CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
These lecture notes are intended as an outline for both student and instructor; for much more detailed
exposition on the topics contained herein, the student should consult Chapters 12-17 of our standard
textbook Calculus: Early Transcendentals by Jon Rogawski.
Remark 0.1. The reader of these notes should remember that they are intended as an aid to the
author’s verbal lectures rather than as a comprehensive source of information, and thus most definitions
and theorems are stated for a particular dimension of Euclidean space (e.g. for R2 , or R3 , etc.) rather
than in their fullest generality. For instance, when we state the chain rule for paths (Theorem 15.1) we
work in R2 as this is the setting we encounter most commonly; conversely when we talk about vector
fields (Definition 25.1) we will most commonly work in R3 . As a general (but not absolute) rule, the
reader may infer that what theorems and definitions make sense for R2 or R3 , mostly likely make sense
for any Rn where n ≥ 2 with a few minor adaptations. (One major exception to this is cross products
which are strictly a 3-dimensional notion.)
1
Preliminaries
Definition 1.1. The symbol R denotes the set of all real numbers. Geometrically we think of R as the
set of all points on the number line, or as 1-dimensional space.
2
Vectors in the Plane
Definition 2.1. The plane, or Cartesian plane, or xy -plane, is the full set of all ordered pairs of
the form (x, y) where x and y are both real numbers. We denote the plane by R2 . We think of R2 as
2-dimensional space.
The following definition is non-rigorous but well captures our intuitive notion of what a “vector”
should be, and will be helpful when we consider applications of calculus to physical problems. We will
give a more clear, i.e. purely mathematical, definition shortly. (Students who have taken a previous
course in linear algebra may be familiar with a certain very general definition of a vector; the definitions
we will end up working with here in fact a special case of the definition they have already learned.)
Definition 2.2 (Naive Definition of a Vector). A vector is an object which consists of both a (strictly
positive) length (or magnitude) and a direction. We typically denote vectors as lower-case letters
with an arrow hat, e.g. ~u, ~v , w,
~ ~a, ~b, etc. We denote the length of a vector ~v by ||~v ||. We consider two
vectors ~u and ~v to be equal if they have both the same length and the same direction, and we write
~u = ~v . (For equality, note that we do not require ~u and ~v to have the same location!) We also allow the
existence of a unique zero vector, denoted ~0, which we consider to have 0 length and no direction.
We may represent a vector ~v pictorially by drawing an arrow. We refer to pointy part at the end of the
arrow as the head, and the base of the arrow as the tail. If P and Q are two points in (two-dimensional
or three-dimensional) space, then we denote by P~Q the vector which has its tail at P and its head at
~ .)
Q. (Note that unless P = Q, we always have P~Q 6= QP
A scalar is just a magnitude, with no direction. In other words, a scalar is just a real number (a
member of the set R).
Example 2.3. Using the intuitive definition above, classify the following physical quantities as either a
vector or a scalar.
(a) A wind blowing southwest at 22 miles per hour.
(b) The mass of an apple.
(c) The force exerted by gravity on the apple.
(d) The temperature in Fargo, ND at 2pm today.
(e) The velocity of a parked car at rest.
1
2
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
Definition 2.4 (Naive Definition of Scalar Multiplication). Let ~v be a vector and let c be a scalar. The
scalar product of c and ~v , denoted c~v , is defined as follows:
(1) if c > 0 then c~v is the vector which points in the same direction as ~v , and whose length is c times
the length of ~v .
(2) if c < 0 then c~v is the vector which points in the opposite direction from ~v , and whose length is
|c| times the length of ~v .
(3) if c = 0 then c~v = ~0.
We call the vector c~v a scalar multiple of ~v .
Example 2.5. Draw any non-zero vector ~v . Then draw the vectors 3~v , 21 ~v , −~v = (−1)~v , − 52 ~v , 0~v , and
−π~v .
Remark 2.6. Observe that another way of writing part (b) in the definition above is that for any vector
~v and any scalar ~c,
||c~v || = |c|||~v ||.
This is a helpful identity we will use many times.
Definition 2.7. Two vectors are called parallel if one is a non-zero scalar multiple of the other.
Definition 2.8 (Naive Definition of Vector Addition and Subtraction). Let ~u and ~v be vectors. If the
tail of ~v is placed at the head of ~u, then the vector sum of ~u and ~v , denoted ~u + ~v , is the vector whose
tail coincides with the tail of ~u and whose head coincides with the head of ~v . (Picture helps here.)
The vector difference of ~u and ~v , denoted ~u − ~v , is defined to be the vector sum ~u + (−1)~v .
Example 2.9. (a) In general is ||~u + ~v || = ||~u|| + ||~v ||?
(b) Is the inequality ||~u + ~v || > ||~u|| + ||~v || possible?
Fact 2.10 (Triangle Inequality). Let ~u and ~v be any vectors. Then
||~u + ~v || ≤ ||~u|| + ||~v ||.
Next we will give a more rigorous definition of vectors and vector operations by coordinatizing them
in R2 .
Definition 2.11. A vector in the plane is an ordered pair ~v = (v1 , v2 ) in R2 . (We think of the tail of
~v as being the origin (0, 0) and the head as the point (v1 , v2 ).) Two vectors ~u = (u1 , u2 ) and ~v = (v1 , v2 )
are considered equal if both u1 = v1 and u2 = v2 , in which case we write ~v = ~u. The number v1 is
called the x-component of ~v and v2 is called the y-component of ~v .
The magnitude ||~v || of a vector ~v = (v1 , v2 ) is the number
p
||~v || = v12 + v22 .
Of course by the Pythagorean theorem, the magnitude of ~v is exactly its genuine geometric length.
If c is a scalar, we define the scalar product of c and ~v to be the vector c~v = (cv1 , cv2 ). We define
the vector sum of ~u and ~v to be the vector ~u + ~v = (u1 + v1 , u2 + v2 ), and the vector difference to
be the vector ~u − ~v = (u1 − v1 , u2 − v2 ).
Remark 2.12. The student should note that our definitions given above (and below) differ slightly
from those of Rogawski. There is more than one way to rigorously define a vector in the plane; we have
chosen the above since it is consistent with the definitions the student might eventually see in linear
algebra or other future math courses. The slight change in definition should present no great difficulty
on the homework.
Also note that according to our definiton, the collection R2 is both the “set of all points in the plane”
as well as the “set of all vectors in the plane.” This duality is intentional. Given a pair (x, y) in R2 , its
geometric “role” as either a point or a vector must be distinguished from the context in which its given.
Definition 2.13. Let P = (x1 , y1 ) and Q = (x2 , y2 ) be two points in R2 . By the notation P~Q, we mean
the vector
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
3
P~Q = (x2 − x1 , y2 − y1 ).
Example 2.14. Let P = (x1 , y1 ) and Q = (x2 , y2 ) be two points in R2 .
(a) Sketch a picture of P~Q.
(b) Compute the magnitude ||P~Q||.
Definition 2.15. Let ~u = (u1 , u2 ) and ~v = (v1 , v2 ) be vectors in R2 and let c be a scalar. Define the
vector sum ~u + ~v , the vector difference ~u − ~v , and the scalar product c~v as follows:
~u + ~v = (u1 + v1 , u2 + v2 );
~u − ~v = (u1 − v1 , u2 − v2 );
c~v = (cv1 , cv2 ).
Example 2.16. Let ~u = (−1, 2) and ~v = (2, 3).
(a) Evaluate ||~u + ~v ||.
(b) Write 2~u − 3~v as a single vector.
(c) Find two distinct vectors half as long as ~u and parallel to ~u.
Definition 2.17. A unit vector is a vector of length 1. In particular we single out the coordinate
unit vectors, which we permanently denote by ~i = (1, 0) and ~j = (0, 1).
Note that if ~v = (v1 , v2 ) is any vector, then we may write
~v = (v1 , v2 ) = (v1 , 0) + (0, v2 ) = v1 (1, 0) + v2 (0, 1) = v1~i + v2~j.
So in general ~v = (v1 , v2 ) = v1~i + v2~j; this gives us another notation for writing vectors.
Fact 2.18. If ~v is a non-zero vector, then the vector
magnitude 1. In other words
~
v
||~
v ||
~
v
||~
v ||
=
1
v
||~
v || ~
has the same direction as ~v , and
is the unique unit vector which points in the same direction as ~v .
Example 2.19. Let P = (1, −2) and Q = (6, 10) be two points in the plane.
(a) Find P~Q and two distinct unit vectors parallel to P~Q.
(b) Find two distinct vectors of length 3 parallel to P~Q.
Example 2.20. Assume the water in a river moves southwest at 4 mi/hr. If a motorboat is traveling
due east at 15 mi/hr relative to the shore, determine the speed of the boat and its heading relative to
the moving water.
Example 2.21. A child pulls a wagon with a force of |F | = 20 lb at an angle of θ = 40◦ to the horizontal.
Find the force vector F .
Example 2.22. A 400-lb engine is suspended from two chains that form 60◦ angles with a horizontal
ceiling. How much weight must each chain withstand?
3
Vectors in Three Dimensions
Definition 3.1. The set of all ordered triples (x, y, z) where x, y, and z are all real numbers is called
xyz-space or three-dimensional space, and denoted permanently by R3 .
Example 3.2.
(1) Plot the points P = (3, 4, 5) and Q = (−2, −3, 4) in xyz-space.
(2) Compute the distance between P and Q.
Definition 3.3. A vector in three dimensions is an ordered triple in R3 . The magnitude of a
vector ~v = (v1 , v2 , v3 ) in three dimensions is the quantity
p
||~v || = v12 + v22 + v32 .
The real numbers v1 , v2 , and v3 are called the x-component, y-component, and z-component
of ~v respectively. The notions of being parallel and of vector addition, vector subtraction, and
scalar multiplication are defined analogously to the two-dimensional case.
Definition 3.4. When working in R3 , the unit coordinate vectors are the following three distinguished vectors:
4
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
~i = (1, 0, 0);
~j = (0, 1, 0);
~k = (0, 0, 1).
Note that for any vector ~v = (v1 , v2 , v3 ) we have ~v = v1~i + v2~j + v3~k.
Example 3.5. Let ~u = (2, −4, 1) and ~v = (3, 0, −1).
(a)
(b)
(c)
(d)
Find −4~u + 2~v .
Find ||~u − ~v ||.
Find the unique unit vector with the same direction as ~u − ~v .
Write the unit vector from part (c) as a sum of scalar multiples of ~i, ~j, and ~k.
4
Dot Products
Definition 4.1. Given two vectors ~u = (u1 , u2 , u3 ) and ~v = (v1 , v2 , v3 ), we define the dot product,
denoted ~u · ~v , to be the scalar given by
~u · ~v = u1 v1 + u2 v2 + u3 v3 .
√
√
Example 4.2. (a) Let√~u = ( 3, 1, 0) and √
~v = (1, 3, 0). Compute ~u · ~v .
(b) Let ~u = (2, −1, −2 3) and ~v = (2, −2, 3). Compute ~u · ~v .
Fact 4.3 (Properties of the Dot Product). For any vectors ~u, ~v , and w,
~ and any scalar c,
~
~
(i) 0 · ~v = ~v · 0 = 0;
(ii) ~u · ~v = ~v · ~u (commutative property);
(iii) (c~u) · ~v = ~u · (c~v ) = c(~u · ~v );
(iv) ~u · (~v + w)
~ = ~u · ~v + ~u · w
~ (distributive property of the dot product over vector addition); and
(v) ~v · ~v = ||~v ||2 .
The usefulness of the dot product emerges in the upcoming Theorem 4.5; in order to prove Theorem
4.5, we need to recall the following fact about triangles.
Fact 4.4 (Law of Cosines). If a triangle has angle measures A, B, and C and corresponding opposite
side lengths a, b, and c, then the following equalities hold:
c2 = a2 + b2 − 2ab cos C; b2 = a2 + c2 − 2ac cos B; a2 = b2 + c2 − 2bc cos A.
The following theorem says that we can use dot products to compute the angles between given vectors.
Theorem 4.5. Let ~u = (u1 , u2 , u3 ) and ~v = (v1 , v2 , v3 ) be non-zero vectors, and let θ be the angle
between ~u and ~v with 0 ≤ θ ≤ π. Then
~u · ~v = ||~u||||~v || cos θ.
Proof. Consider the triangle which has ~u and ~v for two of its sides. The third side is equal as a vector
to ~u − ~v , and hence the Law of Cosines (previous fact) implies that
||~u − ~v ||2 = ||~u||2 + ||~v ||2 − 2||~u||||~v || cos θ.
Now solving for ||~u||||~v || cos θ in the above, we get:
||~u||||~v || cos θ = 21 (||~u − ~v ||2 − ||~u||2 − ||~v ||2 ).
p
Now by the definition of magnitude in R3 , we have ||~u||2 = ( u21 + u22 + u23 )2 = u21 +u22 +u23 . Similarly
we have ||~v ||2 = v12 + v22 + v32 and
||~u − ~v ||2 = (u1 − v1 )2 + (u2 − v2 )2 + (u3 − v3 )2 .
