PRELIMINARIES
Some Useful Notations
∃
∀
⇒
⇔
∈
there exists
for all
implies
equivalent to (iff)
belongs to
{x|q(x)}
set specification
A⊂B
A is a subset of B
A∪B
the union of A and B
A∩B
the intersection of A and B
Points and Sets in Rn
What is Rn ?
Rn is the set {(x1, x2 , ..., xn)|xi are real numbers}. An element, or a point, of Rn is an n-tuple
(x1 , x2, ..., xn).
Rn is the Cartesian product of n Rs.
Example R2 = R × R = {(x, y)| x, y belong to R}
Recall the following way to construct the product of two sets
Definition The Cartesian product of two sets X and Y is the set
X × Y = {(x, y)| x belongs to X and y belongs to Y }.
A rectangular region in Rn can be constructed as the Cartesian product of n intervals.
Example [a, b] × [c, d] = {(x, y)| a ≤ x ≤ b & c ≤ y ≤ d} It is a closed subset of R2 .
Rectangular (Cartesian) Coordinates
In Analytic Geometry, the location of a point in three-dimensional space can be determined
by its coordinates with respect to a coordinate system.
Usually the system is rectangular; it means that the coordinate axes are perpendicular to each
other.
Arrangement of the axes in 3D is according to the “Right-Hand Rule”.
Important Open Sets
In the one dimensional case, Open neighborhoods
Nδ (x0 ) = {x| |x − x0 | < δ}
and deleted open neighborhoods
N δ (x0 ) = {x| 0 < |x − x0 | < δ}
are needed in defining ’limit’.
1
Definition We say that ‘the number l is the limit of f (x) as x approaches x0 ’ and write
lim f (x) = l if for any ε > 0 ∃ δ > 0 such that x ∈ N δ (x0 ) ⇒ f (x) ∈ Nε (l).
x→x0
The corresponding open sets in two dimensions are:
(i) An open disk in R2 , centered at (x0 , y0 ) and with radius δ, is the set
Dδ (x0 , y0 ) = {(x, y)|
p
(x − x0 )2 + (y − y0 )2 < δ}.
(ii) A deleted open disk centered at (x0 , y0 ) with radius δ is the set
Dδ (x0 , y0 ) = {(x, y)| 0 <
p
(x − x0 )2 + (y − y0 )2 < δ}.
Elementary Functions
xn , sin x and cos x, ex (exp x) and ln x
They are continuous and differentiable (infinitely many times) in their natural domains.
2
PART I: VECTORS AND VECTOR-VALUED FUNCTIONS
(plus elements of solid & analytic geometry)
Vectors and Scalars
A vector is described by two things: a direction and a magnitude.
Example What are supposed to be vectors?
Displacement, velocity, force,
angular velocity, ...
−−→
−→
~ or Ā. Its magnitude (or length) is denoted by |−
~ or
A vector is denoted by P Q or A,
P Q| or |A|,
simply A. P is the initial point; Q is the terminal point.
Abstract vectors discard the exact “locations” of the initial/terminal points. A vector can be
concieved as a pointed stick which can be translated freely as long as the direction and length are
unchanged.
A scalar is simply a real number; it is usually denoted by a lower-case alphabet.
Displacement vectors (or pointed sticks) can be used to illustrate the basic properties of general
vectors.
Fundamental Properties & Operations of Vectors
~=B
~
1. Equality: A
iff they have the same magnitude & direction
locations of the initial points are irrelevant
~kB
~
2. Parallel vectors: A
iff they have the same (or opposite) direction
~
3. Negative vector: −A
~ but opposite direction
has the same magnitude as A,
~ +B
~
4. Addition: A
defined by the triangle rule
~ −B
~
5. Subtraction: A
~ + (−B)
~
defined as = A
~−A
~ = ~0 (zero vector)
therefore, A
~
6. Multiplication by a scalar: mA
~ and direction the same or opposite to A
~ according as m is
is a vector with magnitude |m||A|
> 0 or < 0.
~ = ~0.
If m = 0, mA
~ and B
~ =
~kB
~ iff ∃ m 6= 0 such that mA
~ = B.
~
Suppose that A
6 ~0. A
Note that ~0 is not parallel to any other vector.
3
The Algebra of Vectors
~ B,
~ C
~ are vectors, and m, n are scalars, then:
If A,
~+B
~ =B
~ +A
~
1. A
commutative law for vector addition
~ + (B
~ + C)
~ = (A
~ + B)
~ +C
~
2. A
associative law for vector addition
~ = (mn)A
~ = n(mA)
~
3. m(nA)
associate law for scalar multiplication
~ = mA
~ + nA
~
4. (m + n)A
distributive law for scalar multiplication
~ + B)
~ = mA
~ + mB
~
5. m(A
distributive law for scalar multiplication
Unit Vectors
— vectors with unit length
Normalizing a vector
~
A
~
If |A| =
6 0, Â =
is the unit vector along the same direction as A.
~
|A|
Rectangular Unit Vectors

~i k x−axis 
~j k y−axis ⊥ to each other
~k k z−axis 
Similar to the coordinate axes, ~i, ~j, ~k form a right-handed system.
They are called the basic unit vectors.
Correspondence Between Vectors and Points in Rn
~ and B
~ are considered equal iff they have the same magnitude and
Recall that two vectors A
direction, regardless of the initial point.
~ in 3 (or n) dimensions can be represented with initial point placed at the origin
Any vector A
O of a rectangular coodinate system. The terminal point will then fall on some point with label
(A1 , A2 , A3) in the coordinate system. The numbers A1 , A2 , A3 are called the x, y, z components
~ they describe A
~ completely and uniquely as long as the coordinate system is fixed.
of A;
The basic unit vectors can be specified as:
~i = (1, 0, 0),
~j = (0, 1, 0),
4
~k = (0, 0, 1).
~ can be written as
A
~ = A1~i + A2~j + A3~k.
A
~
where A1~i, A2~j, and A3~k are vector components of A.
~ can be computed as
The magnitude of A
~ =
|A|
q
A21 + A22 + A23 .
Vector Operations in Terms of Components
Vector addition
~ +B
~ = (A1 + B1 )~i + (A2 + B2 )~j + (A3 + B3 )~k
A
Scalar multiplication
~ = (mA1 )~i + (mA2 )~j + (mA3 )~k
mA
Directed Line Segment
A ‘directed line segment’ is a straight line segment joining two points in space. The direction is
from the initial point P = (P1 , P2 , P3 ) to the terminal point Q = (Q1 , Q2, Q3 ). The ‘directed line
segment vector’ is
−−→
P Q = (Q1 − P1 )~i + (Q2 − P2 )~j + (Q3 − P3 )~k.
−−→
Example 1. Find P Q if P = (1, 2, 3), Q = (4, 5, 6)
−−→
2. Find a unit vector k P Q
~ kB
~
3. Check if A
√
√
√
~ = ~i + ~j + ~k, B
~ = 2~i + 2~j + 2~k
(a) A
√
√
~ = ~i + ~j + ~k, B
~ = 2~i + 2~j
(b) A
Dot (or Scalar, or Inner) Product
~ B
~ is the real number
Definition The dot product of two vectors A,
~ B
~ = |A||B| cos θ
A·
0 ≤ θ ≤ π.
~ and B.
~
θ is the angle between A
Projection
~ Bk.
~ The scalar projection of A
~ onto B
~ is
Let B̂ = B/k
~ · B̂ = (A
~ · B)/k
~
~
A
Bk.
~ on a line L along B.
~
It is the ‘shadow’ of A
~ onto B
~ is the vector
The vector projection of A
~ = (A
~ · B̂)B̂.
projB~ A
5
~ multiplied by the scalar projection of A
~ onto B.
~
It is the unit vector along B
Basic properties of the dot product
~ B
~ = B·
~ A
~
1. A·
commutative
~ (B
~ + C)
~ = A·
~ B
~ + A·
~ C
~
2. A·
distributive
~ C
~ onto A)
~
(pf: use projections of B,
~ B)
~ = (mA)·
~ B
~ = A·
~ (mB)
~
3. m(A·
m a scalar
4. ~i· ~i = ~j· ~j = ~k· ~k = 1
~i· ~j = ~j· ~k = ~k· ~i = 0
~ = A1~i + A2~j + A3~k, B = B1~i + B2~j + B3~k, then A·
~ B
~ = A1 B1 + A2 B2 + A3 B3
5. If A
Example Find the dot product of
~i + ~j + ~k and ~i + 2~j − 3~k.
~ B
~ 6= ~0, then A ⊥ B
6. If both A,
iff
~ B
~ =0
A·
~ are scalar projections of A
~ on the coordinate unit vectors.
Example The components of A
~ = ~i + ~j + ~k and B
~ = ~i + ~j. Find the scalar projection of A
~ on B.
~ What about
Example Let A
o
~ on A?
~ Plot the vectors and note their positions. What is θ? (35.3 )
B
Cross (or Vector) Product
~ B
~ and such that A,
~ B,
~
Definition Let ~u be a unit vector perpendicular to the plane containing A,
u
~ form a right-handed system.
~ ×B
~ can be defined as
Then the cross product A
~×B
~ = |A||B| sin θ u
A
~
~ and B.
~
where θ is the angle between A
Basic properties of the cross product
~ ×B
~ ⊥ both A
~ and B
~
1. A
~ × B|
~ = area of a parallelogram with sides A
~ and B.
~
2. |A
~ ×B
~ = −B
~ ×A
~
3. A
~ ×A
~ = ~0
=⇒ A
not commutative
~ × (B
~ + C)
~ =A
~×B
~ +A
~×C
~
4. A
distributive
[use parallelograms in 2d]
~ × B)
~ = (mA)
~ ×B
~ =A
~ × (mB)
~
5. m(A
m a scalar
~ B
~ 6= 0, then A
~kB
~ iff A
~ ×B
~ = ~0
6. If both A,
6
0≤θ≤π
7. ~i × ~i = ~j × ~j = ~k × ~k = 0
~i × ~j = ~k, ~j × ~k = ~i, ~k × ~i = ~j
How to remember cross products of the basic unit vectors ?
