Parameterizing a cylinder is similar to parameterizing a circle. If the height of the
cylinder is given by a ≤ z ≤ b and the radius is R we can parameterize it with G(θ, z) =
hR cos(θ), R sin(θ), zi with a ≤ z ≤ b and 0 ≤ θ ≤ 2π. If we only want part of the
cylinder, we can restrict θ to a smaller range.
– To parameterize a sphere of radius R we can use our spherical coordinates functions. Let
θ and φ be our parameters, so our parameterization is G(θ, φ) = hR cos(θ) sin(φ), R sin(θ) sin(φ), R cos(φ)i
with 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π. Depending on the part of our sphere we want, we may
restrict the range on θ and φ.
• Line Integrals: A line integral is the integral of a scalar function or a vector field over a 1D
curve c. Suppose we have parameterized our curve as r(t) with a ≤ t ≤ b.
Zb
Z
– Scalar Line Integrals: If f (x, y, z) is a scalar function, then
f (x, y, z) ds =
c
a
Zb
Z
F (x, y, z) · dr =
– Vector Line Integrals: If F (x, y, z) is a vector field, then
c
f (r(t))kr 0 (t)k dt.
F (r(t)) · r 0 (t) dt.
a
Ex: If c is r(t) = h3 cos(t), 3 sin(t), 4i with 0 ≤ t ≤ π2 (the circle from before, oriented
counterclockwise) then r 0 (t) = h−3 sin(t), 3 cos(t), 0i and kc0 (t)k = 3. The integral of
π
Z2
f (x, y, z) = x2 +z 2 over c is (9 cos2 (t) + 16)(3) dt. The integral of F (x, y, z) = hy, z, 0i
0
π
Z2
over C is
−9 sin2 (t) + 12 cos(t) dt.
0
• Surface Integrals: A surface integral is the integral of a scalar function or a vector field
over a 2D surface S. Suppose we have parameterized our surface as G(u, v) with (u, v) in
domain D.
• Tangent vectors to this surface are given by
∂G
get a normal vector N (u, v) = ∂G
∂u × ∂v .
∂G
∂u
and
∂G
∂v .
If we take their cross product, we
– Scalar Surface Integrals: If f (x, y, z) is a scalar function, then
ZZ
ZZ
f (x, y, z) dS =
f (G(u, v))N (u, v) dA
S
D
– Vector Surface Integrals: If F (x, y, z) is a vector field, then
ZZ
ZZ
F (x, y, z) · dS =
F (G(u, v)) · N (u, v) du dv
S
D
Ex: If S is the cylinder G(u, v) = h3 cos(u), 3 sin(u), vi then the normal vector is
N (u, v) = h3 cos(u), 3 sin(u), 0i and kN (u, v)k = 3. The integral of f (x, y, z) = x2 + z 2
over S is
ZZ
(9 cos2 (u) + v 2 )(3) dA
D
The integral of F (x, y, z) = hy, z, 0i over S is
ZZ
9 cos(u) sin(u) + 3v sin(u) dA
D
For both of these, you would need to set up bounds for the domain D. Look back to
Chapter 15 for help on setting up double integral bounds.
• The Fundamental Theorem of Calculus in multiple dimensions: Conservative fields,
Green’s Theorem, Stokes’ Theorem and the Divergence Theorem.
• Besides just setting up line and surface integrals with parameterizations, we also have a
number of theorems for transforming these integrals. In truth these are all versions of the
fundamental theorem of calculus. All of these theorems deal with vector integrals.
• Terminology and notation: If c is aI closed curve we put a circle on the integral symbol to
show this, and we sometimes refer to F · dr as the circulation of F around C.
C
For
Z Z any surface S and vector field F we can refer to our usual vector surface integral
F · dS as the flux of F across S.
S
Remember: The curl of a vector field, ∇ × FZ Zis itself a vector field, so we can also talk
(∇ × F ) · dS.
about the flux of the curl of F which would be
S
• Conservative Fields
– A vector field F is conservative if there exists a scalar function f , called the potential of
F , such that F = ∇(f ). For a 3D vector field F (x, y, z) = hP, Q, Ri this would mean
∂f
∂f
that P = ∂f
∂x , Q = ∂y and R = ∂z .
Ex: The function f (x, y, z) = x2 + y 2 + z 2 is a potential for the vector field F (x, y, z) =
h2x, 2y, 2zi.
– To check if a vector field is conservative, take its curl. If ∇ × F = h0, 0, 0i then F , and
F has continuous partial derivatives on a simply connected domain is conservative on
that domain. If the curl is not zero, the field is not conservative. For 2D vector fields,
∂P
all we have to check is that ∂Q
∂x − ∂y = 0.
∂P
Ex: The vector field F (x, y) = hx+y, x−yi is conservative, because ∂Q
∂x = 1 and ∂y = 1.