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
5
In that case, expanding out our equality from earlier and simplifying, we get:
1
(|~u − ~v |2 − ||~u||2 − ||~v ||2 )
2
1
= ((u1 − v1 )2 + (u2 − v2 )2 + (u3 − v2 )2 − u21 − u22 − u23 − v12 − v22 − v32 )
2
1
= (2u1 v1 + 2u2 v2 + 2u3 v3 )
2
= u1 v1 + u2 v2 + u3 v3
||~u||||~v || cos θ =
= ~u · ~v .
This proves the equality in the theorem.
Corollary 4.6. Let ~u and ~v be non-zero vectors, and let θ be the angle between ~u and ~v with 0 ≤ θ ≤ π.
Then
~u · ~v
cos θ =
.
||~u||||~v ||
√
√
Example 4.7. (a) Let√~u = ( 3, 1, 0) and √
~v = (1, 3, 0). Compute the angle θ between ~u and ~v .
(b) Let ~u = (2, −1, −2 3) and ~v = (2, −2, 3). Compute the angle θ between ~u and ~v .
Corollary 4.8. If θ is the angle between two vectors ~u and ~v , then θ =
~u · ~v = 0.
π
2
(90 degrees) if and only if
Definition 4.9. Two vectors ~u and ~v are called orthogonal if ~u ·~v = 0. (Orthogonal and perpendicular
are synonyms in two dimensions.)
Definition 4.10. Given two vectors ~u and ~v , define the othogonal projection of ~u onto ~v , denoted
proj~v ~u (Note: Rogawski uses ~u|| here), to be the vector component of ~u which lies in the direction of ~v .
We have already observed that ||~~vv|| is the unit vector which points in the same direction as ~v . Elementary geometric considerations show that the magnitude of proj~v ~u is ||~u|| cos θ, where θ is the angle
formed by ~u and ~v . So we immediately have
proj~v ~u = ||~u|| cos θ ||~~vv|| .
Theorem 4.11. For any two vectors ~u and ~v ,
proj~v ~u =
~u · ~v
~v · ~v
~v .
Example 4.12. Find the projection of ~u = (5, 1, −3) onto ~v = (4, 4, 2).
5
Cross Products
Definition 5.1. Let ~u and ~v be vectors in R3 . Define the cross product ~u × ~v to be the vector with
magnitude given by
||~u × ~v || = ||~u||||~v || sin θ,
where θ is the angle between ~u and ~v (0 ≤ θ ≤ π), and with direction given by the right-hand rule:
when you put the vectors tail to tail and let the fingers of your right hand curl from ~u to ~v , the direction
of ~u × ~v is the direction of your thumb, orthogonal to both ~u and ~v . (Note that the right-hand rule only
makes sense if ~u and ~v are not parallel; if they are parallel and θ = 0 or θ = π, check that the given
magnitude is 0 and hence ~u × ~v = ~0.)
√
Example 5.2. Find the direction and magnitude of ~u × ~v , where ~u = (1, 1, 0) and ~v = (1, 1, 2).
Example 5.3. (a) In general is ~u · ~v = ~v · ~u?
(b) In general is ~u × ~v = ~v × ~u?
6
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
Fact 5.4 (Properties of the Cross Product). Let ~u, ~v , and w
~ be any vectors in R3 and let c be any
scalar. The following properties all hold.
(a) ~u × ~v = −(~v × ~u) (anticommutative property);
(b) (c~u) × ~v = ~u × (c~v ) = c(~u × ~v );
(c) ~u × (~v + w)
~ = (~u × ~v ) + (~u × w)
~ (left distributive property of cross product over vector addition); and
(d) (~u + ~v ) × w
~ = (~u × w)
~ + (~v × w)
~ (right distributive property of cross product over vector addition).
Example 5.5. Evaluate all possible cross products of the coordinate vectors ~i, ~j, and ~k.
Example 5.6. Let ~u = (u1 , u2 , u3 ) and ~v = (v1 , v2 , v3 ) be any two vectors in R3 . Find a closed formula
for the coordinates of ~u × ~v .
Theorem 5.7 (The Cross Product as a Determinant). Let ~u = (u1 , u2 , u3 ) and ~v = (v1 , v2 , v3 ). Then


~k
~i
~j
u2 u3 ~
u1 u3 ~
u1 u2 ~


~u × ~v = det u1 u2 u3 = det
i − det
j + det
k.
v2 v3
v1 v 3
v1 v2
v1 v2 v3
Remark 5.8. Note that the “matrix” written in the formula above is not really a matrix in the sense
that students will have seen previously, as some its entries are vectors rather than real numbers. Hence
the formula above should be read as a formal determinant; that is, equality holds if one ignores
the unusual context and just performs the usual determinant computation (using the Laplace cofactor
expansion), pretending that ~i, ~j, and ~k are real numbers, and treating each scalar multiplication as
though it were just multiplication of real numbers.
Example 5.9. Find a vector orthogonal to ~u = −~i + 6~k and ~v = 2~i − 5~j − 3~k.
6
Vector-Valued Functions
Definition 6.1. A vector-valued function is a function which takes a real number for input, and
outputs a vector (typically in R2 or R3 ). We will typically denote vector-valued functions with the arrow
hat notation, e.g. ~r(t), f~(t), etc., where t is regarded as the input variable. If ~r(t) outputs vectors in
R3 , we may write
~r(t) = (x(t), y(t), z(t)),
where x, y, and z are all real-valued functions. In this way we regard ~r(t) as an ordered triple of
real-valued parametric functions. The graph of a vector-valued function ~r is the set of all possible
outputs ~r(t). (Note that this definition differs from our usual definition of a graph in previous math
courses... the input variable is not pictorially represented in the graph!) The graph of a R3 -valued
function is often called a space curve.
Example 6.2. Sketch the graphs of the following vector-valued functions.
(a) ~r(t) = (2t + 1, t − 2, 3t)
(b) f~(t) = (cos t, sin t, 5)
(c) ~g (t) = (− sin t, cos t, t)
Fact 6.3. A vector-valued function ~r(t) whose graph is the line in three-dimensional space passing
through the point (x0 , y0 , z0 ) in the direction of the vector ~v = (a, b, c) is ~r(t) = r~0 + t~v , is given by
~r(t) = (x0 , y0 , z0 ) + t(a, b, c), for −∞ < t < ∞.
Equivalently, the parametric equations of the line for ~r(t) = (x(t), y(t), z(t)) are
x(t) = x0 + at; y(t) = y0 + bt; z(t) = z0 + ct.
7
Calculus of Vector-Valued Functions
~ as t
Definition 7.1. Let a be a real number. A vector-valued function ~r(t) approaches the limit L
approaches a, written
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
7
~
lim = L,
t→a
~ = 0.
provided lim ||~r(t) − L||
t→a
The function ~r(t) is continuous at a provided ~r(a) exists, lim ~r(t) exists, and lim ~r(t) = ~r(a). The
t→a
t→a
function ~r(t) is simply called continuous if it is continuous at every point in its domain.
Fact 7.2. Let ~r(t) = (x(t), y(t), z(t)) be a vector-valued function. If lim x(t) = L1 , lim y(t) = L2 , and
t→a
t→a
lim z(t) = L3 , then
t→a
lim ~r(t) = (L1 , L2 , L3 ).
t→a
Also, ~r(t) is continuous at a point a provided x(t), y(t), and z(t) are all continuous at a.
Example 7.3. Consider the function
~r(t) = cos πt~i + sin πt~j + e−t~k for t ≥ 0.
(a) Evaluate lim ~r(t).
t→2
(b) Evaluate lim ~r(t).
t→∞
(c) At what points is ~r continuous?
Definition 7.4. Let ~r(t) = (x(t), y(t), z(t)) be a vector-valued function. We define the derivative of
~r(t), denoted ~r0 (t), to be
~r0 (t) = lim
h→0
~r(t + h) − ~r(t)
,
h
if the limit exists. If the derivative exists at a point t then we say ~r is differentiable at t. We also
d
denote the derivative by dt
~r(t).
Note that the derivative ~r0 is a vector-valued function just like ~r. For any given t, the vector ~r0 (t)
points in the same direction as the curve given by ~r(t); for this reason ~r0 (t) is called a tangent vector
at ~r(t). We regard ~r0 (t) as the rate of change of the function ~r(t); for example, if ~r(t) is a position
function in three-dimensional space, then ~r0 (t) is the associated velocity function.
Fact 7.5. Let ~r(t) = x(t)~i + y(t)~j + z(t)~k, where x, y, and z are all differentiable functions of t. Then
~r0 (t) = x0 (t)~i + y 0 (t)~j + z 0 (t)~k.
Example 7.6. Compute the derivative of the following functions.
t3
(a) ~r(t) = (t3 , 3t2 , )
6√
(b) ~r(t) = e−t~i + 10 t~j + 2 cos(3t)~k
Example 7.7. Observe the behavior of ~r = (0, t2 , t3 ) and ~r0 at t = 0.
Definition 7.8. A vector-valued function ~r(t) is called smooth on an interval if it is differentiable on
that interval, and also ~r0 (t) 6= 0 on that interval.
Let ~r(t) be a smooth vector-valued function on some interval [a, b]. The unit tangent vector for
~r(t) on [a, b] is
~r0 (t)
T~ (t) = 0
.
||~r (t)||
Example 7.9. Find the unit tangent vectors for the following functions.
(a) ~r(t) = (t2 , 4t, ln t) for t > 0.
(b) ~r(t) = (10, 3 cos t, 3 sin t) for 0 ≤ t ≤ 2π.
Fact 7.10 (Derivative Rules). Let ~u(t) and ~v (t) be differentiable vector-valued functions and let f (t) be
a differentiable scalar-valued function. Let ~a be a constant vector.
8
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
= ~0
(1)
d
a
dt ~
(2)
d
u(t)
dt (~
(3)
d
u(t))
dt (f (t)~
(4)
d
u(f (t))
dt ~
(5)
d
u(t)
dt (~
· ~v (t)) = ~u0 (t) · ~v (t) + ~u(t) · ~v 0 (t)
(6)
d
u(t)
dt (~
× ~v (t)) = ~u0 (t) × ~v (t) + ~u(t) × ~v 0 (t)
+ ~v (t)) = ~u0 (t) + ~v 0 (t)
= f 0 (t)~u(t) + f (t)~u0 (t)
= f 0 (t)~u0 (f (t))
(Product Rule)
(Chain Rule)
(Dot Product Rule)
(Cross Product Rule)
Example 7.11. Compute the following derivatives, where ~u(t) = t~i + t2~j − t3~k and ~v (t) = sin t~i +
2 cos t~j + cos t~k.
d
~v (t2 )
(a) dt
(b)
d 2
v (t)]
dt [t ~
(c)
d
u(t)
dt [~
· ~v (t)]
Example 7.12. Compute the first, second, and third derivatives of ~r(t) = (t2 , 8 ln t, 3e−2t ).
~ for which R
~ 0 = ~r. If
Definition 7.13. An antiderivative of a vector-valued function ~r is a function R
~
~r(t) = x(t)~i + y(t)~j + z(t)~k, then R(t)
= X(t)~i + Y (t)~j + Z(t)~k, where X, Y , and Z are antiderivatives
~ of ~r is called the indefinite integral
of x, y, and z respectively. The collection of all antiderivatives R
of ~r, and denoted
R
~r(t)dt.
As is the case with real-valued functions, any two antiderivatives of ~r differ only by some constant
~ So if R
~ is any antiderivative of ~r, we may write
vector C.
R
~ + C,
~
~r(t)dt = R(t)
~ is an arbitrary constant.
where C
R
t ~
−3t
~
~
√
Example 7.14. Evaluate
i + e j + (sin 4t + 1)k .
t2 + 2
Example 7.15. Find ~r(t) such that ~r0 (t) = (e2 , sin t, t) and ~r(0) = ~j.
Definition 7.16. Let ~r(t) = f (t)~i + g(t)~j + h(t)~k be a vector-valued function, where f , g, and h are
integrable on the interval [a, b]. We define the definite integral of ~r(t) across [a, b] to be the vector
Rb
Rb
Rb
Rb
~r(t)dt = [ a f (t)dt]~i + [ a g(t)dt]~j + [ a h(t)dt]~k.
a
Rπ
Example 7.17. Compute 0 [~i + 3 cos( 2t )~j − 4t~k]dt.
Definition 7.18. We say that a function ~r(t) or moves on a sphere, if ||~r(t)|| = r for some fixed
radius r ≥ 0, for every t in the domain of ~r. (In other words the graph of ~r(t) is entirely contained in
the sphere of radius r.)
Example 7.19. An object moves on a trajectory described by ~r(t) = (3 cos t, 5 sin t, 4 cos t), for 0 ≤ t ≤
2π.