Use cyclic ordering of ~i, ~j, ~k
~ = A1~i + A2~j + A3~k,
8. If A
~ = B1~i + B2~j + B3~k,
B
then
~i
~ ×B
~ = A1
A
B1
~j
A2
B2
~k A3 B3 You don’t need to use the determinant if you do the cross products of the basic unit vectors
directly.
Example Let Ā = 4ī − j̄ + 3k̄ and B̄ = 2ī + j̄ − k̄. Find a vector perpendicular to Ā and B̄.
[−2~
i+10~
j +6~
k]
Triple Products
Scalar triple product
A1
~
~
~
A· (B × C) = B1
C1
A2
B2
C2
A3 B3 C3 ~ (B
~ × C)
~ = B·
~ (C
~ × A)
~ = C·
~ (A
~ × B)
~
A·
~ (B
~ × C)|
~ = volume of a parallelepiped with A,
~ B,
~ and C
~ as edges
|A·
Vector triple product
~ × (B
~ × C)
~ = (A·
~ C)
~ B
~ − (A·
~ B)
~ C
~
A
Example: ~k × (~j × ~k)
—– Problem Set 1 —–
Lines in Space
Vectors that connect any pair of points on a line l are k.
Vector description of a line
Consider the locus of the terminal points of the vectors (first in 2D)
~
tA
−∞ < t < ∞
~
It is a straight line. This line can be displaced to describe any straight line k to A.
7
The general vector expression for a line in space is:
~
~r = ~r0 + tA
−∞< t < ∞
~ is a vector k to the line.
where ~r0 is an arbitrary point on the line and A
Example Find a vector equation of the line that contains (−1, 3, 0) and is parallel to 2ī − 3j̄ − k̄
Example Find a vector equation of the line that
passes through the two points: (1, 2, 3) and (4, 5, 6).
~ is. What
Note ~r(t) gives the location of a point on the line. It is not parallel to the line. A
~ and Q
~ are any two different points on the line, then
it means is the following: If P
~
P~Q k A.
Parametric equations
x = x0 + tA1
y = y0 + tA2
z = z0 + tA3
Example Find a set of parametric equations for the line of the previous example.
Analytic geometry (AG) description of a line
a1 x + b1y + c1 z = d1
a2 x + b2y + c2 z = d2
Example Find an AG description of the line in the previous example.
Degrees of Freedom / Dimensions
≡ number of free parameters used to describe the geometrical object
~ be a vector in space, the object tA
~ (−∞ < t < ∞) has one free parameter,
Example Let A
therefore the object has dimension 1.
~ and B
~ are not parallel, and both 6= ~0, what is the object
Example If A
~ + sB
~
tA
−∞< t < ∞ −∞< s < ∞ ?
Planes in 3D Space
Basic geometrical facts
~ B
~ respectively) define a plane
(i) Two different intersecting lines l1 , l2 (directions given by A,
through them.
8
(ii) There is a unique line l3 that passes through the point of intersection and is ⊥ to both lines.
~ ·A
~=N
~ ·B
~ = 0)
~ k l3 defines a normal to the plane that contains l1 , l2. (N
A vector N
(iii) There is only one plane that contains a given point and is ⊥ to a given nonzero vector.
(i) can be viewed as the definition of a plane; (ii) & (iii) come from the three-dimensional nature of
space.
Vector description of a plane
A plane through two intersecting lines l1 and l2 can be generated as:
~ + sB
~
~r = ~r0 + tA
− ∞ < t < ∞, −∞ < s < ∞
~ B
~ are vectors along l1 , l2 respectively.
where ~r0 is the point of intersection of l1 , l2, and A,
~ (tA
~ + sB)
~ = 0.
A normal is perpendicular to all directed line segments on the plane as N·
Recall (ii) above for the definition of a normal. A vector is a normal to a plane iff it is perpendicular
to all vectors associated with the directed line segments on the plane.
Parametric description of a plane:
Decompose the vector equation into components, as (linear) functions of the parameters.
x = x0 + tA1 + sB1
y = y0 + tA2 + sB2
z = z0 + tA3 + sB3
This approach is not used much since it needs three vectors (vs two for the AG description).
However, it is used for general surfaces. The coordinates of a point on a surface can be specified
as:
x = f1 (t, s)
y = f2 (t, s)
z = f3 (t, s)
which are components of a function
f : R2 → R3 .
AG description of a plane:
~ be a normal. Then
Let P be a fixed point on the plane, Q be any point on the plane, and N
~ · P~Q = 0
N
~ = (x, y, z), and N
~ = (a, b, c), one has
Writing P~ = (x0 , y0, z0 ), Q
(a, b, c)· [(x, y, z) − (x0 , y0 , z0)] =
9
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0.
The resultant equation is of the form
ax + by + cz = d.
A normal to the plane can be found easily as
~ = a~i + b~j + c~k.
N
Example: Find an equation of the plane that contains the point (−2, 4, 5) and has normal vector
7~i − 6~k
Example: Find a unit normal to the plane
x + y + z = 1.
Example: Find the plane through the three points
(0, 0, 1), (1, 0, 1), (0, 1, 1).
—– Problem Set 2 —–
Vector-Valued Functions
Definition: A vector-valued function consists of two
parts: a domain, which is a subset of R, and a rule, which assigns to each number in
the domain one and only one vector.
The rule is usually given as a formula. The formula of the function can be expressed as
~r(t) = f1 (t)~i + f2 (t)~j + f3 (t)~k
where t is in the common domain of f1 , f2 , f3 (single-variable functions called the component
functions).
The range (or image) of a vector-valued function is a curve in space.
~ = (1 + 2t)~i + 3t~j + 1 ~k
Example: A(t)
t
√
~
~
~
B(t) = 1 + 2t i + 4tk
~
C(t)
= sin t ~i + t~k
t ∈ [0, 2π]
~ excludes the point t=0]
[domain of A
~
[domain of B={t|
t≥− 1 }]
2
~
[plot C(t)]
10
{~
p| p
~=~
r (t) t∈ domain}
Limits of Vector-Valued Functions
~ be a vector-valued function defined at each point in some open interval conDefinition Let A
~ is the limit of A(t)
~
taining t0 , except possibly at t0 itself. A vector L
as t approaches
t0 if for every > 0 ∃ a δ > 0 such that
~ − L|
~ <ε
|A(t)
whenever
0 < |t − t0 | < δ.
~ = L.
~
We write lim A(t)
t→t0
Detailed discussion of ’limit’ will be postponed to the “several-variable functions” section.
~ = A1 (t)~i + A2 (t)~j + A3 (t)~k. Then A
~ has a limit at t0 if and only if A1 , A2 ,
Theorem Let A(t)
and A3 have limits at t0 . In that case
~ = [ lim A1 (t)]~i + [ lim A2 (t)]~j + [ lim A3 (t)]~k.
lim A(t)
t→t0
t→t0
t→t0
Example Find lim (2 cos t ~i +
t→0
t→t0
sin t ~
j + t2~k).
t
Useful Rules
~
~
Assuming that lim A(t),
lim B(t),
lim f (t) exist, then
t→t0
t→t0
t→t0
~ ± lim B(t)
~
~ ± B)(t)
~
1. lim (A
= lim A(t)
t→t0
t→t0
t→t0
~
~
2. lim [f A(t)]
= lim f (t) lim A(t)
t→t0
t→t0
t→t0
~ B)(t)
~
~
~
3. lim (A·
= lim A(t)·
lim B(t)
t→t0
t→t0
t→t0
~ × B)(t)
~
~ × lim B(t)
~
4. lim (A
= lim A(t)
t→t0
t→t0
t→t0
Continuity
~ is continuous at a point t0 in its domain if
Definition A vector-valued function A
~ = A(t
~ 0 ).
lim A(t)
t→t0
~ is continuous at t0 if and only if each of its component
Theorem: A vector-valued function A
functions is continuous at t0 .
~ = 2 cos t ~i + sin t ~j + t2~k is discontinuous at t = 0.
Example: A(t)
t
~
Theorem If A(t)
is continuous at t0 and lim g(s) = t0 , then
s→s0
~
~ lim g(s)) = A(t
~ 0 ).
lim (A(g(s))
= A(
s→s0
s→s0
11
Derivatives
~ If
Definition Let t0 be a number in the domain of a vector-valued function A.
lim
t→t0
~ − A(t
~ 0)
A(t)
t − t0
~ 0 ).
~ at t0 and write it as A
~ 0 (t0 ) or d A(t
exists, we call this limit the derivative of A
dt
~ is differentiable at t0 .
We say that A
Theorem
~ 0 (t0 ) = A0 (t0 )~i + A0 (t0 )~j + A0 (t0 )~k
A
1
2
3
Example Find the derivative of
~ = et cos t~i − ln(t)~j
A(t)
Useful Formulas for Differentiation
~ B,
~ and f be differentiable at t0 , and let g be differentiable at s0 with g(s0 ) = t0 . Then
Let A,
~ ± B)
~ 0 (t0 ) = A
~ 0 (t0 ) ± B
~ 0 (t0 )
1. (A
~ 0 (t0 ) = f 0 (t0 )A(t
~ 0 ) + f (t0 )A
~ 0 (t0 )
2. (f A)
~ B)
~ 0 (t0 ) = A
~ 0 (t0 )· B(t
~ 0 ) + A(t
~ 0 )· B
~ 0 (t0 )
3. (A·
~ × B)
~ 0 (t0 ) = A
~ 0 (t0 ) × B(t
~ 0 ) + A(t
~ 0) × B
~ 0 (t0 )
4. (A
5.
d ~
~ 0 (g(s0))g 0(s0 ) = A
~ 0 (t0 )g 0 (s0 )
A(g)(s0) = A
ds
Curves in Space
We will generally use the symbol C to denote a curve. If the range of a continuous vectorvalued function ~r is C, we say that ~r is the parameterization of C, and C is parameterized by
~r.
A curve C is closed if it has a parameterization whose domain is a closed interval [a, b] such
that ~r(a) = ~r(b).
A vector-valued function ~r(t) defined on an interval I is smooth if ~r has a continuous derivative
on I and ~r0 (t) 6= ~0 for each interior point t. A curve C is smooth if it has a smooth parameterization.
Example Consider the curve parameterized by
~r(t) = t3~i + t2~j.