– Sometimes we can find a potential function just by looking at F . If you are able to do
this for a conservative field on the exam, at least verify that it has the correct partial
derivatives.
Ex: The vector field F (x, y) = hx + y, x − yi has a potential because F is conservative.
∂f
A potential for F is f (x, y) = 12 x2 + xy − 21 y 2 because ∂f
∂x = x + y and ∂y = x − y.
– If you cannot find a potential function “by observation” then you can try the following
process for finding potential f . These instructions are for a 3D vector field F (x, y, z) =
hP, Q, Ri. The procedure for 2D vector fields is similar, but requires less steps.
R
R
R
(i) Integrate P dx, Q dy and R dz.
(ii) Build a potential function f (x, y, z) which is the sum of all unique terms in these
three integrals. In other words, put each term into the potential, but only once.
Ex: The vector field F (x, y, z) = hy 2 + yz 2 , 2xy + xz 2 + ey , 2zyx + 1i is conservative (you
can check that ∇ × F = 0 and this is a nice smooth vector field). To find a potential,
first integrate
Z
y 2 + yz 2 dx = xy 2 + xyz 2
Z
2xy + xz 2 + ey dy = xy 2 + xyz 2 + ey
Z
2zyx + 1 dz = xyz 2 + z
Our final version of the potential is
f (x, y, z) = xy 2 + xyz 2 + ey + z
– The main reason we want a potential function is to evaluate line integrals of our vector
field. The reason for this is the following version of the fundamental theorem of calculus.
If F = ∇(f ) (so f is a potential for F ) and c is a curve going from P to Q, then
Z
F · dr = f (P ) − f (Q)
c
– This means that if you recognize a field is conservative, and find a potential function for
it, you can skip the parameterization process and evaluate line integrals just by plugging
into the potential.
Ex: Let c be the curve r(t) = h2t + 1, t, 3ti with 0 ≤ t ≤ 4 and F (x, y, z) = hy 2 +
yz 2 , 2xy + xz 2 + ey , 2zyx + 1i (from the previous example). The potential f (x, y, z)
for this field was found in the previous example. This curve starts at r(0) = h1, 0, 0i
and ends at r(4) = h9, 4, 12i. By the fundamental theorem for conservative fields given
above,
Z
F · dr = f (9, 4, 12) − f (1, 0, 0)
c
= (9(42 ) + (9)(4)(122 ) + e4 + 12) − (0 + 0 + e0 + 0)
= 5339 + e4
– If c is a closed curve, the initial and end pointsH are the same. Without even calculating
the potential, we can say that the circulation F · dr = 0 for a conservative field.
c
• Green’s Theorem
– Green’s Theorem applies to the circulation around a closed curve c in 2D. Call the region
inside the curve D. We sometimes use the notation ∂D = C for the boundary of D.
Assume that c has a parameterization which is continuous with continuous derivatives.
Also assume that it does not intersect itself.
– Green’s Theorem says that if F is continuous on D and ∂D is oriented with exterior
boundaries counterclockwise and interior boundaries clockwise then
I
ZZ
∂Q ∂P
F · dr =
−
dA
∂y
D ∂x
∂D
– It may be the case that I would tell you I want you to use Green’s theorem on a particular
problem, but it may also be the case that I don’t tell you to use Green’s even though
you should. This applies to all of the integral theorems.
– For this reason, you should be familiar with the types of setups and situations where
these theorems apply.
– With Green’s theorem, here is what to look for. If you are asked about the circulation
of F around a closed curve, and F is not conservative, be suspicious that I want you to
∂P
use Green’s theorem. In particular, if F is a nasty vector field but ∂Q
∂x − ∂y is simple,
or if ∂D is complicated but setting up a double integral over D would be easy, it’s time
for Green’s.
2
Ex: Suppose we want to know the circulation of F (x, y) = hx arctan(x) + y, ey + x2
around the triangle c with corners at (0, 0), (0, 2) and (1, 0) oriented ccw. This is a
terrible vector field, and the curve has to be parameterized in three pieces. On the other
∂P
hand, ∂Q
∂x − ∂y = 2x + 1.
To set up bounds for the inside of the triangle, notice that the hypotenuse is the line
between (0, 2) and (1, 0). This is y = 2 − 2x. So by Green’s theorem we could compute
the circulation as
I
Z Z
1
2−2x
F · dr =
c
2x + 1 dy dx
0
0
– Since Green’s (and several other topics) require you to set up a multiple integral, this
may be something good to look back to Chapter 15 and review as well.
• The last two integral theorems apply to vector fields in 3D.
• Stokes’ Theorem:
– Stokes’ theorem applies to vector surface integrals of the curl on surfaces with boundary.