(a) Show that the object moves on a sphere, and find the radius of the sphere.
(b) Find functions which model the velocity and speed of the object.
Theorem 7.20. Suppose ~r(t) is differentiable and moves on a sphere, i.e. ||~r(t)|| = r for some fixed r,
for all t in the domain of ~r. Then the position vector ~r(t) and the velocity vector ~r0 (t) are orthogonal at
every t in the domain of ~r.
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
Proof. First notice that since ||~r(t)||2 = ~r(t) · ~r(t) = r2 is a constant function, we have
So, applying the dot product rule for derivatives, we have:
9
d
r(t)
dt ~
· ~r(t) = 0.
d
~r(t) · ~r(t)
dt
= ~r0 (t) · ~r(t) + ~r(t) · ~r0 (t)
0=
= 2~r0 (t) · ~r(t).
So we must have ~r0 (t) · ~r(t) = 0, i.e. ~r0 and ~r are orthogonal.
8
Length of Curves
Fact 8.1. Let ~r(t) = (x(t), y(t), z(t)) be a vector-valued function, where x0 , y 0 , and z 0 are continuous.
The graph of ~r(t) is a space curve in R3 . Assume that the curve is traversed only once for a ≤ t ≤ b.
The arc length s of the curve between the points ~r(a) and ~r(b) is
Rb
Rbp
s = a ||~r0 (t)||dt = a x0 (t)2 + y 0 (t)2 + z 0 (t)2 dt.
We will omit the construction of the above formula here, but please see the intro to Section 8.1 in
Rogawski for a plausibility argument (in two dimensions).
Example 8.2. Compute the circumference of a circle of radius a.
Example 8.3. Find the length of the hypocycloid parametrized by ~r(t) = (cos3 t, sin3 t), where 0 ≤ t ≤
2π.
Example 8.4. An eagle rises at a rate of 100 vertical ft/min on a helical path given by ~r(t) =
(250 cos t, 250 sin t, 100t),
√ where ~r is measured in feet and t is measured in minutes. How far does it
travel in 10 minutes? ( 2502 + 1002 ≈ 269.)
Example 8.5. Let f~(t) = () (REPAIR THIS) and let g(t) = ().
(1) Graph both f~ and ~g and compare the graphs.
(2) Compute the arc length of the graphs of both f~ and ~g traversed from t = 0 to t = 4 and compare.
(3) Compute the derivatives of f~ and ~g and compare.
Definition 8.6. Let ~r(t) be a smooth vector-valued function for all t ≥ a. Define the arc length
function (based at a) by
Rt
s(t) = a ||~r0 (u)||du.
We say that ~r is parametrized by arc length if s(t) = t − a for all t ≥ a. (Keep in mind we will
usually define ~r in such a way that a = 0.)
Keep in mind for the following fact, we usually define ~r in such a way that a = 0.
The next theorem says that a function is parametrized by arc length if and only if it moves at a
constant speed of 1.
Theorem 8.7. Let ~r(t) be a smooth vector-valued function for all t ≥ a, and let s(t) be the associated
arc length function. Define the velocity function to be ~v (t) = ~r0 (t), and the speed function to be
||~v (t)||. Then ~r is parametrized by arc length if and only if ||~v (t)|| = 1 for all t ≥ a.
Proof. Suppose ~r(t) is parametrized by arc length, so for each t ≥ a we have
Rt
s(t) = a ||~v (u)||du = t − a.
Taking derivatives on both sides above and applying the fundamental theorem of calculus, we get
||~v (t)|| = 1,
10
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
as claimed. Conversely, suppose that ||~v (u)|| = 1 for all u ≥ a. Then we have for each t ≥ a:
Rt
Rt
s(t) = a ||~v (u)||du = a 1du = [u]ta = t − a.
This proves the theorem.
Example 8.8. Consider the helix parametrized by ~r(t) = (cos 4t, sin 4t, 3t) for t ≥ 0.
(1) Explicitly find the arc length function s(t) for t ≥ 0. Is ~r parametrized by arc length?
(2) Find a new function ~r1 (t) which has the same graph as ~r(t), but which is parametrized by arc
length.
Theorem 8.9. Let ~r(t) be a vector valued function for a ≤ t ≤ b, and let s(t) be the associated arc
length function. Suppose that s is an increasing function of t, so there exists an inverse function s−1 (t)
for which s(s−1 (t)) = s−1 (s(t)) = t. Define a new function ~r1 by
~r1 (t) = ~r(s−1 (t)).
Then ~r1 has the same graph as ~r for s(a) ≤ t ≤ s(b), and is parametrized by arc length.
Proof. It is clear that ~r has the same outputs for a ≤ t ≤ b as ~r1 does for s(a) ≤ t ≤ s(b). So the two
functions have the same graph on these intervals. So to prove the theorem, it remains only to show that
~r1 is parametrized by arc length.
To that end, observe that by the chain rule and the formula for derivatives of inverse functions, we
have the following:
~r10 (t) =
d
~r(s−1 (t))
dt
d −1
s (t)
dt
1
= ~r0 (s−1 (t)) · 0 −1
.
s (s (t))
= ~r0 (s−1 (t)) ·
Note that s0 in the above expression is just ||~r0 || by the fundamental theorem of calculus. Combining
the lines from above, we get:
||~r10 (t)|| = ~r0 (s−1 (t)) ·
1
||~
r 0 (s−1 (t))||
= 1.
This shows ~r1 has constant speed 1, and hence is parametrized by arc length by Theorem 8.7.
9
Curvature
Definition 9.1. Recall from Definition 7.8 that the unit tangent vector associated to a smooth vector~r0 (t)
.
valued function ~r(t) is the function T~ (t) = 0
||~r (t)||
The curvature of a smooth parametrized curve, denoted by κ, is the magnitude of the rate of change
of T~ with respect to arc length s. (A picture helps here.) In other words, if s denotes arc length,
~
κ(s) = || ddsT ||.
We wish to have a method of computing curvature which works for any parameter t, and not just for
the arc length parameter s as above. The following theorem gives us such a method.
Theorem 9.2. Let ~r(t) be a smooth parametrized curve, where t is any parameter. If ~v = ~r0 is the
velocity and T~ is the unit tangent vector, then the curvature is
κ(t) =
~
1
dT
||~
v || || dt ||
=
~ 0 (t)||
||T
||~
r 0 (t)|| .
Proof. Let s denote arc length of ~r. Note that the relationship between the derivatives of T~ with respect
to s and to t is given by the chain rule:
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
~
dT
dt
=
~
dT
ds
·
ds
dt .
Now by the Fundamental Theorem of Calculus, we have
~
dT
ds
=
1
||~
v ||
·
11
ds
dt
= ||~v ||, and hence, solving for
=
~ 0 (t)||
||T
||~
r 0 ||
~
dT
ds ,
we get
~
dT
dt .
So by our original definition of curvature, we have
~
κ(t) = || ddsT || =
~
1
dT
||~
v || || dt ||
as claimed.
Example 9.3. Which has greater curvature: A circle with a large radius or a circle with a small radius?
Example 9.4. Compute the curvature function κ(t) for ~r(t) = (1, et , t).
Theorem 9.5. Let ~r be a smooth vector-valued function. Let ~v = ~r0 be the velocity function and let
~a = ~v 0 = ~r00 be the acceleration function. Then the curvature κ of ~r is
κ=
Proof. Since T~ =
~
v
||~
v || ,
||~a × ~v ||
.
||~v ||3
write
~v = ||~v ||T~ .
Differentiate both sides above using the product rule.
~
d
~a = ( dt
||~v ||)T~ + ||~v || ddtT .
Next compute the cross product ~a × ~v :
"
#
~
d
d
T
~a × ~v =
||~v || T~ + ||~v ||
× [||~v ||T~ ]
dt
dt
d
dT~
~
=
||~v ||T × ||~v ||T~ + ||~v ||
× ||~v ||T~
dt
dt
!
~
d
T
2
= ~0 + ||~v ||
× T~
dt
!
dT~
2
~
= ||~v ||
×T .
dt
~
Now notice that T~ and ddtT are orthogonal since T~ moves in a sphere! (See Theorem 7.20.) So the
~
~
angle between T~ and ddtT is π2 , and hence the cross product ddtT × T~ has magnitude
~
~
~
~
|| ddtT × T~ || = || ddtT ||||T~ || sin π2 = || ddtT || · 1 · 1 = || ddtT ||.
Putting everything together, we have
12
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
dT~
× T~
dt
||~a × ~v || = ||~v ||2
= ||~v ||2
dT~
dt
= ||~v ||3 ·
1
||~v ||
dT~
dt
= ||~v ||3 κ.
So κ =
||~a × ~v ||
as claimed.
||~v ||3
Example 9.6. Find the curvature of the parabola ~r(t) = (t, at2 ).
Example 9.7. Find the curvature of the helix ~r(t) = (a cos t, a sin t, bt), where a, b > 0 are real numbers.
10
Motion in Space
Example 10.1. Consider the two-dimensional motion given by the position vector
~r(t) = (3 cos t, 3 sin t) for 0 ≤ t ≤ 2π.
(a) Sketch the trajectory of the object.
(b) Find the velocity of the object.
(c) Find the speed of the object.
(d) Find the acceleration of the object.
(e) Sketch the position, velocity, and acceleration vectors for t = 0, t = π2 , t = π, and t =
3π
2 .
Example 10.2. Find ~r(t) if ~a(t) = 2~i + 12t~j, ~v (0) = 7~i, and ~r(0) = 2~i + 9~k.
Fact 10.3.
(1) Force (vector) equals mass (scalar) times acceleration (vector), i.e. F~ = m~a.
(2) Gravity accelerates objects downward at a rate of approximately 32 ft/s2 , or 9.8 m/s2 .
Example 10.4. A baseball is hit from 3 ft above home plate with an initial velocity in ft/s of ~v (0) =
(80, 80).
(a) Find functions ~v (t) and ~r(t) which model the position and velocity of the ball between the time it is
hit and the time when it first hits the ground. (Neglect all forces acting on the ball except gravity.)
(b) Show that the trajectory of the ball is a segment of a parabola.
(c) Assuming a flat playing field, how far does the ball travel horizontally? Plot the trajectory of the
ball. (Answer: t = 5.04s, x = 403ft.)
(d) What is the maximum height of the ball? (Answer: y = 103ft.)
(e) Does the ball clear a 20-ft fence that is 380 ft from home plate (directly under the path of the ball)?
(t = 4.75s, y − 22ft.)
Example 10.5. A bullet is fired from the ground at an angle of 60◦ above the horizontal. What initial
speed v0 must the bullet have in order to hit a point 150 m high on a tower located 250 m away?
Example 10.6. A small projectile is fired over horizontal ground in an easterly direction with an initial
speed of |~v0 | = 300 m/s at an angle of α = 30◦ above the horizontal. A crosswind blows from south to
north producing an acceleration of the projectile of 0.36 m/s2 to the north.
(a) Where does the projectile land? (t = 30.6 s, (x, y, z) = (7953, 169, 0) m.)
(b) In order to correct for the crosswind and make the projectile land due east of the launch site, at
what angle from due east must the projectile be fired? (−1.21◦ .)
~ is
Definition 10.7. Let ~r(t) be a smooth vector-valued function. The unit normal vector N
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
~ (t) =
N
13
~ 0 (t)
T
~ 0 (t)|| .
||T
Intuitively, the unit normal vector points in the direction the graph of ~r is curving toward.
Example 10.8. Find the unit normal vector at t =
π
4
to the helix ~r(t) = (a cos t, a sin t, bt).
Fact 10.9. Let ~r be a smooth vector-valued function with unit tangent vector T~ and principal unit normal
~ . Then T~ and N
~ are orthogonal.
vector N
~
Proof. Since T~ moves in a sphere, it is orthogonal to its own derivative T~ 0 by Theorem 7.20. Since N
0
~
~
~
points in the same direction as T by definition, T and N are orthogonal.
Theorem 10.10. Let ~r be a smooth vector-valued function describing the motion of an object through
space. Then the acceleration of the object ~a = ~r00 has a unique representation as the sum of its tangential
component aT~ and its normal component aN~ :
~ + a ~ T~ ,
~a = aN~ N
T
where aN~ = κ||~v ||2 =
||~a × ~v ||
and aT~ = s00 .
||~v ||
Proof. We begin with the fact that T~ = ||~~vv|| and hence ~v = ||~v ||T~ = s0 · T~ . Differentiating both sides
and using the product rule and chain rule respectively, we get
~a = ~v 0
d 0 ~
(s · T )
dt
= s0 · T~ 0 + s00 · T~
=
= ||~v ||T~ 0 + s00 · T~ .
~ . Also ||T~ 0 || = κ||~v ||, so T~ 0 = κ||~v ||N
~ . Substituting this
~ = T~0 , we have T~ 0 = ||T~ 0 ||N
Now since N
~ 0 ||
||T
equality into that above, we end up with
~ + s00 · T~
~a = κ||~v ||2 N
and we are done.