[plot with the substitution s=t3 ]
Higher-Order Derivatives
~0
~0
~ 00 (t0 ) = lim A (t) − A (t0 )
A
t→t0
t − t0
12
Geometrical Interpretation of Derivatives
Tangent vector to a point on a curve
Velocity and acceleration
Position: ~r(t) = x(t)~i + y(t)~j + z(t)~k
Velocity: ~v (t) =
speed:
dx
dy
dz
d
~r(t) = ~r˙ = ~i + ~j + ~k
dt
dt
dt
dt
|~v(t)| =
Acceleration: ~a(t) =
r
(
dx 2
dy
dz
) + ( )2 + ( )2
dt
dt
dt
d~v
d2~r
= 2 = ~r¨
dt
dt
Example Show that the velocity vector of a particle moving on a sphere
perpendicular to the radial vector.
(radius = r0 )
is always
Example Show that the projection of the acceleration vector of a particle moving on a sphere
|~r˙ |2
on the radial direction is −
r̂.
|~r|
—– Problem Set 3 —–
13
PART II: MULTI-VARIABLE CALCULUS
Functions of Several Variables
Definition A function of several variables consists of two parts: a domain, which is a set of points
in the plane or in space, and a rule, which assigns to each member of the domain one
and only one real number.
Example Area of a rectangle
[note domain limitation]
Example f (x, y) = 2x2 y 2 + 4x2 − 7y 2 + 2y +
are of the form cxm y n ).
√
3 is an example of a polynomial formula (terms
Note A formula (like the f (x, y) above) is not a function; it is only a rule of assignment.
The natural domain of a formula for n-independent variables consists of n-tuples that can be
properly evaluated by the formula (to produce real numbers).
Example Find the natural domain of
p
f (x, y, z) = xyz 2 .
Generating New Functions from Known Functions
Algebraic combination of functions
(f ± g)(x, y) = f (x, y) ± g(x, y)
(f · g)(x, y) = f (x, y)g(x, y)
f
f (x, y)
(x, y) =
for g(x, y) 6= 0
g
g(x, y)
Note The domain of the new function is (a subset of) the intersection of those of the original
functions.
Composite function
If g is a single variable function and f is a multivariable function (say of two variables),
(g ◦ f )(x, y) = g(f (x, y))
is the formula of g ◦ f .
There are two conditions that restrict the domain of g ◦ f, Dg◦f : (i) (x, y) ∈ Df , (ii) f (x, y) ∈
Dg . Therefore,
Dg◦f = Df ∩ {(x, y)| f (x, y) ∈ Dg }.
Example Consider the domain of g ◦ f where g(x) =
some f (x,y) f all outside domain of g
√
x and f (x, y) = y.
(−∞,∞)×[0,∞)
Example Consider the domain of g ◦ f where g(x) = x2 and f (x, y) =
natural domain of g(f (x,y)) bigger than domain of f
14
√
xy.
Example Let f (x, y) = ln(4 − x2 − y 2 ) and g(x) =
√
x. Find the domain of g ◦ f .
Graph and Level Curves of a two-variable function
Definition The graph of a function f of two variables is the collection of points (x, y, f (x, y)) for
which (x, y) is in the domain of f
Definition The set of points (x, y) in the xy plane such that f (x, y) = c is a level curve of f .
Note Different level curves (i.e. different c) do not intersect.
Example Let f (x, y) = sin y. Sketch the graph.
Example Find the domain, draw some level curves (e.g. f = 0, 1/2, 1, 2), and sketch the
graph of f .
r
x2
y2
f (x, y) = 1 − 2 − 2 .
a
b
(an example of quadric surfaces)
Level surfaces of a three-variable function
Definition The set of points (x, y, z) in space such that f (x, y, z) = c is a level surface of f .
y2
x2
−
− z2 .
a2
b2
Sketch the level surfaces g(x, y, z) = c for c = −1, 0, 1, 2.
Example Let g(x, y, z) = 1 −
Quadric Surfaces
Definition: A quadric surface is a surface which contains points that satisfies the 2nd degree
equation (AG description)
Ax2 + By 2 + Cz 2 + Dxy + Eyz + F zx + Gx + Hy + Iz + J = 0
where A, ..., J are constants.
Quadric surfaces fall into nine major classes. Examples of each are provided as following (a, b, c >
0):
x2
y2
z2
+
+
=1
a2
b2
c2
x2
y2
z2
Hyperboloid of one sheet
+ 2 − 2 =1
2
a
b
c
2
2
x
y
z2
Hyperboloid of two sheets
−
−
=1
a2
b2
c2
x2
y2
z2
Elliptic double cone
+
−
=0
a2
b2
c2
x2
y2
z
Elliptic paraboloid
+
=
2
2
a
b
c
1. Ellipsoid
2.
3.
4.
5.
15
x2
y2
z
−
=
2
2
a
b
c
x2
y2
7. Elliptic cylinder
+ 2 =1
a2
b
x2
y2
8. Hyperbolic cylinder
−
=1
a2
b2
z
9. Parabolic cylinder x2 =
c
6. Hyperbolic paraboloid
Limits and Continuity
Definition In the xy-plane, the open disk centered at (x0 , y0 ) with a radius δ is the set
Dδ (x0 , y0 ) = {(x, y)| |(x, y) − (x0 , y0 )| < δ}
The deleted open disk D δ (x0 , y0 ) is
{(x, y)| 0 < |(x, y) − (x0 , y0 )| < δ}.
Note In 3D, we use the open ball Bδ (x0 , y0 , z0) & deleted open ball B δ (x0 , y0 , z0 ).
If A is a subset of the domain of f , the image of A can be written as
f (A) = {y| y = f (x), x ∈ A}.
Definition Let f (x, y) be defined throughout an open disk centered at (x0 , y0 ), except possibly
at (x0 , y0 ) itself, and let l be a number. Then l is the limit of f at (x0 , y0 ) if for every
ε > 0 ∃ a δ > 0 such that if
p
0 < (x − x0 )2 + (y − y0 )2 < δ, then
|f (x, y) − l| < ε.
In this case we write
lim
(x,y)→(x0 ,y0 )
f (x, y) = l,
and say that
lim
(x,y)→(x0 ,y0 )
f (x, y) exists.
Another way to write the requirement is:
∀ε > 0, ∃δ > 0 such that f (D δ (x0 , y0 )) ⊂ Nε (l).
Example Show that
lim
x = x0
(x,y)→(x0 ,y0 )
and
lim
(x,y)→(x0 ,y0 )
y = y0
Limit Formulas
If
lim
(x,y)→(x0 ,y0 )
1.
f (x, y) and
lim
g(x, y) exist, then
(f ± g)(x, y) =
(x,y)→(x0 ,y0 )
lim
(x,y)→(x0 ,y0 )
2.
lim
(x,y)→(x0 ,y0 )
lim
f (x, y) ±
lim
(x,y)→(x0 ,y0 )
g(x, y)
(f g)(x, y) =
(x,y)→(x0 ,y0 )
lim
(x,y)→(x0 ,y0 )
f (x, y)
lim
(x,y)→(x0 ,y0 )
g(x, y)
16
3.
f
(x, y) =
g
(x,y)→(x0 ,y0 )
lim
if
lim
(x,y)→(x0 ,y0 )
lim
f (x, y)
lim
g(x, y)
(x,y)→(x0 ,y0 )
(x,y)→(x0 ,y0 )
g(x, y) 6= 0
Example Show that
a.
b.
lim
x3 + y 3
=1
x2 + y 2
lim
x3 + y 3
=0
x2 + y 2
(x,y)→(0,1)
(x,y)→(0,0)
[use x=r cos θ, y=r sin θ, and |a+b|≤|a|+|b|]
[Note: lim(x,y)→(0,0) ≡limr→0 ]
Theorem If a limit exists, it is a unique number.
[pf by contradiction]
Theorem Suppose that
lim
(x,y)→(x0 ,y0 )
f (x, y) = l.
Let ~r : (a, b) → R be a vector-valued function so that lim− ~r(t) = (x0 , y0 ) but
t→b
~r(t) 6= (x0 , y0) for any t ∈ (a, b), then
lim (f ◦ ~r)(t) = l.
t→b−
These theorems are useful for showing the non-existence of limits with:
Example Show that
Example Find
lim
(x,y)→(0,0)
lim
(x,y)→(0,0)
lim
along path 1
6=
x2 − y 2
does not exist.
x2 + y 2
x3 y
.
2x6 + y 2
[y=mx, y=x3 ]
x2 − y 2
?
(x,y)→(0,0) x + y
Note the subtlety of the definition of limit.
Example What about
lim
Substitution Rule (for composite functions)
If
lim
(x,y)→(x0 ,y0 )
f (x, y) = L and g is a single-variable function continuous at L, then
lim
(x,y)→(x0 ,y0 )
g(f (x, y)) = g(L).
In other words,
lim
(x,y)→(x0 ,y0 )
g(f (x, y)) = g(
[Note that f need not be def ined at (x0 ,y0 )]
17
lim
(x,y)→(x0 ,y0 )
f (x, y)).
lim
.
along path 2
p
sin x2 + y 2
Example Find
lim
ln( p
)
(x,y)→(0,0)
x2 + y 2
Example Let
g(x) =
and f (x, y) = xy, show that
lim
(x,y)→(0,0)
n
0
1
x<0
x≥0
g ◦ f (x, y) does not exist.
Continuity
Definition Suppose f is a function of two variables defined throughout an open disk about
(x0 , y0 ). Then f is continuous at (x0 , y0 ) iff
lim
(x,y)→(x0 ,y0 )
f (x, y) = f (x0 , y0).
Definition If a function f is continuous at each point in its domain, then it is said to be a
continuous function.
Note A composite of continuous functions is continuous.
Example
F (x, y) = sin
xy
1 + x2 + y 2
is a continuous function.
—– Problem Set 4 —–
Limit and Continuity on a Set (general case)
Open and close sets
Let R be a set in the plane. Then for each point P in the plane, one and only one of the
following possibilities holds:
1. ∃ an open disk centered at P and contained totally in R. In this case P is an interior points
of R.
2. ∃ an open disk centered at P and containing no points of R. In this case P is an
exterior point of R.
3. Every open disk centered at P contains a point in R and a point outside of R. In this case
P is a boundary point of R.
Definition The collection of boundary points of R is the boundary of R.