Generally we say S is the surface and ∂S is the boundary curve of S. Assume both S
and ∂S have parameterizations which are continuous with continuous partial derivatives.
Also assume that they do not intersect themselves.
– To apply Stokes’ we need ∂S to be oriented in a particular way with respect to the
normal vectors on S. For the correct orientation, we want the surface to be on the left
as we move around the boundary facing in the direction of the normal vector. You can
imagine “walking” your fingers around the boundary. Make a finger-person with the
index and middle fingers on your right hand. If your finger-person stands lined up with
the normal vectors and walks around the boundary, the surface should be on the same
side as your thumb.
– Stokes’ theorem says that if F is continuous and has continuous partial derivatives on
S, and ∂S is oriented as described above, then
ZZ
I
(∇ × F ) · dS = F · dr
S
∂S
– In words you can say that the flux of the curl of F over S is equal to the circulation of
F around the boundary.
RR
– When do you want to use this? Any time I ask you for
(∇ × F ) · N dS, and ∂S is
S
easy to parameterize, you should try Stokes’ theorem. If you use this theorem, you don’t
even need to compute ∇ × F because it is just F that goes into the line integral.
You may also use this to transform line integrals into a surface integral, much like Green’s
Theorem.
Ex: Let S be the cylinder x2 + y 2 = 10 with −1 ≤ z ≤ 3. Suppose this cylinder has a
cap at the upper end but not at the lower end, and that its normal vectors
RR point out of
2
2
the cylinder. Let F = hy , y , zi. We can use Stokes’ theorem to compute (∇×F )·dS
S
as a line integral. That way we don’t have to parameterize both the sides and top of the
cylinder, and we don’t have to compute the curl.
The boundary of S is the circle at the bottom of the cylinder. We want this circle to
be oriented counterclockwise to use Stokes’ theorem. Imagine walking your right hand
around the base of the cylinder. Your thumb should point up toward the surface
of
√
the cylinder, so your fingers walk counterclockwise. This
circle
has
radius
10
and
√
√
sits at height z = −1, so we parameterize it as r(t) = h 10 cos(t), 10 sin(t), −1i with
0 ≤ t ≤ 2π. This matches the correct orientation.
From here we need to set up the line integral. So we plug r(t) into F and compute r 0 (t)
and obtain
F (r(t)) = h10 sin2 (t), 10 sin2 (t), −1i
√
√
r 0 (t) = h− 10 sin(t), 10 cos(t), 0i
ZZ
(∇ × F ) · dS =
S
Z2π
I
F · dr =
∂S
0
√
10 10 sin2 (t)(cos(t) − sin(t)) dt
• Divergence Theorem
– The Divergence theorem applies to closed surfaces. These could be whole spheres or
boxes, cylinders with both caps, or the surface surrounding any volume. The only
assumption we need about orientation is that the normal vectors on S are oriented out
of the volume inside S.
– The Divergence theorem says that if S is a closed surface (with nice parameterization
as usual) and F is continuous on the volume V inside S then
ZZ
ZZZ
F · dS =
∇ · F dV
S
V
– Generally, any time you have a surface integral over a closed surface it is going to be
easier to use the Divergence theorem. This is because setting up the triple integral is
usually easier than messing around with parameterizations.
– If we think of the divergence ∇ · F as the instantaneous flux, this theorem takes on a
nice interpretation. It says the flux across the surface S is equal to the net instantaneous
flux inside S.
Ex: Let S be the surface of the region above the graph z = 2 + x2 + y 2 and under the
2
plane
RR z = 6. Let F (x, y, z) = hxy, yz, xy i. We can compute the flux across S, which is
F ·dS by setting up a triple integral over this region. The region is bounded between
S
2 + x2 + y 2 ≤ z ≤ 6. Where these surfaces intersect, 6 = 2 + x2 + y 2 so this volume sits
over top of the circle x2 + y 2 = 4 in the xy-plane. The divergence is ∇ · F (x, y, z) = y + z.
While it is not totally necessary, this integral is easier to set up in cylindrical coordinates
as
Z
Z Z
2π
2
6
r sin(θ) + z dz dr dθ
0
0
2+r2
• Last Notes: Here are a few more miscellaneous tips. The integral of 1 over any geometric
object (like a surface, curve or volume) gives us the measure of that object (so for curves it
gives us the length, for surfaces the surface area, etc.). So if I ask for the length of a curve C
I want
Z
1 ds
C
Or if I ask for the surface are of S I want
ZZ
1 dS
S
These are both scalar integrals, so in both cases you should have the magnitude of something
inside the integral. This also works the other direction. If you wind up with the integral of a
constant, pull the constant out of the integral and now it represents the measure of something.
For example
ZZ
ZZ
5 dA = 5
D
1 dA = 5Area(D)
D
Use this fact to make evaluating integrals easier!