Example 10.11. The driver of a car follows the parabolic trajectory ~r(t) = (t, t2 ) for −2 ≤ t ≤ 2
through a sharp bend in the road. Find the tangential and normal components of the acceleration of
the car.
Theorem 10.12. Let aT~ and aN~ be the tangential component and normal component of ~a for some
vector-valued function ~r, as in the previous theorem. Then
·~
v
aT~ T~ = proj~v ~a = ~~av·~
v
v ~
and
~ = ~a − a ~ T~ = ~a −
aN~ N
T
Proof. The first equality is because
~
a·~
v
~
v ·~
v
~v .
14
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
~ ) · T~
~a · T~ = (aT~ T~ + aN~ N
~ · T~ )
= aT~ (T~ · T~ ) + aN~ (N
= aT~ ||T~ ||2 + aN~ (0)
and hence aT~ T~ = (~a · T~ )T~ =
11
~
a·~
v
||~
v ||
~
v
||~
v ||
= aT~ ,
~
a·~
v
= ||~
v = proj~v ~a. Then the second equality is immediate.
v ||2 ~
Functions of Several Variables
Definition 11.1. A function in two variables f (x, y) is a function which assigns a unique (real)
output to each input pair (x, y) from a particular set D in R2 . The set D is called the domain of f
and is often implicitly (rather than explicitly) described. The range of f is the set of all real numbers
z = (x, y) which are assumed as (x, y) ranges over the domain D.
Definition 11.2 (Set-Builder Notation). To facilitate the next example and the homework problems,
we recall for the student the use of set-builder notation. For convenience we will describe informally
using examples, rather than give a formal definition. For an example, suppose we wish to formally
describe the set D of all ordered pairs (x, y) for which x is twice y. Then we may write
A = {(x, y) : x = 2y}.
The above notation should be read as The set of all (x, y) in R2 such that x = 2y, which clearly and
precisely defines our set A. For another example, we could write
Y = {x : 5 ≤ x < 10},
which reads The set of all x in R such that 5 is less than or equal to x and x is strictly less than 10.
The student should easily verify that, using the interval notation, Y = [5, 10). In general, given a set A
and a precise mathematical sentence P (x) about a variable x, the set-builder notation should be read
as follows.
{
x
:
P (x)}
“The set of all elements x such that sentence P (x) is true for the element x.
Note that for our present purposes we implicitly assume that variables x range over real numbers in
R, pairs (x, y) range over R2 , triples (x, y, z) range over R3 , etc.
p
Example 11.3. Let g(x, y) = 4 − x2 − y 2 . Find the domain and range of g.
Remark 11.4. Notice that for an R3 -valued function ~r(t), our definition of the graph of ~r is
{(x, y, z) : there exists a t such that ~r(t) = (x, y, z)},
i.e. the graph of ~r equals the range of ~r.
Conversely, for a real-valued function f (x) like those the student will have worked with in previous
courses, our definition of the graph of f is
{(x, y) : f (x) = y}.
Definition 11.5. Let f (x, y) be a function in two variables. Define the graph of f to be the set
{(x, y, z) : f (x, y) = z}.
p
Example 11.6. Sketch the graph of g(x, y) = 4 − x2 − y 2 .
Definition 11.7. Given a fixed point P0 = (x0 , y0 , z0 ) in R3 and a non-zero vector ~n = (a, b, c) in R2 ,
the set of all points P = (x, y, z) in R3 for which P − P0 is orthogonal to ~n is called a plane. (The
vector ~n is called the normal vector to the plane.)
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
15
Example 11.8. Given a point P0 and a normal vector ~n as in the above definition, find an equation in
three variables whose graph is the plane determined by P0 and ~n.
Fact 11.9. The plane passing through P0 = (x0 , y0 , z0 ) with normal vector ~n = (a, b, c) is the set of all
points (x, y, z) which satisfy the equation
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0.
The plane is the graph of a function if and only if c 6= 0, which happens if and only if ~n is not
orthogonal to ~k = (0, 0, 1).
Definition 11.10. Working in R3 : the xy-plane is the plane passing through (0, 0, 0) with normal
vector ~k = (0, 0, 1); the xz-plane is the plane passing through (0, 0, 0) with normal vector ~j = (0, 1, 0);
and the yz-plane is the plane passing through (0, 0, 0) with normal vector ~i = (1, 0, 0). These three are
called the coordinate planes in R3 .
Example 11.11. Find an equation of the plane that passes through the points (2, −1, 3), (1, 4, 0), and
(0, −1, 5).
Definition 11.12. Two distinct planes are called parallel if their respective normal vectors are parallel.
Two planes are orthogonal if their respective normal vectors are orthogonal.
Example 11.13. Which of the following distinct planes are parallel and which are orthogonal?
Q: 2x − 3y + 6z = 12
R: −x + 23 y − 3z = 14
S: 6x + 8y + 2z = 1
T : −9x − 12y − 3z = 7
Example 11.14. Sketch the graph of f (x, y) = 2x2 + 5y 2 .
Definition 11.15. If f (x, y) is a function in two variables, then a trace of the graph of f is the
intersection of the graph with any plane parallel to a coordinate plane. The traces which lie in the
coordinate planes are called the xy-trace, the xz-trace, and the yz-trace respectively.
Example 11.16. Use traces to sketch the graphs of the following functions.
(a) f (x, y) = x − y 2
(b) f (x, y) = x sin y
Definition 11.17. Let f (x, y) be a function. Given any fixed real number z0 , the level curve of f at
z0 is the set of all (x, y) in R2 for which f (x, y) = z0 .
Example 11.18. Sketch some of the level curves of the following functions.
(a) f (x, y) = y − x2 − 1
2
2
(b) f (x, y) = e−x −y
12
Limits and Continuity for Functions of Several Variables
Definition 12.1. A disk in R2 is the set {(x, y) : ||(x, y) − (a, b)|| < r} for some fixed point (a, b) and
some fixed radius r.
Let R be a subset of R2 . A point P in R is called an interior point if there is some disk about P
which is contained entirely in R.
A point Q is called a boundary point if every disk containing Q contains both a point in R and a
point not in R.
A region R is open if it consists entirely of interior points. A region is closed if it contains all its
boundary points.
Example 12.2. Are the following sets open? Closed? (Both? Neither?)
(a) Any disk in R2 .
(b) The set R = {(x, y) : 0 < x < 1, 0 < y < 1}.
(c) The set R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.
16
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
(d) The set R = {(x, y) : 0 ≤ x < 1, 0 ≤ x < 1}.
Definition 12.3. A punctured disk in R2 is a disk with its center point removed. (Note that any
punctured disk is still open.)
Definition 12.4. Let f (x, y) be a function in two variables. The function f has the limit L as (x, y)
approaches (a, b), denoted
lim
f (x, y) = L,
(x,y)→(a,b)
if for every > 0, there exists a punctured disk P centered at (a, b) such that
|f (x, y) − L| < whenever (x, y) is in P .
A function f (x, y) is continuous at a point (a, b) provided f (a, b) exists,
lim
f (x, y) exists,
(x,y)→(a,b)
and
lim
f (x, y) = f (a, b). We say f is continuous if f is continuous at every point in its domain.
(x,y)→(a,b)
Example 12.5. Evaluate
lim
(3x2 y +
√
xy), if it exists.
(x,y)→(2,8)
Example 12.6. Evaluate
lim
3x−2 y 2 , if it exists.
(x,y)→(0,0)
Example 12.7. Evaluate
xy − 4y 2
√
√ , if it exists.
x−2 y
(x,y)→(4,1)
Example 12.8. Evaluate
(x + y)2
, if it exists.
(x,y)→(0,0) x2 + y 2
13
lim
lim
Partial Derivatives
Definition 13.1. Let f (x, y) be a function in two variables. The partial derivative of f with respect
to x is
f (x + h, y) − f (x, y)
δ
,
fx (x, y) = δx
f (x, y) = lim
h→0
h
provided the limit exists. The partial derivative of f with respect to y is
fy (x, y) =
δ
δy f (x, y)
f (x, y + h) − f (x, y)
,
h→0
h
= lim
provided the limit exists.
Example 13.2. Let f (x, y) = x2 − y 2 + 4.
δf
(a) Compute δf
δx and δy .
(b) Evaluate each derivative at (2, −4).
Example 13.3. Compute the partial derivatives of the following functions.
(a) f (x, y) = sin xy
(b) g(x, y) = x2 exy
Definition 13.4. Let f (x, y) be a function in two variables. The second-order partial derivatives
of f are the following four functions (written with both available notations):
δf
δ2 f
δ
δx δx = δx2 of (fx )x = fxx ;
δf
δ2 f
δ
δy δy = δy 2 of (fy )y = fyy ;
δf
δ2 f
δ
δx δy = δxδy of (fy )x = fyx ;
δf
δ2 f
δ
δx δy = δyδx of (fx )y = fxy .
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
17
The latter two derivatives above are called mixed partial derivatives.
Example 13.5. Compute the second-order partial derivatives of f (x, y) = 3x4 y − 2xy + 5xy 3 .
Example 13.6 (Special Example!). Let

2
2
 xy(x − y ) , if (x, y) 6= (0, 0);
f (x, y) =
x2 + y 2

0,
if (x, y) = (0, 0).
(a) Is f continuous?
(b) Compute fxy (0, 0) and fyx (0, 0). Are the mixed partial derivatives equal to one another?
To answer the above questions we need the following quick lemma, which follows from L’Hospital’s
rule.
Lemma 13.7. Let f (x) be a function of one variable, differentiable everywhere except possibly at a single
point x = a. Suppose that lim f 0 (x) exists. Then f is differentiable at x = a and f 0 (a) = lim f 0 (x).
x→a
x→a
Proof. By the definition of the derivative, and an application of L’Hospital’s rule (applied to the variable
h) we have:
f (a + h) − f (a)
h→0
h
f 0 (a + h) − 0
= lim
h→0
1
= lim f 0 (a + h)
f 0 (a) = lim
h→0
= lim f 0 (x).
x→a
Since the latter limit exists by our hypothesis, this proves the lemma.
Solution to Example 13.6. (a) First we prove that f is indeed continuous. Since f is a rational function
with positive denominator at every point (x, y) 6= (0, 0), it is clear that f is continuous at every point
except possibly the origin. So we just need to check that f is continuous at (0, 0). By our definition of
xy(x2 − y 2 )
continuity, this amounts to verifying that
lim
= 0 = f (0), i.e. that the function value
x2 + y 2
(x,y)→(0,0)
agrees with the function’s limit at (0, 0).
To see this, we consider approaching (0, 0) along any straight-line path through the origin. The line
through the origin at an arbitrary angle θ through (0, 0) is parametrized by
~r(t) = (t cos θ, t sin θ).
Now compute the values of f along the graph of ~r:
f (~r(t)) = f (t cos θ, t sin θ)
t cos θt sin θ(cos2 θ − sin2 θ)
cos2 θ + sin2 θ
t2 ( 12 sin 2θ)(cos 2θ)
=
1
1 2
= t sin 2θ cos 2θ.
2
1 2
This means that |f (~r(t))| = | 2 t sin 2θ cos 2θ| ≤ 21 t2 | sin 2θ|| cos 2θ| ≤ 12 t2 (1)(1) = 12 t2 (regardless of
the choice of θ!). So let > 0. Choose P to be a punctured disk of radius R so small that 12 R2 < .
Then for every (x, y) in P we have (x, y) = ~r(t) for some θ and some t < R, and hence
=
18
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
|f (x, y) − 0| = |f (x, y)| = |f (~r(t))| ≤ 21 t2 ≤ 12 R2 < .
This shows that
lim
f (x, y) = 0 = f (0, 0) as we intended. So f is continuous.
(x,y)→(0,0)
(b) Our next step is to compute the partial derivatives. We start by finding fx . For all (x, y) 6= (0, 0),
we just use the quotient rule:
δ x3 y − xy 3
δx
x2 + y 2
(x2 + y 2 )(3x2 y − y 3 ) − (x3 y − xy 3 )(2x)
=
(x2 + y 2 )2
3x4 y − x2 y 3 + 3x2 y 3 − y 5 − 2x4 y + 2x2 y 3
=
(x2 + y 2 )2
x4 y + 4x2 y 3 − y 5
=
.
(x2 + y 2 )2
fx (x, y) =
To find the value of fx (0, 0), we first compute
lim
(x,y)→(0,0)
fx (x, y). As in part (a), consider the approach
along the parametrized path ~r(t) = (t cos θ, t sin θ) (where θ is arbitrary). We have
fx (~r(θ)) =
t5 cos4 θ sin θ + 4t5 cos2 θ sin3 θ − t5 sin5 θ
(cos2 θ + sin2 θ)2
= t5 (cos4 θ sin θ + 4 cos2 θ sin3 θ − sin5 θ).