Definition If a set R contains its boundary, then R is closed.
Definition If a set R contains only interior points, then R is open.
Example The open disk Dε (x0 , y0 ) =
p
{(x, y)| (x − x0 )2 + (y − y0 )2 < ε}
is open.
Example The set {(x, y)| |x| < 1, |y| ≤ 1} is neither open nor close.
18
Limit and continuity at a boundary point
Definition Let (x0 , y0 ) be on the boundary of R. A number L is the limit of f restricted to R
at (x0 , y0 ) if for every ε > 0, ∃ a number δ > 0 such that if (x, y) ∈ R and
p
0 < (x − x0 )2 + (y − y0 )2 < δ, then
|f (x, y) − L| < ε. Then, we write
lim
(x,y)R →(x0 ,y0 )
f (x, y) = L.
Example Let R be the natural domain of
x2 − y 2
f (x, y) =
.
x+y
Consider
lim
f (x, y).
(x,y)R →(0,0)
Continuity at a boundary point (x0 , y0 ) can then be defined by the condition
lim
(x,y)R →(x0 ,y0 )
f (x, y) = f (x0 , y0 ).
Continuity on a set (general situation)
Definition If f is continuous at every interior point of R and
lim
(x,y)R →(x0 ,y0 )
f (x, y) = f (x0 , y0)
for every boundary point (x0 , y0 ) in R, we say that f is continuous on R.
Theorem Let R be a close, bounded set in the plane, and let f be continuous on R. Then f
has both a maximum and a minimum on R.
Definition A set is ‘bounded’ if there exists a number M so that the distances of all its points
to the origin are less than M .
Example Consider f (x, y) = xy on [0, 1] × [0, 1] − {(1, 1)}
[pf by contradiction.
x<1 or y<1]
Partial Derivatives
Use graph, mention the tangent lines (plane) to motivate
Definition Let f be a function of two variables, and let (x0 , y0 ) be in the domain of f . The
partial
derivative of f with respect to (w.r.t.) x
at (x0 , y0 ) is defined by
fx (x0 , y0 ) = lim
∆x→0
f (x0 + ∆x, y0 ) − f (x0 , y0 )
∆x
provided that this limit exists.
The notations
∂
f , ∂x f are also in use.
∂x
19
The tangent line to the curve (x, y0 , f (x, y0)) through the point (x0 , y0 , z0) where z0 = f (x0 , y0 ) is
described by the vector equation (x, y, z) = (x0 , y0 , z0) + (x − x0 )(1, 0, fx). Similarly the tangent
line to the curve (x0 , y, f (x0, y)) can be found. The plane that contains these two lines is given by
−fx (x − x0 ) − fy (y − y0 ) + (z − f (x0 , y0 )) = 0.
The partial derivative of f w.r.t. y at (x0 , y0 ) can be
similarly defined.
Example Let z = sin(xy 2 ). Find
∂z
∂z
and
.
∂x
∂y
Formulas
(f ± g)x = fx ± gx
(f g)x = fx g + f gx
fx g − f gx
(f /g)x =
g2
for g 6= 0
Similarly for ( )y
Higher-Order Partial Derivatives
2nd order
(fx )x
fxx
(fx )y
fxy
(fy )x
fyx
(fy )y
fyy
∂ 2f
∂x2
∂ 2f
∂y∂x
∂ 2f
∂x∂y
∂ 2f
∂y 2
mixed partials
mixed partials
Example Let f (x, y) = sin(xy 2 ). Find all 2nd order partial derivatives of f (illustrate that
fxy = fyx ).
[fxy =−2xy 3 sin(xy 2 )+2y cos(xy 2 )]
Theorem Let f be a function of two variables, and assume that fxy and fyx are continuous at
(x0 , y0 ). Then
fxy (x0 , y0 ) = fyx (x0 , y0 ).
Since polynomials, trigonometric functions, exponential and logarithmic functions are continuously differentiable
everywhere (in their domains), mixed partials of their composites
can have the orderings switched under most circumstances.
Though a necessary requirement, the existence of fx and fy does not guarantee differentiability.
It does not even guarantee continuity.
Example Consider
f (x, y) =
(
0 on the x and y axes
1 otherwise
20
Differentiability
Definition Let 4x = x − x0 and 4y = y − y0 . If f is such that 4f = f (x, y) − f (x0 , y0 ) can
be expressed in the form 4f = fx 4x + fy 4y + 1 4x + 2 4y where 1 , 2 → 0 as
4x, 4y → 0, we call f differentiable at (x0 , y0).
Consider the plane though the two first-partial tangents.
Note This definition is equivalent to the regular definition in the single-variable situation.
Example Consider f (x, y) = xy. At (0, 0), f = 0, fx = 0 and fy = 0. As ∆x = x and ∆y = y,
one can pick, for example, 1 = y/2 and 2 = x/2.
Theorem If fx and fy are continuous at (x0 , y0 ), f is differentiable at the point.
Theorem differentiability ⇒ continuity
Definition A function f is called
differentiable on a region R if it is differentiable at each point of R.
Tangent Plane
If f (x, y) is differentiable, a tangent plane to its
graph at the point (x0 , y0 , f (x0, y0 )) can be defined. The equation
z = f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0)(y − y0 ).
describes the tangent plane that passes through the point. A normal to this plane is
(fx (x0 , y0), fy (x0 , y0 ), −1).
p
Example Let f (x, y) = 1 − x2 − y 2 , find the tangent plane to the graph of this function at
(0, 0, 1). Note: normal k position vector.
Tangent Plane Approximation
When f is differentiable at (x0 , y0 ), the value of f at a point (x, y) near (x0 , y0 ) can be
approximated by
f (x, y) − f (x0 , y0 ) ≈ fx (x0 , y0 )4x + fy (x0 , y0 )4y.
i.e. f (x, y) ≈ z while (x, y, z) is a point on the tangent plane through (x0 , y0 , f (x0, y0 )). The exact
value of f (x, y) is not necessarily z, but it can be closely approximated by z (the error terms are
1 4x + 2 4y).
(graphical illustration of the tangent plane approximation).
Example A rectangular cardboard box has outer dimensions 30, 30, and 20 cm. If the cardboard
is 3 mm thick, estimate the volume of cardboard.
[∆x=6mm; ∆V ≈1260cc]
21
Differentials
As 4x, 4y become very small, the approximation above can be written in the form
df = fx (x, y)dx + fy (x, y)dy.
It is called the differential of f at (x, y). dx and dy are called the differentials of x and y, respectively.
—– Problem Set 5 —–
Differentiation of Composite Functions
Chain rules: Assuming that f, r1 , r2, h1 , h2 are differentiable:
1. Let z = f (x, y), x = r1 (t) and y = r2 (t). Then z = f (r1 (t), r2(t)), and
dz
∂z dx ∂z dy
=
+
dt
∂x dt
∂y dt
(diff. along a curve)
pf use defn of differentiability
Note ~r(t) = r1 (t)~i + r2 (t)~j is a vector-valued function that traces a curve on the (x, y)
plane.
z(t) = (f ◦ ~r)(t)
2. Let z = f (x, y), x = h1 (u, v) and y = h2 (u, v). Then z = f (h1 (u, v), h2(u, v)), and
∂z
∂z ∂x ∂z ∂y
=
+
∂u
∂x ∂u ∂y ∂u
∂z
∂z ∂x ∂z ∂y
=
+
∂v
∂x ∂v
∂y ∂v
(use directed graphs)
Example Let z = x2 ey , x = sin t, and y = t3 . Find
dz
.
dt
Example Let z = xln(y), x = u2 + v 2 , and y = u2 − v 2 . Find ∂z/∂u and ∂z/∂v.
Example (implicit differentiation)
a. Given y + sin(yx2 ) = 1, find dy/dx.
[−2xy cos x2 y/(1+x2 cos x2 y)]
b. Assuming that f (x, y) = 0 defines a differentiable function y = g(x) of x, so that
f (x, g(x)) = 0. Find g 0 in terms of the partial derivatives of f .
Directional Derivatives
– differentiation along a direction different from the x-, y-axes.
Definition Let f be a function defined on a set containing an open disk centered at (x0 , y0 ), and
let û = u1~i + u2~j be a unit vector. Then the directional derivative of f at
22
(x0 , y0 ) in the direction of û, denoted
Dû f (x0 , y0 ), is defined by
f (x0 + su1 , y0 + su2 ) − f (x0 , y0 )
s→0
s
provided that this limit exists.
Dû f (x0 , y0 ) = lim
Note s is a distance parameter along the direction of û. |s| measures the distance to (x0 , y0 ).
Theorem Let f be differentiable at (x0 , y0 ). Then f has a directional derivative at (x0 , y0 ) in
every direction.
Moreover, if û = u1~i + u2~j,
Dû f (x0 , y0 ) = fx (x0 , y0 )u1 + fy (x0 , y0 )u2 .
Pf: Let g1 (s) = x0 + su1 & g2 (s) = y0 + su2 and use the Chain Rule.
Example Let f (x, y) = xy 2 and ~u = ~i − 2~j. Find the directional derivative of f at (−3, 1) in
√
the direction of ~u.
[13/ 5]
The Gradient
Definition Grad f (x0 , y0 )
~ (x0 , y0 ) = fx (x0 , y0 )~i + fy (x0 , y0 )~j
= 5f
~ (x0 , y0)
Theorem a. Dû f (x0 , y0) = û· 5f
~ (x0 , y0 )|| cos φ
b. Dû f (x0 , y0 ) = ||5f
~ )
(φ is the angle between ~u and 5f
~ (x0 , y0 ) is a vector.
Note Dû f (x0 , y0 ) is a scalar while 5f
~ ||, and this value is obtained when û points in the
The largest value of Dû f is ||5f
~ .
direction of 5f
Example Let f (x, y) = 6 − 3x2 − y 2 . Determine the directions in which f increases/decreases
most rapidly at (1, 2) and find the maximal value of the directional derivative.
√
~ (1,2)=(−6,−4), ||5f
~ (1,2)||= 52]
[5f
The Gradient as a Normal Vector
Let C be a level curve f (x, y) = c of a function f . Let (x0 , y0 ) be a point on C, and assume
~ (x0 , y0 ) 6= ~0, then 5f
~ (x0 , y0 ) is normal
that f is differentiable at (x0 , y0 ). If C is smooth and 5f
to C at (x0 , y0 ).
Pf.: Let ~r(t) = x(t)~i + y(t)~j be a smooth parameterization of C, then
dx
dy
d~r ~
d
f (x(t), y(t)) = fx
+ fy
=
· 5f = 0.
dt
dt
dt
dt
Note The first version of the Chain Rule can be written as
d(f ◦ ~r)(t)
~ · d~r
= 5f
dt
dt
23
Example Find a unit vector ⊥ to the curve x2 − xy + 3y 2 = 5 at (1, −1).
[(1/
√
58)(3,−7)]
Theorem Let S be a level surface of a function f and (x0 , y0 , z0 ) is a point on S. If f is
~ (x0 , y0, z0 ) 6= ~0, then 5f
~ (x0 , y0 , z0) is ⊥ to the
differentiable at (x0 , y0, z0 ) and 5f
tangent vector at (x0 , y0 , z0 ) of any smooth curve lying on S and passing through this
~ (x0 , y0 , z0 ) is a normal to the tangent plane of S at (x0 , y0 , z0).
point. Therefore, 5f
√
Example Find an equation of the plane tangent to the sphere x2 + y 2 + z 2 = 4 at (−1, 1, 2).
Now suppose that f is a function of two variables that is differentiable at (x0 , y0). In order to
obtain an equation of the plane tangent to the graph of f at (x0 , y0 ), one can think of the graph
of f as the level surface
g(x, y, z) = 0 where g = f (x, y) − z. Then
~
~
~ ~
5g(x
0 , y0 , z0) = fx (x0 , y0 )i + fy (x0 , y0 )j − k
provides a normal to the tangent plane of the graph at (x0 , y0 ).
Example Find an equation for the tangent plane to the graph of
f (x, y) = 6 − 3x2 − y 2 at (1, 2, −1).
—– Problem Set 6 —–
Taylor Series
Can f (x, y) = f (x0 +∆x, y0 +∆y) be expressed in terms of f (x0 , y0 ) and the partial derivatives
of f at (x0 , y0 )?
Let (x0 , y0 ) + sû be the parameterization of a line that passes through (x0 , y0) and (x, y), with û
being the unit vector
1
u1~i + u2~j ≡ p
(∆x~i + ∆y~j).
2
∆x + ∆y 2
The parameter s is the distance from (x0 , y0 ) to a point on the line.
The function f (x0 + su1 , y0 + su2 ) can be considered as a single-variable function of s. From the
discussion of directional derivative, we know that
d
∂
∂
f = Du f = (u1
+ u2 )f
ds
∂x
∂y
in which Dû is written as a differential operator.
With û fixed, Dû f can be considered as a function of (x, y). Suppose that f is sufficiently differentiable, then
d2
f = Du2 f , ......
ds2
Considered as a function of s, the Taylor series of
f (x0 + su1 , y0 + su2 ) can be written as
= f (x0 , y0 ) + sDu f +
1 2 2
1
s Du f + · · · + sn Dun f + · · ·
2!
n!
24
p
∆x2 + ∆y 2 , one obtains
∂
∂
+ ∆y )f ,
sDu f = (∆x
∂x
∂y
∂
∂
s2 Du2 f = (∆x
+ ∆y )2 f ,
∂x
∂y
By choosing s =
· · · · · ·,
and
∂
∂
+ ∆y )f
∂x
∂y
1
∂
∂
1
∂
∂
+ (∆x
+ ∆y )2 f + · · · + (∆x
+ ∆y )n f + · · ·
2!
∂x
∂y
n!
∂x
∂y
f (x0 + ∆x, y0 + ∆y) = f (x0 , y0 ) + (∆x
assuming that all relevant derivatives exist.
[Usef ul f or determining max/min.]
Extreme Values
Definition Let f be a function of two variables, R a set contained in the domain of f , and (x0 , y0 )
a point in R. Then f has a maximum value (respectively, a minimum value) on R at
(x0 , y0 ) if f (x, y) ≤ f (x0 , y0) (respectively, f (x, y) ≥ f (x0 , y0 )) ∀(x, y) in R. If R is
the domain of f , we say that f has a maximum value (respectively, a minimum value)
at (x0 , y0 ).
Definition f (x, y) has a relative maximum value (respectively, a relative minimum value)
at (x0 , y0 ) if there is an open disk D centered at (x0 , y0 ) and contained in the domain
of f such that f (x0 , y0 ) is the maximum value (respectively, the minimum
value) on D.
Extremum ≡ maximum or minimum.
Theorem If f (x, y) has a relative extreme value at
(x0 , y0 ) and its first partials exist at
(x0 , y0 ), then
fx (x0 , y0 ) = fy (x0 , y0 ) = 0
~ (x0 , y0 ) = ~0.
or equivalently, 5f
Definition A point (x0 , y0 ) in the interior of the domain of f is a critical point of f if either
(i) fx (x0 , y0 ) = fy (x0 , y0 ) = 0,
or
(ii) any of the first partial derivatives does not exist.
The above theorem can be restated as: f has relative extreme values only at critical points in
its domain.
Example Let f (x, y) = 3 − x2 + 2x − y 2 − 4y. Find all critical points of f .
Example Consider the critical points of f (x, y) = |x| + y 2 .
25
[y−axis]
[(1,−2)]
Example Let f (x, y) = y 2 − x2 . Show that the origin is the only critical point but f (0, 0) is
not an extremum.
Definition If f is a function for which fx (x0 , y0) = fy (x0 , y0 ) = 0, we say that f has a
saddle point at (x0 , y0 ) if ∃ a disk centered at (x0 , y0 ) such that f assumes its maximum value on one diameter of the disk only at (x0 , y0 ) and assumes its minimum
value on another diameter of the disk only at (x0 , y0 ).
The Second Partials Test
Theorem Assume that f has a critical points at
(x0 , y0 ) and that f has continuous second partial derivatives in a disk centered at
(x0 , y0 ). Let
D(x0 , y0) = fxx (x0 , y0 )fyy (x0 , y0 ) − [fxy (x0 , y0 )]2
a. If D(x0 , y0 ) > 0 and fxx (x0 , y0 ) < 0 (or fyy (x0 , y0 ) < 0), then f has a relative
maximum value at (x0 , y0).
b. If D(x0 , y0 ) > 0 and fxx (x0 , y0 ) > 0 (or fyy (x0 , y0 ) > 0), then f has a relative
minimum value at (x0 , y0 ).
c. If D(x0 , y0 ) < 0, then f has a saddle point at (x0 , y0 ).
d. If D(x0 , y0 ) = 0, the nature of the critical point is unknown.
[T aylor series expansion is only f or a special case]
[g(m)=fxx +2fxy m+fyy m2 can be a parabola or a straight line.]
Example Let f (x, y) = x2 − 2xy + 31 y 3 − 3y. Determine locations of relative extrema and saddle
points.
[D=4(y−1)]
[(3,3) rel. min., (−1,−1) saddle pt.]
Example Consider f (x, y) = x3 and g(x, y) = x4 + y 4 .
Recall the following theorem:
The Maximum-Minimum Theorem
Theorem Let R be a close, bounded set in the plane, and let f be continuous on R. Then f
has both a maximum value and a minimum value on R.
Procedure to find the maximum & minimum:
1. Find the critical points of f in the interior of R, and compute the values of f at these
points.
2. Find the extreme values of f on the boundary of R.
3. The maximum value of f on R will be the largest of the values computed in steps 1 and 2,
and the minimum value of f on R will be the smallest of those values.
Example Find the maximum value of f (x, y) = xy on the close triangle bounded by the lines:
x = 0, y = 0, and x + 2y = 2.
26
[f (1,1/2)=1/2]
Optimization Subject to Constraints
Example (for motivation) Suppose heavy-duty tape is to be applied on the bottom and side
edges of a rectangular carton. If 288 cm of tape are available, find the maximum
volume of the carton.
[48×48×24]
Lagrange Multipliers
Theorem Let f, g be differentiable at (x0 , y0 ). Let C be a smooth curve defined by the constraint
~ 6= ~0 at any point on the curve. If f has an extreme value on C
g(x, y) = c, and 5g
at (x0 , y0 ), then there is a number λ such that
~ (x0 , y0 ) = λ5g(x
~
5f
0, y0 ).
pf: Let r(t) be a smooth parameterization of g(x...)=c, then df(r(t))/dt=0.
The three-variable case is similar. The constraint defines a surface. The equation containing
gradients looks the same.
The method of determining extreme values by means of Lagrange multipliers proceeds as
follows:
1. Assume that f has an extreme value on the curve g(x, y) = c.
2. Solve the equations
(
~ (x, y) = λ5g(x,
~
5f
y)
g(x, y) = c
~ and λ = 0 is possible.
Note λ must be put on the side with ∇g,
3. Calculate the value of f at each point (x, y) that arises in step 2. If f has a maximum value
on the curve g(x, y) = c, it will be the largest of the values computed; Similarly, the minimum
can be found.
Example Let f (x, y) = x2 + 4y 3 . Find the extreme values of f on the ellipse x2 + 2y 2 = 1, and
the points at which they occur.
[f (0,±1/
√
2)=±
√
2, f (±1,0)=1, f (±
√
7/9,1/3)=25/27]
4. To find the extrema on a closed and bound region, include values at interior critical points
for comparison.
Example Let f (x, y) = 3x2 + 2y 2 − 4y + 1. Find the extreme values of f on the close disk
x2 + y 2 ≤ 16.
[boundary: f (0,±4)=17, 49
f (±
√
12,−2)=53 (max)]
[interior: f (0,1)=−1 (min)]
—– Problem Set 7 —–
27
Double Integrals
Motivation
Double integral as volume: Consider a region R in the xy-plane, and a function f that is nonnegative
and continuous on R. What is the volume of the solid region between the graph of f and R? [sketch]
There are two steps in the consideration:
1. Area of the base region
First, consider a convex region R on the xy-plane (so that a straight line segment connecting
any two points in R lies inside R). How to find its area?
We use the standard divide and add strategy to estimate the area. Apply a rectangular mesh (over
[a, b] × [c, d]) composed of uniform sub-rectangles Rk to cover the region R. Each sub-rectangle has
an area
b−a
d−c
.