It follows from the line above that |fx (~r(θ))| ≤ 6|t5 |. So by a similar argument to the one above in part
(a), we can check that
lim
fx (x, y) = lim 6|t5 | = 0. Now Lemma 13.7 implies that fx (0, 0) = 0. So
(x,y)→(0,0)
t→0
we have computed:
 4
2 3
5
 x y + 4x y − y ,
2
2
2
(x + y )
fx (x, y) =

0,
if (x, y) 6= (0, 0);
if (x, y) = (0, 0).
An extremely similar computation shows that:
 5
3 2
4
 x − 4x y − xy , if (x, y) 6= (0, 0);
(x2 + y 2 )2
fy (x, y) =

0,
if (x, y) = (0, 0).
Now we are equipped to finish the problem. Check via plug and chug that
fx (0, 0 + h) − fx (0, 0)
h
−h5 /h4
= lim
= −1
h→0
h
fxy (0, 0) = lim
h→0
but
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
fy (0 + h, 0) − fy (0, 0)
h
h5 /h4
= lim
= 1.
h→0
h
This shows that the two mixed partial derivatives fxy and fyx do not agree at the origin (0, 0).
19
fyx (0, 0) = lim
h→0
The above example shows that in general it is possible for fxy to not be equal to fyx . However, the
example given above may seem somewhat “pathological,” and we hope there is a chance that most natural
functions we run across on a day-to-day basis have the nice property that fxy = fyx . The following
famous theorem says that this is indeed the case; if the second derivatives are merely continuous, then
we have equality of mixed partials. The proof is not terribly difficult, but is beyond the scope of our
immediate goals and is omitted.
Theorem 13.8 (Clairaut’s Theorem). Assume that f is defined on an open set D of R2 , and fxy and
fyx are continuous at every point of D. Then fxy = fyx at every point of D.
14
Differentiability and Tangent Planes
In this section we wish to give a meaningful definition of a function in multiple variables being
differentiable. This notion is a bit more demanding in the multivariable (Cal III) setting than it was
in the single-variable (Cal I/II) setting.
As a warmup, let’s consider an alternative definition of differentiability for single-variable functions,
different from what the student would have learned in Cal I.
Definition 14.1. A single-variable function f (x) is differentiable at the point a if there exists a tangent line L to f at a. To be precise, L is a function of the form
L(x) = m(x − a) + f (a),
so the graph of L is a line of slope m passing through (a, f (a)), which is tangent to f in the sense that
the secant lines connecting (a, f (a)) and (a + h, f (a + h)) get closer to the slope m as h goes to 0, i.e.
lim
h→0
f (a + h) − f (a) − mh
f (a + h) − f (a)
− m = lim
= 0.
h→0
h
h
In this case we set f 0 (x) = m.
It is not too hard to check that the previous definition is equivalent to the one the student has already
f (a + h) − f (a)
seen, i.e. the tangent line exists at a if and only if f 0 (x) = lim
exists.
h→0
h
Now consider a multivariable function f (x, y). We want f to be differentiable at a point (a, b)
only if there exists a tangent plane L to f at (a, b). The tangent plane at (a, b) should give a good
approximation to the secant lines connecting (a, b, f (a, b)) and (a + h1 , b + h2 , f (a + h1 , b + h2 )) for all
choices of (small) ~h = (h1 , h2 ). (A picture helps here.) Notice that the slope of this secant line is given by
f (a + h1 , b + h2 ) − f (a, b)
.
||~h||
So we want our tangent plane to be a planar function passing through (a, b, f (a, b)), i.e. a function
of the form
L(x, y) = m1 (x − a) + m2 (x − b) + f (a, b)
and with the property that the secant lines connecting (a, b, f (a, b)) and (a + h1 , b + h2 , f (a + h1 , b + h2 ))
get closer and closer to their corresponding ones on L, as ~h = (h1 , h2 ) goes to (0, 0) (for all choices of
20
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
~h = (h1 , h2 )). Now given any particular ~h, check that the slope of the associated secant line on L is
L(a + h1 , b + h2 ) − L(a, b)
m1 h1 + m2 h2
=
.
~
||h||
||~h||
Check that if such an approximation happens, it must be the case that fx (a, b) and fy (a, b) both
exist, and that m1 = fx (a, b) and m2 = fy (a, b). So we arrive at the following definition, which will be
our permanent one.
Definition 14.2. Let f (x, y) be a function in two variables. We call f differentiable at a point (a, b)
if fx (a, b) and fy (a, b) exist, and in addition
f (a + h1 , b + h2 ) − f (a, b) − (fx (a, b)h1 + fy (a, b)h2 )
= 0.
||(h1 , h2 )||
(h1 ,h2 )→0
lim
If this is the case the tangent plane to f at (a, b) is the graph of the function
L(x, y) = fx (a, b)(x − a) + fy (a, b)(y − b) + f (a, b).
As was the case in Clairaut’s theorem, we get nice results if the partial derivatives are continuous on
an open set.
Theorem 14.3. Let f (x, y) be a function in two variables. If fx and fy exist and are continuous at
every point of some open set D, then f (x, y) is differentiable at every point of D.
Example 14.4. Show that f (x, y) = 5x + 4y 2 is differentiable, and find the equation of the tangent
plane at (2, 1).
15
The Chain Rule for Paths
Theorem 15.1 (Chain Rule for Paths). Let f be a differentiable function in two variables and let
~r(t) = (x(t), y(t)) be a differentiable vector valued function. Then the composition f (~r(t)) is differentiable, and has derivative
d
r(t))
dt f (~
= fx (~r(t))x0 (t) + fy (~r(t))y 0 (t).
Written in shorthand with the Leibniz notation, we have
d
δf dx δf dy
f (~r(t)) =
+
.
dt
δx dt
δy dt
Example 15.2. Let f (x, y) = x2 − 3y 2 + 20 and let ~r(t) = (2 cos t, 2 sin t). Find
it at t = π4 .
Example 15.3. Compute
16
d
c(t)),
dt f (~
d
r(t))
dt f (~
and evaluate
where f (x, y, z) = xy + z 2 ) and ~c(t) = (cos t, sin t, t).
Directional Derivatives and the Gradient
Definition 16.1. Let f be differentiable at (a, b) and let ~u = (cos θ, sin θ) be a unit vector in R2 . The
directional derivative of f at (a, b) in the direction of ~u is
f (a + h cos θ, b + h sin θ) − f (a, b)
D~u f (a, b) = lim
,
h→0
h
provided the limit exists.
Theorem 16.2. Let f be differentiable at (a, b) and let ~u = (u1 , u2 ) be a unit vector in R2 . Then
D~u f (a, b) = (fx (a, b), fy (a, b)) · (u1 , u2 ).
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
21
Proof. Define a new function g(s) (real inputs and real outputs) by the rule
g(s) = f (a + su1 , b + su2 ).
Geometrically, the graph of g is a “cross-section” of the graph of f passing through the point (a, b)
parallel to vector ~u. It is clear from the definition of the directional derivative that
D~u f (a, b) = g 0 (0).
Now setting x(s) = a + su1 and y(s) = b + su2 (so g(s) = (x(s), y(s)) and applying the chain rule for
paths, we get
D~u f (a, b) = g 0 (0)
= fx (x(0), y(0))x0 (0) + fy (x(0), y(0))y 0 (0)
= fx (a, b)u1 + fy (a, b)u2 )
= (fx (a, b), fy (a, b)) · (u1 , u2 ).
Example 16.3. Consider the paraboloid z = f (x, y) = 41 (x2 +2y 2 )+2 and the unit vectors ~u = ( √12 , √12 )
and ~v = ( 12 , −
√
3
2 ).
(a) Find the directional derivative of f at (3, 2) in the directions of ~u and ~v .
(b) Graph the surface and interpret the directional derivatives.
Definition 16.4. Let f (x, y) be differentiable. The gradient of f at (x, y) is the function
∇f (x, y) = (fx (x, y), fy (x, y)).
Remark 16.5. Observe that ∇f is our first example of a vector-valued function in two variables, i.e. it
takes vectors in R2 for input and returns vectors in R2 as output. We will study functions of this type
more closely when we look at vector fields later in the course.
Fact 16.6 (Rephrasing of the Chain Rule for Paths). If f is a differentiable function of two variables
and ~r is a differentiable R2 -valued function, then f (~r(t)) is differentiable and
d
r(t))
dt f (~
= ∇f (~r(t)) · ~r0 (t).
Example 16.7. Find ∇f and ∇f (3, 2) for f (x, y) = x2 + 2xy − y 3 .
Example 16.8. Let f (x, y) = 3 −
x2
10
+
xy 2
10 .
(a) Compute ∇f (3, −1).
(b) Compute D~u f (3, −1) where ~u = ( √12 , √12 ).
(c) Compute the directional derivative of f at (3, −1) in the direction of the vector (3, 4).
Theorem 16.9. Let f be differentiable at (a, b).
(1) f has its maximum rate of increase at (a, b) in the direction of the gradient ∇f (a, b). The rate
of increase in this direction is |∇f (a, b)|.
(2) f has its maximum rate of decrease at (a, b) in the direction of −∇f (a, b). The rate of decrease
in this direction is −|∇f (a, b)|.
(3) If ~u is orthogonal to ∇f (a, b), then D~u f (a, b) = 0.
22
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
Proof. For any unit vector ~u, we have
D~u = ∇f (a, b) · ~u
= ||∇f (a, b)||||~u|| cos θ
= ||∇f (a, b)|| cos θ,
where θ is the angle between ∇f (a, b) and ~u. Then cos θ is maximized when θ = 0 and minimized when
θ = π, which proves statements (1) and (2) above. If ∇f (a, b) and ~u are orthogonal then θ = π2 and
hence cos θ = 0; this shows statement (3).
Example 16.10. Consider the bowl-shaped paraboloid z = f (x, y) = 4 + x2 + 3y 2 .
(a) If you are located at the point (2, − 21 , 35
4 ) on the paraboloid, in which direction should you move in
order to ascend the surface at the maximum rate? How quickly will you ascend?
(b) If you are at the point (3, 1, 16), in which directions may you walk in order to neither gain nor lose
height?
17
Optimization in Several Variables
Definition 17.1. Let f be a function in two variables. We say f has a local maximum at (a, b) is
there is some disk D containing (a, b) such that f (a, b) ≥ f (x, y) for all (x, y) in D. We say that f has
a local minimum at (a, b) if there is some disk D containing (a, b) such that f (a, b) ≤ f (x, y) for all
(x, y) in D. In either case, we say that f has a local extremum at (a, b).
A point (a, b) is a critical point of f if either
(1) fx (a, b) = fy (a, b) = 0 or
(2) one (or both) of fx and fy does not exist at (a, b).
Fact 17.2. If f has a local maximum or minimum at (a, b), then (a, b) is a critical point of f .
Example 17.3. Find the critical points of f (x, y) = xy(x − 2)(y + 3).
Definition 17.4. Let f be a function in two variables. We say that f has a saddle point at (a, b)
if (a, b) is a critical point, but for every disk D containing (a, b) there are points (x, y) in D for which
f (x, y) > f (a, b) and points (x, y) in D for which f (x, y) < f (a, b) (in other words f has neither a min
nor a max at (a, b)).
Theorem 17.5 (Second Derivative Test). Suppose that the second partial derivatives of f (x, y) are
continuous in a disk containing (a, b), where fx (a, b) = fy (a, b) = 0. Set
2
D(x, y) = [fxx fyy − fxy
](x, y).
(1)
(2)
(3)
(4)
If
If
If
If
D(a, b) > 0 and fxx (a, b) < 0, then f has a local maximum value at (a, b).
D(a, b) > 0 and fxx (a, b) > 0, then f has a local minimum value at (a, b).
D(a, b) < 0, then f has a saddle point at (a, b).
D(a, b) = 0, the test is inconclusive.
Definition 17.6. The quantity D(x, y) in the above
Theorem 17.5 is called the discriminant of f .
fxx fxy
D(x, y) is the determinant of the Hessian matrix
.
fyx fyy
Example 17.7. Classify all the critical points of f (x, y) = x2 + 2y 2 − 4x + 4y + 6.
Example 17.8. Classify all the critical points of f (x, y) = xy(x − 2)(y + 3).
Example 17.9. A shipping company handles rectangular boxes provided the sum of the length, width,
and height of the box does not exceed 96 in. Find the dimensions of the box that meets the condition
at has the largest volume.
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
23
Definition 17.10. If f (x, y) ≤ f (a, b) for all (x, y) in the domain of f , then f has an absolute
maximum at (a, b). If f (x, y) ≥ f (a, b) for all (x, y) in the domain of f , then f has an absolute
minimum at (a, b).
Example 17.11. Find the absolute maximum and minimum values, if they exist, of f (x, y) = 4−x2 −y 2
on the open disk R = {(x, y) : x2 + y 2 < 1}.
Fact 17.12 (Extreme Value Theorem). If f is continuous on a closed bounded region R in R2 , then f
obtains an absolute maximum and absolute minimum value on R.