∆Ak = ∆x∆y =
M
N
Count and add the areas of the sub-rectangles inside R:
X
4Ai .
Ri ⊂R
This will give an approximate value, but the approximation will become better as the mesh is
refined (i.e. ∆x, ∆y → 0).
2. The solid region between the graph of f and R
The volume underneath the graph of f (x, y) over the region R can be approximated by the
sum
X
f (ξi , ηi)4Ai
(the Riemann sum for f on R).
Ri ⊂R
where (ξi , ηi) ∈ Ri (note the non-uniqueness of choice). f (ξi , ηi) represents an approximate height
of the slender rectangular sub-solid over Ri.
Next, try to take the limit
lim
∆x, ∆y→0
X
f (ξi , ηi)4Ai .
Ri ⊂R
If this limit exists, it is the volume of the solid region.
The limit of the Riemann sum, denoted by
ZZ
f (x, y) dA,
R
is called the double integral of f over R. This definition is used even when f is negative somewhere
and the ‘volume’ interpretation is not used.
Procedure to compute the volume/double integral
First, consider a rectangular region R = [a, b] × [c, d]. Divide the region into M × N subrectangles (M, N are integers) with sides ∆x = (b − a)/M and ∆y = (d − c)/N . The interval [a, b] is
28
partitioned into M equal subintervals [x0 , x1], ..., [xM −1 , xM ], and the interval [c, d] is partitioned
into N equal subintervals [y0 , y1 ], ..., [yN −1 , yN ].
The Riemann sum can now be rewritten as a double sum
N X
M
X
f (ξi , ηj )∆x∆y.
j=1 i=1
where ξi is a point in the subinterval [xi−1 , xi] and ηj is a point in [yj−1 , yj ]. For example, if one
picks the point to be the upper-right corner of the subrectangle, (ξi , ηj ) = (xi , yj ). Now try to take
the limit of this sum for ∆x → 0, ∆y → 0.
Theorem (Fubini’s Theorem) If f is continuous, this limit always exists. The result is
Z
c
d
Z
!
b
f (x, y)dx dy.
a
Next, consider slightly more complex regions.
Vertically and Horizontally Simple Regions
Definition 1. A plane region R is vertically simple if there are two continuous functions g1 and
g2 on an interval [a, b] such that g1 (x) ≤ g2 (x) for a ≤ x ≤ b and such that R is the
region between the graphs of g1 and g2 on [a, b].
2. A plane region R is horizontally simple if there are two continuous functions h1
and h2 on an interval [c, d] such that h1 (y) ≤ h2 (y) for c ≤ y ≤ d and such that R is
the region between the graphs of h1 and h2 on [c, d].
3. A plane region R is simple if it is both vertically simple and horizontally simple.
All these regions can be approximated as composed of rectangular slices.
Cover the region with a larger rectangle. The double sum for case 1 can be written as
M N
2 (i)
X
X
f (xi, yj )∆y∆x
where N1 , N2 depend on i.
i=1 j=N1 (i)
Evaluation of Double Integrals
Theorem Let f be continuous on a region R in the xy plane.
1. If R is the vertically simple region between the graphs of g1 and g2 on [a, b], then
f is integrable on R, and
ZZ
f (x, y)dA =
Z bZ
a
R
(called iterated integral)
29
g2 (x)
g1 (x)
f (x, y)dydx.
2. If R is the horizontally simple region between the graphs of h1 and h2 on [c, d],
then
ZZ
Z dZ h2 (y)
f (x, y)dA =
f (x, y)dxdy.
c
R
h1 (y)
3. If R is simple, then
ZZ
f (x, y)dA =
Z bZ
a
R
g2 (x)
f dydx =
g1 (x)
Z dZ
c
Example Let f (x, y) = 1,
h2 (y)
f dxdy.
h1 (y)
R = [a, b] × [c, d]. Evaluate
ZZ
f dA
ZZ
1dA
R
Note The area of a plane region is given by
R
Example Let R = [0, 1] × [2, 3]. Evaluate
ZZ
x2 ydA
5/6
R
Example Let f (x, y) = 1 − 2y and R be the
RR triangular region between the graph of y = 1 − x
and the x axis on [−1, 1]. Find f dA. −2/3
R
Example Evaluate
RR
f dA of the above example by reversing the order of integration. Then
R
compare results.
Z 1Z y p
Example Evaluate
x y 2 − x2 dxdy .
0
1/12
0
What about reversing the order of integration?
R√
√
√
2
u2 −a2 du=(u/2)
u2 −a2 −(u /2) ln |u+
u2 −a2 |+C
Double Integrals in Polar Coordinates
Simple Regions
Suppose that h1 (θ) and h2 (θ) are continuous on an interval [α, β], and that
0 ≤ h1 (θ) ≤ h2 (θ)
∀θ ∈ [α, β]
Let R be the closed region in the (r, θ) plane bounded by the lines θ = α and θ = β and by
the polar graphs of r = h1 (θ) and r = h2 (θ). We say that R is the (simple) region between
the polar graphs of h1 and h2 on [α, β].
30
Theorem Let R be the region between the graphs of continuous functions h1 and h2 on [α, β].
If f is continuous on R, then
ZZ
f (x, y)dA =
Z βZ
α
R
h2 (θ)
f (r cos θ, r sin θ)rdrdθ
h1 (θ)
Consider integrating rf over a rectangular region of (r, theta)
Example Suppose R is the region bounded by the circles r = 1 and r = 2 and the lines θ = 0
and θ = π2 (R is a quarter-ring). Express
ZZ
(3x + 8y 2 )dA
R
as an integral in polar coordinates and evaluate the integral.
[sin 2 θ=(1−cos 2θ)/2; 7+15π/2]
Example Let D be the solid region bounded above by the paraboloid z = 4 − x2 − y 2 and below
by the xy plane. Find the volume of D.
[8π]
Surface Area of a Graph
First, consider the relationship between the area of a tilted rectangle (S = a b) and the area
of its projection (A = a b cos θ) on a plane.
S = ab =
A
cos θ
Let n̂ be the upward pointing (z component positive) unit normal vector to the rectangle. If the
plane is the xy-plane, then
cos θ = n̂· ~k
and
S=
A
A
=
.
n̂ · ~k
|n̂ · ~k|
Let Σ be the graph of a two-variable function f on a region R, then the surface area of Σ is
ZZ q
ZZ
dS =
fx2 + fy2 + 1 dA.
Σ
R
Pf: A normal to the graph at (x, y, f (x, y)) is fx~i + fy~j − ~k. The upward pointing unit normal
P
to
is
−fx~i − fy~j + ~k
n̂ = q
fx2 + fy2 + 1
Therefore n̂· ~k = q
1
fx2 + fy2 + 1
X
, and
4S =
X 4A
n̂· ~k
=
Xq
fx2 + fy2 + 1 4A.
Note The following expression works for ˆl = ~i, ~j, or ~k (different projections), and for either
direction of n̂:
dA = |n̂· l̂| dS.
31
The formula
ZZ
dS =
Σ
ZZ
R
dA
|n̂· l̂|
can be used for other projections.
Example Let R be the rectangular region [0, 3] × [0, 2] and f (x, y) = 32 x3/2 . Find the surface
area of the graph of f over R.
[28/3]
Example Find the surface area of the portion of the plane x + 2y + 3z = 6 inside the cylinder
√
x2 − 4x + y 2 = 0.
[4 14π/3]
Surface Integrals
n
X
k=1
g(xk , yk , zk )4sk →
ZZ
gdS
Σ
Let Σ be the graph of a function f having continuous partial derivatives and defined on a
region R in the xy plane that is composed of a finite number of vertically or horizontally simple
regions. Let g be continuous on Σ. The surface integral of g over Σ is
ZZ
g(x, y, z)dS
Σ
=
ZZ
R
Example Evaluate
ZZ
q
g(x, y, f (x, y)) [fx (x, y)]2 + [fy (x, y)]2 + 1 dA
z 2 dS where Σ is the portion of the cone z =
Σ
x2 + y 2 ≤ 4.
[15π/
√
p
x2 + y 2 for which 1 ≤
2]
—– Problem Set 8 —–
Triple Integrals
n
X
k=1
f (xk , yk , zk )4Vk →
ZZZ
f (x, y, z)dV
D
Theorem Let D be the solid (3-dimensional) region between the graphs of two continuous
functions F1 and F2 on a vertically or horizontally simple region R in the xy plane,
and let f be continuous on D. Then
ZZZ
D
f (x, y, z)dV =
ZZ
R
Z
F2 (x,y)
f (x, y, z)dz dA.
F1 (x,y)
Example Let D be the solid rectangular region
32
!
5
[2, ] × [0, π] × [0, 2] and
2
f (x, y, z) = zx sin xy. Evaluate
ZZZ
f dV
D
[1− 2 ]
π
Triple Integrals in Cylindrical Coordinates
Cylindrical Coordinates
(
(
x = r cos θ
y = r sin θ
z=z
r 2 = x2 + y 2
y
tan θ =
x
dV = dzrdrdθ
This coordinate system is best for problems with axial symmetry (e.g. the region is bounded by a
cylinder, and the integrand can be converted to depend on r and z only).
Example Evaluate the mass of a cylindrical rod with radius a, length l, and density f (x, y, z) =
1 + z where z is the distance from one end of the rod.
[πa2 l(1+l/2)]
Example Let D be the solid region bounded above by the plane y + z = 4, below by the xy
plane, and on the sides by the cylinder x2 + y 2 = 16. Evaluate
ZZZ p
x2 + y 2 dV
D
[44 2π/3]
Triple Integrals in Spherical Coordinates
Spherical Coordinates
(
x = r cos θ = ρ sin φ cos θ
y = r sin θ = ρ sin φ sin θ
z = ρ cos φ
ρ 2 = x2 + y 2 + z 2
dV = ρ2 sin φ dρ dφ dθ
This coordinate system is suitable for problems in which the solid region D can be described in
terms of the coordinates in a simple way (especially if the problem has spherical symmetry).
33
Note that in some texts, the symbols φ and θ are
switched.
Example Evaluate the mass of a solid ball with radius a and density 1+kρ2 .
[4π/3a3 (1+3ka2 /5)]
Example Let D be the solid region between the
spheres ρ = 1 and ρ = 2, and inside the cone φ = π/4. Evaluate
ZZZ
√
z 2 dV
[π(31/15)(2−1/ 2)]
D
Change of Variables in Multiple Integrals
Single integral
A substitution of x by g(u) converts an integral as following
Z
a
b
f (x)dx =
Z
g −1 (b)
f (g(u))g 0(u)du.
g −1 (a)
Double integral
Consider a transformation T from the uv-plane to the xy-plane with the parametric equations
x = x(u, v) &
y = y(u, v).
With v = v0 held fixed, the curve (x(u, v0), y(u, v0)) describes a “coordinate line” of constant v on
the xy-plane.
Example Consider polar coordinates in 2D.
A short secant vector of this line with initial point at (u0 , v0 ) is
~a = (x(u0 + ∆u, v0) − x(u0 , v0 ), y(u0 + ∆u, v0 ) − y(u0 , v0 ))
∂y
∂x
≈ ( ∂u
∆u, ∂u
∆u).
∂y
Similarly, a short secant vector of the v coordinate line can be approximated by ~b ≈ ( ∂x
∆v, ∂v
∆v).
∂v
The area of the parallelogram bounded between the two vectors can be obtained as the magnitude of their cross product, which gives
∆A ≈ |J(u, v)|∆u∆v
where
∂x
∂(x, y) ∂u
J(u, v) =
= ∂y
∂(u, v) ∂u
∂x
∂v
∂y
∂v
is called the Jacobian of the transformation T . |J(u, v)| is the absolute value of J(u, v).
34
Therefore, if the transformation x = x(u, v), y = y(u, v) maps the region S in the uv-plane onto
the region R in the xy-plane, and if the Jacobian ∂(x, y)/∂(u, v) is nonzero and does not change
sign on S, then
ZZ
ZZ
f (x, y)dAxy =
f (x(u, v), y(u, v))|J(u, v)|dAuv
R
S
in which the subscripts are attached to identify the associated variables and
S = T −1 (R) = {(u, v)|(x, y) = T (u, v), (x, y) ∈ R}.
Example Evaluate
ZZ
exy dA
R
where R is the region in the first quadrant enclosed by the lines y = 12 x and y = x
and the hyperbolas y = 1/x and y = 2/x.
[ 1 (e2 −e) ln 2]
2
Triple integral
If T is the transformation from uvw-space to xyz-space defined by the equations
x = x(u, v, w), y = y(u, v, w), z = z(u, v, w),
then the volume of the parallelepiped bounded by short secant vectors ~a, ~b, ~c of the coordinate lines
can be obtained by the magnitude of the triple product |(~a ×~b)·~c|, thus ∆V ≈ |J(u, v, w)|∆u∆v∆w.
∂x
∂u
∂y
∂(x, y, z)
J(u, v, w) =
= ∂u
∂(u, v, w)
∂z
∂u
is the Jacobian of T .
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
∂y
∂w
∂z
∂w
An integral is transformed as
ZZZ
f (x, y, z)dVxyz =
D
ZZZ
f (x(u, v, w), y(u, v, w), z(u, v, w))|J(u, v, w)|dVuvw
T −1 (D)
Example Show that the Jacobian of the transformation from cylindrical coordinates to rectangular coordinates is r.
—– Problem Set 9 —–
35
PART III: VECTOR FIELDS
[W e are getting less and less rigorous!]
~ consists of two parts: a collection D of points in space, called the
Definition A vector field F
domain, and a rule, which assigns to each point (x, y, z) in D one and only one vector
F~ (x, y, z).
Physical examples: electric field, gravity field,
fluid flows (water, air)
Plot examples including source & sink, vortex.
The Gradient of a Scaler Function Creates
a Vector Field
~ = fx~i + fy~j + fz ~k
grad f = ∇f
~ is equal to ∇f
~ for some differentiable function f (of several
Definition If a vector field F
~ is called a conservative
variables), then F
vector field, and f is a potential function of F~ .
Note In physics, −f is used as the potential of a force field.
Example Electric field of a point charge q.
~ = q r̂
E
r2
−q
−q
= 2
f=
r
(x + y 2 + z 2 )1/2
r̂ denotes the unit vector pointing along ~r.
Divergence and Curl
Definition Let F~ = F1~i + F2~j + F3~k be a vector field such that ∂F1 /∂x, ∂F2/∂y, ∂F3/∂z exist.
Then the divergence of F~ , denoted
~ or ∇·
~ F
~ , is the scalar function defined by
div F
~ F
~ = ∂F1 + ∂F2 + ∂F3
∇·
∂x
∂y
∂z
~ F
~ = 0, then F~ is said to be divergence-free.
If ∇·
Source and Sink
Let F~ (x, y, z) be a vector field in space (example: a flow field), then
~ F
~ >0
A point (x, y, z) is a source if ∇·
~ F
~ <0
A point (x, y, z) is a sink if ∇·
Example Consider the divergence of the field
36
F~ = ax ~i
for cases with a >, =, and < 0.
Example Consider the field of a uniformly charged sphere
~ =
E
~ E
~ =
∇·


rr̂
3
 a r̂
r2
for r ≤ a
for r > a



3 for r < a
undefined for r = a


0 for r > a
Note The divergence of a radial inverse square field is zero everywhere, except at the center.
Definition Let F~ = F1~i + F2~j + F3~k be a vector field such that the first derivatives of F1 , F2 , F3
all exist. Then the curl of F , denoted
~ × F~ , is defined by
curl F~ or ∇
∂~
∂~
∂~
i+
j+
k × F~
∂x
∂y
∂z
∂F3
∂F2 ~
∂F1
∂F3 ~
∂F2
∂F1 ~
=
−
i+
−
j+
−
k.
∂y
∂z
∂z
∂x
∂x
∂y
~ × F~ (x, y, z) =
∇
~ × F~
Example Let F~ (x, y, z) = xz~i + xy 2 z~j − e2y ~k. Find ∇
[−(2e2y +xy 2 )~
i+x~
j+y 2 z~
k]
~ × F~ = 0, F~ is said to be irrotational.
Definition If ∇
Example Consider the flow
~ = +Ωr θ̂
V
where Ω is a constant (angular velocity).
a. Plot the flow.
~ ×V
~ and ∇·
~ V
~.
b. Find ∇
Note that since θ̂ = − sin θ~i + cos θ~j,
V~ = Ω(−y~i + x~j).
~
Formulas Involving ∇
~ + g) = ∇f
~ + ∇g
~
1. ∇(f
~ (F~ + G)
~ = ∇·
~ F
~ + ∇·
~ G
~
∇·
~ × (F~ + G)
~ =∇
~ × F~ + ∇
~ ×G
~
∇
~ g) = f ∇g + (∇f )g
2. ∇(f
~ (f G)
~ = f ∇·
~ G
~ + (∇f
~ )· G
~
∇·
37
~ × (f G)
~ = f∇
~ ×G
~ + (∇f
~ )×G
∇
Special cases: f = a constant c
~ × (∇f
~ ) = ~0
3. Theorem: ∇
~ (∇
~ × F~ ) = 0
∇·
(curl grad = 0)
(div curl = 0)
Example For r 6= 0 the curl of the electric field
~ = q r̂ is ~0.
E
r2
~ × (∇
~ × F~ ) may not be ~0.
Note ∇
Example Show that the field of the example in the previous subsection is the curl of another
vector field:
2 r
~
~
~
~
V = Ωr θ̂ = Ω(−y i + xj) = Ω∇× − ~k
2
—– Problem Set 10 —–
Line Integrals
~ is a continuous vector field. Approximate
Let C be a smooth, oriented (directed) curve in space. F
P~
C by a collection of small line segments (directed) {4~ri }. Consider the sum
F (ξi)· 4~ri where
i
ξi is a point on the line segment 4~ri (the concept of work), then take the limit |4ri | → 0
X
i
F~ (ξi )· 4~ri →
Z
F~ · d~r
C
This is called the line integral of F~ over C.
If C is not smooth but is piecewise smooth, composed of smooth curves C1 , C2 , · · · Cn , then
Z
~ · d~r =
F
C
n Z
X
i=1 C
~ · d~r.
F
i
Also, with − C defined to be the curve having the same points but opposite orientation of C,
Z
F~ · d~r = −
Z
F~ · d~r.
C
−C
As F~ = F1~i + F2~j + F3~k and d~r = dx~i + dy~j + dz~k, another form to write a line integral is
Z
C
F~ · d~r =
Z
F1 dx + F2 dy + F3 dz.
C
38
Evaluation of Line Integrals
Let ~r(t) = x(t)~i + y(t)~j + z(t)~k be a parameterization of C with domain [a, b], and assume that
the parameterization induces the given orientation
(direction) of C. Then (the formula for evaluation)
Z
~ · d~r =
F
C
Z
b
a
~ ◦ ~r(t)· d~r dt .
F
dt
Example A particle moves upward along the circular helix C, parameterized by ~r(t) = cos t~i +
sin t~j + t~k for 0 ≤ t ≤ 2π under a force given by F~ (x, y, z) = −zy~i + zx~j + xy~k. Find
he work done on the particle by the force (i.e. the line integral).
[2π 2 ]
Example Assume that the particle in the previous example moves under the same force and with
the same initial and terminal points, but along the line segment C2 parameterized by
~r(t) = ~i + t~k
for 0 ≤ t ≤ 2π.
Find the work done on the particle by the force.
Z
Z
Note that
6=
C1
[0]
C2
The Fundamental Theorem of Line Integrals
Theorem Let C be an oriented curve with initial
point (x0 , y0 , z0 ) and terminal point
(x1 , y1 , z1 ). Let f be a function of three variables that is differentiable at every
~ is continuous on C. Then
point on C, and assume that ∇f
Z
~ · d~r = f (x1 , y1, z1 ) − f (x0 , y0, z0 )
∇f
C
pf: Use Chain Rule.
Example Let C be the straight line segment from (0, 2, 0) to (1, 0, 0) and F~ be the electric field
of a point charge q at the origin. Find the work done by F~ on a unit point charge
that traverses C.
[−q/2]
Path Independence
Definition If a vector field F~ has the property that
R
C1
F~ · d~r =
R
C2
~ · d~r for any two oriented
F
curves having the same initial and terminal points, its line integrals are called
path independent.