Example 17.13. Find the absolute maximum and minimum values of f (x, y) = x2 + y 2 − 2x + 2y + 5
on the set R = {(x, y) : x2 + y 2 ≤ 4}.
Example 17.14. Find the absolute maximum and minimum values of f (x, y) = 6 − x2 − 4y 2 on the set
R = {(x, y) : −2 ≤ x ≤ 2, −1 ≤ y ≤ 1}.
18
Optimizing with a Constraint: The Method of LaGrange Multipliers
Fact 18.1 (Method of LaGrange Multipliers). Assume that f (x, y) and g(x, y) are differentiable functions. If f (x, y) has a local minimum or a local maximum on the constraint curve g(x, y) = 0 at the
point (a, b), and if ∇g(a, b) 6= ~0, then there is a scalar λ (called a LaGrange multiplier) such that
∇f (a, b) = λ∇g(a, b).
Definition 18.2. A point (a, b) satisfying the equation ∇f (a, b) = λ∇g(a, b) for some scalar λ is said
to be a critical point for optimizing f (x, y) with respect to the constraint g(x, y) = 0.
Example 18.3. Find the extreme values of f (x, y) = 2x + 5y on the ellipse ( x4 )2 + ( y3 )2 = 1.
Example 18.4. Find the point on the plane
x
2
+
y
4
+
z
4
= 1 which is closest to the origin in R3 .
Fact 18.5. Assume that f (x, y, z), g(x, y, z), and h(x, y, z) are differentiable functions. If f (x, y, z) has
a local minimum or a local maximum at (a, b, c) subject to the constraints g(x, y, z) = 0 and h(x, y, z) = 0,
and if neither ∇g nor ∇h are ~0 at (a, b, c), then there are scalars λ and µ such that
∇f (a, b, c) = λ∇g(a, b, c) + µ∇h(a, b, c).
Example 18.6. The intersection of the plane x + 12 y + 13 z = 0 with the sphere x2 + y 2 + z 2 = 1 is a
circle. Find the point on this circle with the largest x-coordinate.
19
Double Integrals over Rectangular Regions
Definition 19.1. A region R in R2 is called a rectangle or rectangular region if R is of the form
R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}
for some real numbers a ≤ b, c ≤ d. We also denote R by
R = [a, b] × [c, d].
Definition 19.2. Let f be a function in two variables, and let R = [a, b] × [c, d] be a rectangular region
d−c
in R2 . Given any positive integer n, set ∆x = b−a
n and ∆y = n . For each integer k with 0 ≤ k ≤ n,
set
xk = a + k∆y and yk = c + k∆y.
For each pair of integers j, k with 1 ≤ j ≤ n and 1 ≤ k ≤ n, let (xj , y k ) be a point chosen arbitrarily
from the rectangular region [xj−1 , xj ] × [yk−1 , yk ].
We define the double integral, or double definite integral, of f over R to be
24
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
RR
f (x, y)d(x, y) = lim
R
n→∞
n X
n
X
f (xj , y k )∆x∆y,
j=1 k=1
provided the limit exists and is independent of the choices of (xk , y k ). (Note: Our notation differs
slightly from Rogawski here.) If the limit exists we say that f is integrable over R. If f is non-negative
on R, then the double definite integral corresponds to the volume of the solid bounded by the graph of
f over R.
Theorem 19.3 (Fubini’s Theorem). Let f be continuous on the rectangular region R = [a, b] × [c, d].
Then f is integrable, and
RR
RdRb
RbRd
f (x, y)d(x, y) = c a f (x, y)dxdy = a c f (x, y)dydx.
R
Example 19.4. Find the volume of the solid bounded by the surface z = 4 + 9x2 y 2 over the region
R = [−1, 1] × [0, 2]. Use both possible orders of integration.
RR
Example 19.5. Evaluate
xexy d(x, y), where R = [0, 1] × [0, ln 2].
R
RR
1
Example 19.6. Evaluate
d(x, y), where R = [1, 2] × [0, 1].
R (x+y)2
Example 19.7. Calculate
20
R 2 R π/2
0
0
ex cos ydydx.
Double Integrals over General Regions
Fact 20.1. Let g and h be continuous functions in one variable. Suppose R is a region in R2 bounded
below and above by the graphs of y = g(x) and y = h(x) respectively, and the lines x = a and x = b. If
f is continuous on R, then
RR
R b R h(x)
f (x, y)d(x, y) = a g(x) f (x, y)dydx.
R
Alternatively, if R is bounded on the left and right by the graphs of x = g(y) and x = h(y) respectively,
and the lines y = c and y = d, and f is continuous on R, then
RR
R
f (x, y)d(x, y) =
Example 20.2. Compute the integral
y = 3x2 and y = 16 − x2 .
RR
R
R d R h(y)
c
g(y)
f (x, y)dxdy.
2x2 yd(x, y), where R is the region bounded by the parabolas
Example 20.3. Compute the volume of the solid below the surface f (x, y) = 2+ y1 and above the region
R in the xy-plane bounded by the lines y = x, y = 8 − x, and y = 1.
R R y2
Example 20.4. Evaluate
e d(x, y), where R is the triangle bounded by the lines x = 0, y = 21 x,
R
and y = 2.
R √π R √π
sin x2 dxdy.
Example 20.5. Evaluate 0
y
21
Triple Integrals
Definition 21.1. Let D = {(x, y, z) : a ≤ x ≤ b, g(x) ≤ y ≤ h(x), G(x, y) ≤ z ≤ H(x, y)}, where g, h,
G, H are continuous functions. Then
RRR
D
f (x, y, z)d(x, y, z) =
R b R h(x) R H(x,y)
a
g(x)
(Other orders of integration are handled similarly.)
R2 R2Re 2
Example 21.2. Evaluate −2 1 1 xyz dzdxdy.
G(x,y)
f (x, y, z)dzdydx.
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
Example 21.3. Evaluate
R1
R√
0
0
1−x
√
2 R
0
1−x2 −y 2
25
2xzdzdydx.
Example 21.4. A solid box D is bounded by the planes x = 0, x = 3, y = 0, y = 2, z = 0, and z = 1.
The density of the box decreases linearly in the z-direction and is given by f (x, y, z) = 2 − z. Find the
mass of the box. (Hint: Mass is the integral of density over the box.)
Example 21.5. Compute the volume of the region D bounded by the paraboloids y = x2 + z 2 and
y = 16 − 3x2 − z 2 .
22
Integration in Polar Coordinates
Definition 22.1. To each pair (r, θ) of real numbers with r ≥ 0, we (implicitly) associate the pair (x, y)
in R2 where x = r cos θ and y = r sin θ. We refer to (r, θ) as the polar coordinates of (x, y). Notice
that any pair (x, y) has infinitely many polar coordinatizations, as (r, θ + 2πn) gives the same (x, y) for
any choice of integer n.
A polar rectangle is a set of the form {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β}, where a, b, α, β are real
numbers with β − α ≤ 2π.
Theorem 22.2. Let f be continuous on the region R = {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ} (where (r, θ)
corresponds to a point (x, y) = (r cos θ, r sin θ) as in the above definition). Then
RR
R
f (r, θ)d(x, y) =
RβRb
α
a
f (r, θ)rdrdθ.
Example 22.3. Find the volume of the solid bounded by the paraboloid z = 9 − x2 − y 2 and the
xy-plane.
Example 22.4. Find the volume of the region bounded beneath the surface z = xy + 10 and above the
annular region R = {(r, θ) : 2 ≤ r ≤ 4, 0 ≤ θ ≤ 2π}.
Fact 22.5. Let f be continuous on the region R = {(r, θ) : 0 ≤ g(θ) ≤ r ≤ h(θ), α ≤ θ ≤ β}, where g
and h are continuous functions of θ. Then
RR
D
Example 22.6. Compute
23
f (r, θ)d(x, y) =
R 3 R √9−x2 p
0
0
R β R h(θ)
α
g(θ)
f (r, θ)rdrdθ.
x2 + y 2 dydx.
Change of Variables in Multiple Integrals
Definition 23.1. A transformation of two variables is a function T which takes pairs (u, v) in R2 for
input and returns pairs (x, y) = T (u, v) for output. We also call T a map or mapping. If S is the
domain of T (a region of R2 ) then we denote the range of T by T (S). T (S) is another region of R2 ,
called the image of S under T .
Example 23.2. Consider the transformation from polar coordinates to rectangular coordinates given by
T (r, θ) = (r cos θ, r sin θ).
Find the image under this transformation of the rectangle S = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤
π
2 }.
Definition 23.3. A transformation of two variables T on a domain S is called one-to-one, or injective,
if T (u1 , v1 ) = T (u2 , v2 ) only when (u1 , v1 ) = (u2 , v2 ).
Example 23.4. Let T be as in the previous example.
(1) Let S = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ π2 }. Is T one-to-one on S?
(2) Let S = {(r, θ : 1 ≤ r ≤ 2, 0 ≤ θ ≤ π2 }. Is T one-to-one on S?
Fact 23.5. If T is one-to-one on a domain S, then there exists an inverse map T −1 whose domain is
T (S) and whose range is S.
26
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
Definition 23.6. A transformation of two variables T on a domain S is called linear if it has the form
T (u, v) = (Au + Cv, Bu + Dv), where A, B, C, D are constants.
Fact 23.7. If T is a linear transformation and P and Q are two points in R2 , then the image under T
of the line segment connecting P and Q is the line segment connecting T (P ) and T (Q).
Example 23.8. Let S be the triangle with corner points (1, 2), (2, 1), and (3, 4), and let T (u, v) =
(2u − v, u + v). Find and sketch the image T (S).
Definition 23.9. Given a transformation T (u, v) = (x(u, v), y(u, v)), where x and y are differentiable
functions of u and v, the Jacobian determinant (or just Jacobian) J(T ) of T is the function of two
variables defined by:
[J(T )](u, v) = det
xu (u, v) xv (u, v)
yu (u, v) yv (u, v)
= det
δx
δu
δy
δu
δx
δv
δy
δv
.
Example 23.10. Compute the Jacobian of the transformation T (r, θ) = (r cos θ, r sin θ).
Theorem 23.11 (Change of Variables Formula). Let T (u, v) = (x(u, v), y(u, v)) be a transformation of
two variables with a closed bounded domain S. Assume T is one-to-one on the interior of S and that x
and y have continuous first partial derivatives there. Set R = T (S). If f is continuous on R, then
RR
R
f (x, y)d(x, y) =
RR
S
f (T (u, v))|J(u, v)|d(u, v).
RR p
Example 23.12. Evaluate the integral
2x(y − 2x)d(x, y), where R is the parallelogram in the
R
xy-plane with vertices (0, 0), (0, 1), (2, 4), and (2, 5). Use the transformation T (u, v) = (2u, 4u + v).
Z Z r
x−y
d(x, y), where R is the square with vertices (0, 0), (1, −1),
Example 23.13. Compute
x
+
y+1
R
(2, 0), and (1, 1).
Solution. We wish to take substitute u = x − y and v = x + y into our integral to simplify both the
integrand and the domain of integration. We regard u and v as our input variables, so this defines our
inverse map:
T −1 (x, y) = (u, v) = (x − y, x + y).
To find our desired transformation T , we solve for x and y in terms of u and v to find x =
y = v−u
2 . This gives:
u+v
2
and
v−u
T (u, v) = (x, y) = ( u+v
2 , 2 ).
Next we want to find a region S such that T (S) = R. For this we may take S = T −1 (R). Since T −1
is a linear transformation, we can describe S exactly by computing T −1 of the corner points:
T −1 (0, 0) = (0, 0);
T −1 (1, −1) = (2, 0);
T −1 (2, 0) = (2, 2);
T −1 (1, 1) = 0, 2).
So S = T −1 (R) = [0, 2] × [0, 2]. Lastly we compute the Jacobian of T , which turns out to be J = 12 .
So we have:
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
Z Z r
R
x−y
d(x, y) =
x+y+1
Z Z r
S
27
u
1
· d(u, v)
v+1 2
Z Z
1 2 2 1/2
u (v + 1)−1/2 dudv
=
2 0 0
1 2
= · (23/2 − 0) · 2(31/2 − 1)
2√ 3
√
4 6 4 2
=
−
.
3
3
Fact 23.14. If T is a differentiable one-to-one transformation with differentiable inverse T −1 , then
J(T ) =
1
.
J(T −1 )
RR 2
Example 23.15. Compute
y d(x, y), where R is the region bounded by the parabolas x = y 2 ,
R
x = y 2 − 4, x = 9 − y 2 , and x = 16 − y 2 in the first quadrant.