~ =⇒
Clearly, F~ = ∇f
Z
~
F~ · dr
path independent.
C
39
(Fundamental Theorem of Line Integrals)
On the other hand, if the line integrals of F~ are path independent, then a potential function
R
~ where C is an arbitrary curve connecting the origin (or
~ · dr
of F~ can be found as f (x, y, z) = F
C
any fixed reference point) to (x, y, z).
Z
~ path independent =⇒ F~ = ∇f
~ .
Therefore,
F~ · dr
C
[pf : r
~(s)=(s,y0 ,z0 ), s∈[x0 ,x], ∂x f (x0 ,y0 ,z0 )=(d/dx)
Example Let F~ = xy 2~i + x2 y~j.
Rx
x0
F1 (s,y0 ,z0 )ds]
a. Evaluate the line integrals
R
C1
R
C2
~ · dr
~ and
F
~ where C1 consists of the line segments connecting (0, 0) to (x0 , 0) and
~ · dr
F
(x0 , 0) to (x0 , y0 ), and C2 consists of the line segments connecting (0, 0) to (0, y0)
and (0, y0) to (x0 , y0 ).
[x2 y 2 /2]
0 0
b. Find a potential function for F~ .
Important Properties of a Conservative Field
Theorem The following statements are equivalent:
~ for some function f , i.e. conservative.
1. F~ = ∇f
Z
2.
F~ · d~r is independent of path.
C
3.
Z
F~ · d~r = 0 for every closed loop C.
C
~ is a region with no holes, then also
If the domain of F
~ × F~ = 0.
4. ∇
Green’s Theroem
Theorem Let R be a simple region in the xy plane with a piecewise smooth boundary C oriented
counterclockwise. Let F1 and F2 be functions of two variables having continuous
partial derivatives on R. Then
Z
F1 (x, y)dx + F2 (x, y)dy =
C
ZZ
(
∂F2
∂F1
−
)dA
∂x
∂y
R
~ as F1~i + F2~j and considering the situation in 3 dimensions, this equation
Note Writing F
can be expressed as
Z
C
~ · d~r =
F
ZZ
~ F
~ ) · ~kdA
(∇×
R
40
Pf: It is sufficient to show that
Z
Z
ZZ
∂F2
~
dA
F2 (x, y)dy = F2 j · d~r =
∂x
C
and
Z
C
Z
F1 (x, y)dx =
C
C
R
F1~i · d~r = −
ZZ
∂F1
dA.
∂y
R
Note that the separation is done through writing F~ = F~1 + F~2 where F~1 = F1~i and
F~2 = F2~j.
Z b Z g2 (x)
ZZ
∂F1
∂F1
dA = −
dydx
−
∂y
∂y
a
g1 (x)
R
=
Z
a
b
[F1 (x, g1(x)) − F1 (x, g2(x))]dx
R
The first term is the line integral C1 F1~i·d~r on the curve C1 parameterized by ~r1 (x) =
R
x~i + g1 (x)~j. The second term is −
F1~i · d~r where C2 is the curve parameterized
C2
and oriented by ~r2(t) = x~i + g2 (x)~j, x ∈ [a, b].
Z
Example Evaluate xydx + (x3/2 + y 3/2 )dy where C is the square with vertices (0, 0), (1, 0),
C
(1, 1), (0, 1) and oriented counterclockwise.
[1/2]
R
Example Find −x2 ydx + x3 dy where C is the circle x2 + y 2 = 4, oriented conterclockwise.
C
[16π]
—– Problem Set 11 —–
Oriented Surfaces and Flux Integrals
Let Σ be a surface that has a tangent plane at each of its nonboundary points. At such a point
on the surface two unit normal vectors exist, and they have opposite directions.
If it is possible to select one normal at each nonboundary point in such a way that the chosen
normal varies continuously on the whole of Σ, then the surface Σ is said to be orientable, or
two-sided, and the selection of the normal gives an orientation to Σ and thus makes Σ an oriented
surface. In such case, there are two possible orientations. Some surfaces are not orientable. For
example, the Möbus band; it is one-sided.
Induced Orientation
If Σ is an oriented surface bounded by a curve C, then the orientation of Σ induces an orientation for C, based on the Right-Hand-Rule.
41
Flux Integrals
Let F~ be a vector field and n̂ be the unit normal to the oriented surface Σ, the flux integral
over Σ is
ZZ
F~ · n̂ dS.
Σ
This integral gives the net flux through Σ. The field strength (i.e. |F~ |) can be measured as
the amount of flux per unit area perpendicular to the local direction of F~ .
When Σ is the graph of a function f with continuous partials on a region R in the xy plane
that is composed of vertically or horizontally simple regions, and its orientation is chosen to be
directed upward (i.e. the ~k component of the unit normal is positive), then
ZZ
ZZ
~
F · n̂dS =
[−F1 fx − F2 fy + F3 ]dA
Σ
R
for F~ = F1~i + F2~j + F3~k.
Note Another way to remember this is
ZZ
ZZ
~
~
F · n̂dS =
F~ · NdA
Σ
R
~ = −fx~i + −fy~j + ~k is the upward normal to Σ with the z component equal
where N
to 1.
Example Suppose Σ is the part of the paraboloid z = 1 − x2 − y 2 that lies above the xy plane
and is oriented by the unit normal directed upward. Assume that the velocity of a
fluid is ~v = x~i + y~j + 2z~k. Determine the flow rate (volume/time) through Σ.
[2π]
Flux integrals can be defined for a surface Σ composed of several oriented surfaces Σ1 , Σ2, · · · Σn ,
as
ZZ
ZZ
ZZ
~
~
F · n̂dS =
F · n̂dS + · · · · · · +
F~ · n̂dS
Σ
Σ1
Σn
Example Let Σ be the unit sphere x2 + y 2 + z 2 Z=Z 1, oriented with the unit normal directed
outward, and let F (x, y, z) = z~k. Find
F~ · n̂dS.
Σ
[4π/3]
Stokes’ Theorem
Theorem Let Σ be an oriented surface with normal n̂ (unit vector) and finite surface area.
Assume that Σ is bounded by a closed, piecewise smooth curve C whose orientation
is induced by Σ. Let F~ be a continuous vector field defined on Σ, and assume that the
~ have continuous partial derivatives at each nonboundary
component functions of F
point of Σ. Then
Z
ZZ
F~ · d~r =
C
~ × F~ )· n̂dS
(∇
Σ
42
Example Let C be the intersection of the paraboloid z = x2 + y 2 and the plane z = y, and give
C a counterclockwise direction as viewed from the positive z axis. Evaluate
Z
2xydx + x2 dy + z 2 dz.
C
What
about
Z
(2xy − y)dx + x2 dy + z 2 dz ?
C
[0; π/4]
Example Verify Stokes’ Theorem for F~ = 3y~i − xz~j + yz 2~k where Σ is the surface of the
paraboloid 2z = x2 + y 2 bounded by z = 2, and C is its boundary.
[−20π]
Note If the domain of integration R is symmetry with respect to certain thing and the
integrand is anti-symmetric w.r.t. the same thing, then the integral is 0.
For example, if f (−x, y, z) = −f (x, y, z), the function f is anti-symmetric w.r.t. the yzplane.
The Divergence Theorem
Definition A solid region D is called a
simple solid region if D is the solid region between the graphs of two functions f (x, y)
and g(x, y) on a simple region R in the xy plane and if D has the corresponding
properties with respect to the xz plane and the yz plane.
Theorem Let D be a simple solid region whose
boundary surface Σ is oriented by the normal n̂ directed outward from D, and let F~
be a vector field whose component functions have continuous partial derivatives on
D. Then
ZZ
ZZZ
~
~ F
~ dV.
F · n̂dS =
∇·
Σ
pf:
D
Let F~ = F1~i + F2~j + F3~k, then
F1~i· n̂dS =
∂F1
dV
∂x
Σ
ZZZ
ZZ
F2~j· n̂dS =
∂F2
dV
∂y
Σ
ZZZ
ZZ
F3~k· n̂dS =
ZZZ
∂F3
dV
∂z
ZZ
D
D
Σ
Note The interpretation of
RR
Σ
D
F~ · n̂ dS is “net outward flux through the boundary of D”.
Note The theorem can be applied to a ‘not so simple’ solid region by considering the
division of the region into a number of simple solid regions.
43
Example Suppose
Σ is the sphere x2 +y 2 +z 2 = 1, and let F~ (x, y, z) = x~i+3y~j +2z~k. Evaluate
ZZ
F~ · n̂dS.
Σ
[(8π;
both ways)]
Example Let D be the region bounded by the xy plane and the hemisphere x2 + y 2 + z 2 = 4
with z ≥ 0, and let F~ (x, y, z) = 3x4~i + 4xy 3~j + 4xz 3~k.
ZZ
Evaluate
F~ · n̂dS, where Σ is the
Σ
boundary of D.
[0]
q
Example Let F~ = 2 r̂.
r
Show that for any suface Σ enclosing the origin,
ZZ
Σ
—– Problem Set 12 —–
A proof of Stokes’ theorem
~ ×F
~) · N
~ = (∂y F3 − ∂z F2 )(−fx ) + (∂z F1 − ∂x F3 )(−fy )
(∇
+ (∂x F2 − ∂y F1 )
= (∂x F2 + fx ∂z F2 + fy ∂x F3 ) − (∂y F1 + fy ∂z F1 + fx ∂y F3 )
= ∂x (F2 (x, y, f (x, y)) + fy F3 (x, y, f (x, y))
− ∂y (F1 (x, y, f (x, y)) + fx F3 (x, y, f (x, y))
Now apply Green’s theorem on the xy-plane as following.
ZZ ∂
∂
(F2 + fy F3 ) −
(F1 + fx F3 ) dA
∂x
∂y
R
I
= (F1 + fx F3 )dx + (F2 + fy F3 )dy
∂R
=
I
F1 dx + F2 dy + F3 dz.
∂Σ
Therefore,
ZZ
Σ
~ × F~ ) · n̂dS =
(∇
ZZ
R
~ × F~ ) · NdA
~
(∇
=
I
∂Σ
44
~ · d~r.
F
F~ · n̂dS = 4πq.