24
Integration in Cylindrical and Spherical Coordinates
Definition 24.1. Let T be a transformation of three variables, i.e. T (u, v, w) = (x(u, v, w), y(u, v, w), z(u, v, w)),
where x, y, and z are all differentiable. Then the Jacobian J(T ) of T is

xu (u, v, w) xv (u, v, w) xw (u, v, w)
[J(T )](u, v, w) = det  yu (u, v, w) yv (u, v, w) yw (u, v, w) .
zu (u, v, w) zv (u, v, w) zw (u, v, w)

Definition 24.2. To each triple (r, θ, z) with r ≥ 0 we implicitly associate the triple
(x, y, z) = (r cos θ, r sin θ, z)
in R3 . We call (r, θ, z) the cylindrical coordinates of (x, y, z).
Also, to each triple (ρ, θ, φ) with ρ ≥ 0, we implicitly associate the triple
(x, y, z) = (ρ cos θ sin φ, ρ sin θ sin φ, ρ cos φ).
The triple (ρ, θ, φ) is called the spherical coordinates of (x, y, z).
Example 24.3.
(1) Let T (r, θ, z) = (r cos θ, r sin θ, z) be the transformation from cylindrical coordinates to Cartesian coordinates. Compute the Jacobian J(T ).
(2) Let T (ρ, θ, φ) = (ρ cos θ sin φ, ρ sin θ sin φ, ρ cos φ) be the transformation from spherical to Cartesian coordinates. Compute the Jacobian J(T ).
Fact 24.4.
(1) The Jacobian J(T ) of the cylindrical-to-Cartesian transformation T is J(T ) = r .
(2) The Jacobian J(T ) of the spherical-to-Cartesian transformation T is J(T ) = −ρ2 sin φ .
Fact 24.5.
(1) Let f be continuous on the region D = {(r, θ, z) : z1 (r, θ) ≤ z ≤ z2 (r, θ), 0 ≤ r1 (θ) ≤
r ≤ r2 (θ), α ≤ θ ≤ β}, where z1 and z2 are continuous functions of r and θ, and r1 and r2 are
continuous functions of θ. Then
RRR
D
f (x, y, z)d(x, y, z) =
R β R r2 (θ) R z2 (r,θ)
α
r1 (θ)
z1 (r,θ)
f (r, θ, z)rdzdrdθ.
28
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
(2) Let f be continuous on the region D = {(ρ, θ, φ) : 0 ≤ ρ1 (θ, φ) ≤ ρ ≤ ρ2 (θ, φ), 0 ≤ α ≤ θ ≤ β ≤
2π, 0 ≤ γ ≤ φ ≤ δ ≤ π}, where ρ1 and ρ2 are continuous functions of θ and φ. Then
RRR
D
f (x, y, z)d(x, y, z) =
Example 24.6. Integrate f (x, y, z) = z
p
R β R δ R ρ2 (θ.φ)
α
γ
ρ1 (θ,φ)
f (ρ, θ, φ)ρ2 sin φdρdφdθ.
x2 + y 2 over the cylinder x2 + y 2 ≤ 4 for 1 ≤ z ≤ 5.
Example 24.7. Compute the integral of f (x, y, z) = z over the region D within the cylinder x2 + y 2 ≤ 4
where 0 ≤ z ≤ y.
Example 24.8. Compute the integral of f (x, y, z) = x2 + y 2 over the sphere S of radius 4 centered at
the origin.
Example 24.9. Integrate f (x, y, z) = z over the region D lying above the cone x2 + y 2 = z 2 and within
the sphere of radius R centered at the origin.
25
Vector Fields
Definition 25.1. A vector field over R3 is a function F~ which takes for input a vector in R3 and
returns as output a vector in R3 . We will typically write a vector field as
F~ (x, y, z) = (F1 (x, y, z), F2 (x, y, z), F3 (x, y, z)),
where each of F1 , F2 , and F3 is understood to be a real-valued function in three variables. The domain
of F~ is the set of all points in R3 where F~ is defined. For this entire course, we will assume that all
components F1 , F2 , and F3 are infinitely differentiable on the domain of F~ , i.e. they have continuous
partial derivatives of all orders. (Vector fields over R2 , R4 , etc. are defined similarly.)
Example 25.2. Sketch the following vector fields.
(1) F~ = (1, −1, 3).
~ = (1, x).
(2) G
~ = (−y, x).
(3) H
(4) (Vortex Field ) F~ =
−y
x
!
p
,p
.
x2 + y 2
x2 + y 2
Definition 25.3. A unit vector field is a vector field F~ such that ||F~ (x, y, z)|| = 1 for all points
(x, y, z).
A radial vector field is a vector field where each vector F~ (x, y, z) is p
a scalar multiple of (x, y, z), i.e.
~
F (x, y, z) = c(r)(x, y, z), where c(r) depends only on the distance r = x2 + y 2 + z 2 from (x, y, z) to
the origin. (In particular all vectors F~ (x, y, z) point directly away from or directly toward the origin.)
The unit radial vector field, which we will denote permanently by ~er , is the field ~er = 1r (x, y, z).
Definition 25.4. A vector field F~ is called conservative if there exists a differentiable real-valued
function V (x, y, z) such that
F~ = ∇V .
Such a function V is called a potential function for F~ .
Example 25.5. Determine whether or not F~ = (y, x + z 2 , 2yz) is a conservative vector field. If it is,
find a potential function.
Theorem 25.6. If F~ = (F1 , F2 , F3 ) is a conservative vector field, then
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
δF1
δy
=
δF2 δF2
δx , δz
=
δF3 δF3
δy , δx
=
29
δF1
δz .
Proof. Let V be a potential function for F~ , so F~ = (F1 , F2 , F3 ) = ∇V = (Vx , Vy , Vz ). Then, for instance,
δF2
δF1
δ
δ
δy = δy Vx = Vxy , and δx = δx Vy = Vyx ; hence the first equality follows from Clairaut’s theorem. The
other two equalities are similar.
Example 25.7. Show that F~ = (y, 0, 0) is not conservative.
Example 25.8. Show that the vortex field F~ of Example 25.2 (d) satisfies the cross-partial-derivative
condition of Theorem 25.6, but is not conservative.
Definition 25.9. A region D in R3 is called pathwise-connected if any two points in D can be
connected by a path in D. That is, for all (x1 , y1 , z1 ) and (x2 , y2 , z2 ) in D, there is a continuous R3 valued function ~r(t), 0 ≤ t ≤ 1, such that ~r(0) = (x1 , y1 , z1 ), ~r(1) = (x2 , y2 , z2 ), and the graph of ~r is
entirely contained in D.
Theorem 25.10. If F~ is conservative on an open pathwise-connected domain, then any two potential
functions of F~ differ only by a constant.
p
Example 25.11. Show that V (x, y, z) = x2 + y 2 + z 2 is a potential function for the unit radial vector
function ~er .
26
Line Integrals
Definition 26.1. Let C be the graph of a smooth one-to-one R3 -valued function ~c(t) with domain [a, b].
The line integral of a function f (x, y, z) in three variables over C is
Rb
R
f (x, y, z)ds = a f (~c(t))||~c 0 (t)||dt,
C
if this integral exists.
The following theorem implies that the definition of the line integral depends only on the curve C
itself, and not on the choice of function ~c used to parametrize it.
Theorem 26.2. Let f (x, y, z) be any function in three variables. Suppose ~c is a smooth one-to-one
R3 -valued function with domain [a, b]. Let ~c1 be the arc length parametrization of ~c based at a, which
has domain [0, `], where ` is the length of the curve C. Then
Rb
R`
f (~c(t))||~c 0 (t)||dt = 0 f (~c1 (t))dt (if both integrals exist).
a
In other words, the line integral of f over C does not depend on the choice of parametrization of C.
Rt
Proof. Let s(t) = a ||~c 0 (u)||du, i.e. let s be the arc length function for ~c based at a. The function s
has domain [a, b] and range [0, `]. Since ~c is one-to-one, s is one-to-one, and hence the inverse function
s−1 exists, which takes [0, `] onto [a, b]. By Theorem 8.9, ~c(s−1 (t)) is the arc length parametrization of
~c based at a, i.e. ~c(s−1 (t)) = ~c1 (t). Note that s0 (t) = ||~c 0 (t)|| by the fundamental theorem of calculus,
and so by the substitution rule we have
Rb
a
f (~c(t))||~c 0 (t)||dt =
R s(b)
s(a)
f (~c(s−1 (t)))dt =
R`
0
f (~c1 (t))dt.
Example 26.3. Calculate
0 ≤ t ≤ π.
R
C
(x + y + z)ds, where C is the helical graph of ~c(t) = (cos t, sin t, t) for
Fact 26.4. If ~c(t) is a smooth R3 -valued function with domain [a, b] whose graph is C, then
exactly the arc length of C.
R
C
1ds is
Example 26.5. Find the total mass of a wire in the shape of the parabola y = x2 for 1 ≤ x ≤ 4 (in
centimeters) with mass density given by ρ(x, y) = xy g/cm. (Hint: The mass of the wire is the line
integral of the density function over the length of the wire.)
30
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
Definition 26.6 (Informal). Let C be a curve in R3 . We may fix an orientation on C by fixing a
particular parametrization ~c(t) function (with domain [a, b]) whose graph is C. We declare that the
positive direction along C is the direction traversed as t increases from a to b, and the negative
direction to be the opposite direction. We call C an oriented curve.
Definition 26.7. Let F~ be a vector field over R3 . Let C be an oriented curve in R3 , parametrized
by the smooth R3 -valued function ~c(t) with domain [a, b]. Let T~ (t) be the unit tangent vector function associated to ~c(t). The tangential component of F~ is the dot product F~ · T~ . The vector line
R
integral of F~ over C, denoted C F~ · ds, is the line integral of the tangential component function of F~ , i.e.
R
C
F~ · ds =
R
C
(F~ · T~ )ds.
Fact 26.8. For F~ , C, ~c(t) as in the previous definition, we have
F~ · ds =
R
C
Rb
a
F~ (~c(t)) · ~c 0 (t)dt.
Proof. Write down the definition.
R
Example 26.9. Evaluate C F~ · ds, where F~ = (z, y 2 , x) and C is parametrized in the positive direction
by ~c(t) = (t + 1, et , t2 ) for 0 ≤ t ≤ 2.
Example R26.10. Let C be the ellipse parametrized by ~c(θ) = (5 + 4 cos θ, 3 + 2 sin θ) for 0 ≤ θ ≤ 2π.
Calculate C (3, 2y) · ds.
Definition 26.11. Suppose C is a curve in R3 parametrized in the positive direction by the smooth
~ 3 -valued function ~c(t) with domain [a, b]. We permanently denote by −C the same curve with the
R
~ = ~c(b + (a − t))
opposite orientation. (Written out explicitly, −C is the graph of the smooth function d(t)
with domain [a, b].)
Now suppose C1 , C2 , ..., Cn are finitely many oriented curves in R3 . We write C = C1 + C2 + ... + Cn to
denote the union of the paths (with the same orientations). The resulting curve C is called piecewise
smooth.
Fact 26.12. Let C be an oriented curve which is the graph of a smooth R3 -valued function, and let F~
~ be vector fields over R3 . Then
and G
(1)
R
(2)
R
C
~ · ds =
(F~ + G)
−C
F~ · ds = −
R
C
R
C
F~ · ds +
R
C
~ · ds;
G
F~ · ds; and
(3) if C = C1 + ... + Cn for smooth curves C1 , ..., Cn , then
R
R
R
F~ · ds = C1 F~ · ds + ... + Cn F~ · ds.
C
R
Example 26.13. Compute C F~ · ds, where F~ = (ez , ey , x + y) and C is the triangle with corners (1, 0, 0),
(0, 1, 0), and (0, 0, 1) oriented in the counterclockwise direction when viewed from above (in Octant I
facing the origin).
27
Conservative Vector Fields
Definition 27.1. From now on, when C is a curve parametrized by a smooth R3 -valued function ~c(t),
R
R
we will write C F~ · ds = ~c F~ · ds for any vector field F~ .
If ~c(a) = ~c(b) where [a, b] is the domain of ~c, then we call the graph C a closed curve. If this is
the case, then we refer to the line integral as the circulation of F~ around C, and we use the alternate
notation:
H
F~ · ds.
~
c
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
31
A vector field F~ is called path-independent if for any two points P and Q in R3 , the line integrals
of F~ along any path from P to Q are all equal, i.e. the integral depends on the endpoints but not the
choice of path.
Theorem 27.2 (Fundamental Theorem for Conservative Vector Fields). Let F~ be a conservative vector
field, so F~ = ∇V on a domain D. Let P and Q be two points in D.
(1) If ~c(t) is a smooth R3 -valued function with domain [a, b], whose graph lies in D, and ~c(a) = P ,
~c(b) = Q, then
R
~
c
F~ · ds = V (Q) − V (P ).
In particular, F~ is path-independent.
(2) The circulation around any closed curve ~c is zero, i.e.
H
~
c
F~ · ds = 0.
Proof. Proof of (1). By Fact 26.8 we have
R
~
c
F~ · ds =
R
~
c
∇V · ds =
By the chain rule for paths (Theorem 15.1),
theorem of calculus, we get
Z
~
c
d
dt V
F~ · ds =
Rb
a
∇V (~c(t)) · ~c 0 (t)dt.
(~c(t)) = ∇V (~c(t))·~c 0 (t), and hence by the fundamental
Z
a
b
d
V (~c(t))dt
dt
= [V (~c(t))]ba
= V (~c(b)) − V (~c(a))
= V (Q) − V (P ).
(Proof of (2). Now if ~c gives a closed curve, i.e. P = Q, then by the previous part of this proof we
H
have ~c F~ · ds = V (Q) − V (P ) = V (P ) − V (P ) = 0.
Example 27.3. Let F~ = (2xy + z, x2 , x).
(1) Find a potential function V for F~ .
R
(2) Evaluate ~c F~ · ds, where ~c is a path from P = (1, −1, 2) to Q = (2, 2, 3).
R
Example 27.4. Find a potential function for F~ = (2x + y, x) and use it to compute ~c F~ · ds, where ~c
is a path from (1, 2) to (5, 7).
H
Example 27.5. Let V (x, y, z) = xy sin(yz), and evaluate C ∇V · ds, where C is any closed curve of
your choice.
Theorem 27.6. A vector field F~ is path-independent if and only if F~ is conservative.
Proof. If F~ is conservative then it is path-independent by the fundamental theorem for conservative
vector fields. On the other hand suppose F~ is path-independent. For simplicity’s sake let us consider
the case where F~ is a vector field over R2 ; the proof in R3 and other dimensions is analogous but harder
to write down.
Fix any point P0 = (x0 , y0 ) in R2 . For each point P = (x, y) in R2 , define
32
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
V (P ) = V (x, y) =
R
~
c
F~ · ds,
where ~c is any path from P0 to P . (Note this definition makes sense because F~ is path-independent.)
We claim that F~ = ∇V .
2
c1 (t) = (x + t, y) for 0 ≤ t ≤ h,
First let us check that δV
δx = F1 . Let (x, y) in R be arbitrary. Let ~
where h is some small positive number, so the graph of ~c1 is the straight-line path from (x, y) to (x+h, y).
Let ~c + c~1 be the path ~c followed by the path ~c1 . We have
Z
Z
F~ · ds − F~ · ds
~
c+c~
~
c
Z 1
Z
Z
= F~ · ds +
F~ · ds − F~ · ds
~
c1
~
c
Z~c
=
F~ · ds.
V (x + h, y) − V (x, y) =
~
c1
Note ~c1 0 (t) = (1, 0), and hence F~ (~c1 (t)) · ~c1 0 (t) = F1 (x + t, y). So we have shown
V (x + h, y) − V (x, y)
=
h
1
h
Rh
0
F~ (x + t, y) =
1
h
R x+h
x
F1 (u, y)du,
which is the average value of F1 over [x, x + h]. Since F1 is continuous, this converges to F1 (x, y) as
δV
δV
h → 0, and hence
= F1 at (x, y). A similar argument shows that
= F2 , and hence F~ = ∇V . δx
δy
Definition 27.7 (Informal). A subset D of R2 is called simply connected if it does not have any
“holes;” that is, every loop in D can be continuously contracted down to a point without leaving the set
D. (This definition can be made formal but it would be a big unnecessary diversion at this point.) The
definition of simple-connectedness is similar in R3 .
Theorem 27.8. Let F~ be a vector field on a simply connected domain. If F~ satisfies the cross-partial
condition of 25.6, then F~ is conservative.
Example 27.9. Show that F~ = (2xy + y 3 , x2 + 3xy 2 + 2y) is conservative, and find a potential function.
Example 27.10. Find a potential function for F~ = (2xyz −1 , z + x2 z −1 , y − x2 yz −2 ).
28
Parametrized Surfaces and Surface Integrals
Definition 28.1. Let G(u, v) = (x(u, v), y(u, v), z(u, v)) be a R3 -valued function of two variables. The
graph of G is the set {(x, y, z) ∈ R3 : there exists (u, v) with G(u, v) = (x, y, z)} (this definition is
analogous to that of the graph of a vector-valued function). The graph of G is also called a parametrized
surface. The domain D of G is called the parameter domain.
Example 28.2 (Parametrization of a Sphere). Fix any R > 0. Let D = {(θ, φ) : 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π},
and define G(θ, φ) = (R cos θ sin φ, R sin θ sin φ, R cos φ) on D. Sketch the domain D and sketch the graph
of G.
Example 28.3 (Parametrization of the Graph of a Real-Valued Function of Two Variables). Let f (x, y)
be a real-valued function of two variables with domain D. Define G(x, y) = (x, y, f (x, y)) on D. Observe
that the graph of f is equal to the graph of G.
Definition 28.4. Let G(u, v) = (x, y, z) be a continuously differentiable R3 -valued function of two variables. Define the u-tangent vector and the v-tangent vector functions of G, respectively, to be
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
33
T~u = (xu , yu , zu ) and T~v = (xv , yv , zv ).
Assume both T~u and T~v are non-zero. Define the normal vector associated to G to be the function
~n = T~u × T~v .
Example 28.5. Let G(θ, z) = (2 cos θ, 2 sin θ, z) be a parametrization of the cylinder described by
x2 + y 2 = 4.
(1) Compute T~θ , T~z , and ~n.
(2) Find an equation for the tangent plane to the graph of G at ( π4 , 5, G( π4 , 5)).
Definition 28.6. Let G(u, v) = (x, y, z) be a one-to-one continuously differentiable parametrization of a
surface S in R3 with domain D, with non-zero tangent vectors T~u and T~v . Let f (x, y, z) be a real-valued
function of three variables. Define the surface integral of f over S to be
RR
f (x, y, z)dS =
RR
f (G(u, v))||~n(u, v)||d(u, v).
RR
Fact 28.7. If G is as above, then the surface integral
1dS is exactly the surface area of the graph
S
of G.
S
D
Example 28.8. Calculate the
area of the portion S of the cone x2 + y 2 = z 2 within the cylinder
R Rsurface
2
2
2
x + y = 4. Then calculate
x zdS.
S
Fact 28.9. Suppose g(x, y) is a function of two variables let S be the graph of g. Suppose G(x, y) is a
parametrization of S. Then
T~x = (1, 0, gx ) and T~y = (0, 1, gy );
~n =
q
1 + gx2 + gy2 ;
and therefore for any function f with domain S we have
RR
S
f (x, y, z)dS =
Example 28.10. Compute
29
RR
S
RR
D
q
f (x, y, g(x, y)) 1 + gx2 + gy2 d(x, y).
(z − x)dS, where S is the graph of z = x + y 2 , 0 ≤ x ≤ y, 0 ≤ y ≤ 1.
Flux Integrals
Definition 29.1. Let G(u, v) be a parametrization of the surface S in R3 , and let ~n be its associated
normal vector. Let F~ be a vector field over R3 . The normal
component of F~ with respect to G is
RR
~
~
F · ~n. The flux integral or vector surface integral
F · dS of F~ across S is the surface integral
S
~
of the normal component of F , i.e.
RR
S
F~ · dS =
RR
S
(F~ · ~n)dS.
Fact 29.2. Let G, S, ~n, F~ be as in the previous definition, and let D be the domain of G. Then
RR
S
F~ · dS =
RR
D
F~ (G(u, v)) · ~n(u, v)d(u, v).
RR
Example 29.3. Compute
F~ ·dS, where F~ = (0, 0, x) and S is the graph of G(u, v) = (u2 , v, u3 −v 2 )
S
with domain D = [0, 1] × [0, 1].
34
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
Example 29.4. Calculate the flux integral of F~ = (z, x, 1) across the upper unit hemisphere S, with
normal vector ~n facing outward from the origin.
Example 29.5. Calculate the flux of F~ = x2~j through the surface S defined by y = 1 + x2 + z 2 for
1 ≤ y ≤ 5. Orient S with the normal vector pointing in the negative y-direction.
30
Green’s Theorem
Definition 30.1. Let D be a region in R2 . Recall from Definition 12.1 that a point (x, y) is called a
boundary point of D if every open disk about (x, y) intersects both D and the exterior of D. Denote
the set of all boundary points of D by δD. We call δD the boundary of D.
Suppose C may be parametrized by a continuous one-to-one R2 -valued function ~c with domain [a, b],
where ~c(a) = ~c(b). Then we call C a simple closed curve.
If δD is a simple closed curve, then we choose to orient δD in the counterclockwise direction. This is
called the boundary orientation.
Theorem 30.2 (Green’s Theorem). Let D be a domain whose boundary δD is a simple closed curve.
Let F = (F1 , F2 ) be a vector field over R2 . Then
δF2
δF1
−
d(x, y).
δx
δx
δD
D
H
Example 30.3. Verify Green’s theorem for the line integral C (xy 2 , x)ds about the unit circle C.
I
F~ · ds =
Z Z Example 30.4. Compute the circulation of F~ = (sin x, x2 y 3 ) about the path C = δD, where D is the
triangle with vertices (0, 0), (0, 2), and (2, 2).
Corollary 30.5 (Area Formula). Let D be a region in R2 and assume C = δD is a simple closed curve.
Then the area of D is equal to
1
2
Proof.
The expression above is equal to
RR
1d(x,
y), the area of D.
D
1
2
H
C
(−y, x)ds.
RR
D
(1 − (−1))d(x, y) by Green’s theorem, which equals
Example 30.6. Compute the area of the ellipse bounded by the graph of ( xa )2 + ( yb )2 = 1 using a line
integral.
Corollary 30.7 (Additivity of Circulation). Suppose D is a region of R2 whose boundary is a simple
closed curve. If we write D as the union of two disjoint subsets D1 and D2 whose boundaries are also
simple closed curves, then for any vector field F~ ,
H
H
H
F~ · ds = δD1 F~ · ds + δD2 F~ · ds.
δD
31
Stokes’ Theorem
Definition 31.1. Let F~ = (F1 , F2 , F3 ) be a vector field. The curl of F~ is the vector field curl(F~ ) defined
by
δF3
δF2 δF3
δF1 δF2
δF1
~
curl(F ) =
−
,
−
,
−
.
δy
δz δx
δz δx
δy
There are two mnemonic devices we may use to remember this definition. The first is as a symbolic
determinant:


~k
~i
~j
δ
δ
δ 
curl(F~ ) = det  δx
.
δy
δz
F1 F2 F3
CALCULUS III MATH 265 FALL 2013 (COHEN) LECTURE NOTES
35
The second is as a symbolic cross product:
curl(F~ ) = ∇ × F~ .
Example 31.2. Calculate the curl of F~ = (xy, ex , y + z).
Example 31.3. Show that if F~ is conservative, then curl(F~ ) = ~0.
Definition 31.4 (Informal). If S is an oriented surface in R3 , we can fix a boundary orientation on
the boundary δS as follows: If you are the unit normal vector to S standing at a boundary point, then
you traverse the curve δS by walking in such a way that the surface is on your left side.
Theorem 31.5 (Stokes’ Theorem). Let G(u, v) be a smooth one-to-one parametrization of a surface S,
whose boundary δS is comprised of simple closed curves. Then
H
RR
F~ · ds =
curl F~ · dS.
δS
S
The integral on the left is defined relative to the boundary orientation of δS. If δS is empty, then the
surface integral on the right is 0.
Example 31.6. Verify Stokes’ theorem for F~ = (−y, 2x, x + z) and S = {(x, y, z) : x2 + y 2 + z 2 = 1, z ≥
0}.
H
2
Example 31.7. Show that C F~ · ds = 0, where F~ = (sin(x2 ), ey + x2 , z 4 + 2x2 ) and C is the boundary
of the triangle with vertices (0, 0, 1), (0, 2, 0), and (3, 0, 0).
~ then the flux of F~ through a surface S depends
Corollary 31.8. If A is a vector field and F~ = curl(A),
only on the oriented boundary δS of S. Specifically:
RR
H
~ · ds.
F~ · dS = δS A
S
In particular, if S is closed, i.e. it has no boundary, then
32
RR
S
F~ · dS = 0.
Divergence Theorem
Definition 32.1. Let F~ be a vector field. The divergence of F~ is
div(F~ ) =
δF2
δF3
δF1
+
+
.
δx
δy
δz
We can think of div(F~ ) as the symbolic dot product:
div(F~ ) = ∇ · F~ .
Theorem 32.2 (Divergence Theorem). Let W be a region in R3 , whose boundary is the closed surface
S. Assume S is oriented with normal vectors pointing outward to the exterior of W. Let F~ be a vector
field over W. Then
RR
RRR
F~ · dS =
div(F~ )d(x, y, z).
S
W
Example 32.3. Verify the divergence theorem for F~ = (y, yz, z 2 ) and the cylinder W = {(x, y, z) :
x2 + y 2 ≤ 4, 0 ≤ z ≤ 5}.
RR
Example 32.4. Evaluate
(x2 , z 4 , ez ) · dS, where S is the boundary of the box [0, 2] × [0, 3] × [0, 1].
S