Section 18.1 Line Integrals
a. Work Done by Varying Force Over a Curved Path
b. Definition
c. Theorem 18.1.4
d. Piecewise Smoothness
e. Properties of Piecewise Smooth Curves
Section 18.2 Fundamental Theorem for Line Integrals
a. Theorem 18.2.1
b. Corollary and Example
Section 18.3 Work-Energy Formula; Conservation of Mechanical Energy
a. Work-Energy Formula
b. Conservative Field, Potential Energy Functions
c. Conservation of Mechanical Energy
Section 18.4 Another Notation for Line Integrals; Line Integrals with Respect to Arc Length
a. Line Integral with Respect to Arc Length
b. Properties for a Thin Wire
Section 18.5 Green’s Theorem
a. Green’s Theorem, Theorem 18.5.1
b. Type I and Type II Region
c. Area of a Jordan Region
d. Regions Bounded by Two or More Jordan Curves
Chapter 18: Line Integrals and Surface Integrals
Section 18.7 Surface Integrals
a. Mass of a Material Surface
b. Surface Integrals
c. Average Value of H on S
d. Moment of Inertia
e. Flux of a Vector Field
Section 18.8 The Vector Differential Operator ∇
a. Vector Differential Operator
b. Divergence
c. Curl
d. Theoremsfor Divergence and Curl
e. Identities and Properties
f. The Laplacian
Section 18.9 The Divergence Theorem
a. Theorem 18.9.2
b. Divergence as Outward Flux per Unit Volume
c. Solids Bounded by Two or More Closed Surfaces
Section 18.10 Stokes’s Theorem
a. Theorem 18.10.1, Stokes’s Theorem
b. Partial Converse to Stokes Theorem
c. Irrotational Flow
d. Theorem 18.10.3
Line Integrals
The Work Done by a Varying Force over a Curved Path
The work done by a constant force F on an object that moves along a straight line is, by definition, the component of F in the direction of the displacement multiplied by the length of the displacement vector r (Project 13.3):
W = (compd F)||r||.
Line Integrals
Line Integrals
Line Integrals
Such a curve is said to be piecewise smooth.
Line Integrals
If a force F is continually applied to an object that moves over a piecewise- smooth curve C, then the work done by F is the line integral of F over C:
Fundamental Theorem for Line Integrals
Fundamental Theorem for Line Integrals
Example Evaluate the line integral
where C is the square with vertices (0, 0), (1, 0), (1, 1), (0, 1) traversed counterclockwise and
h (x, y) = (3x2 y + xy2 − 1) i + (x3 + x2 y + 4y3) j.
Solution First we try to determine whether h is a gradient. The functions P(x, y) = 3x2 y + xy2 − 1 and Q(x, y) = x3 + x2 y + 4y3
are continuously differentiable everywhere, and
Therefore h is the gradient of a function f . By (18.2.2),
C
( )
⋅∫
h r dr( )
⋅ = ∇f( )
⋅ =0∫
h r dr∫
r dr3 2 2
P Q
x xy
y x
∂ = + = ∂
∂ ∂
Work-Energy Formula; Conservation of Mechanical Energy
This relation is called the work–energy formula.
Work-Energy Formula; Conservation of Mechanical Energy
Conservative Force Fields
In general, if an object moves from one point to another, the work done (and hence the change in kinetic energy) depends on the path of the motion. There is, however, an important exception: if the force field is a gradient field,
then the work done (and hence the change in kinetic energy) depends only on the endpoints of the path, not on the path itself. (This follows directly from the
fundamental theorem for line integrals.) A force field that is a gradient field is called a conservative field.
Since the line integral over a closed path is zero, the work done by a conservative field over a closed path is always zero. An object that passes through a given point with a certain kinetic energy returns to that same point with exactly the same
kinetic energy.
= ∇ f F
Potential Energy Functions
Suppose that F is a conservative force field. It is then a gradient field. Then −F is also a gradient field. The functions U for which are called potential energy functions for F.
∇ = −FU
Work-Energy Formula; Conservation of Mechanical Energy
The Conservation of Mechanical Energy
As an object moves in a conservative force field, its kinetic energy can vary and its potential energy can vary, but the sum of these two quantities remains constant. We call this constant the total mechanical energy.
The total mechanical energy is usually denoted by the letter E. The law of conservation of mechanical energy can then be written
Line Integrals with Respect to Arc Length
Suppose that f is a scalar field continuous on a piecewise-smooth curve C : r (u) = x(u) i + y(u) j + z(u) k, u ∈ [a, b].
If s(u) is the length of the curve from the tip of r (a) to the tip of r (u), then, as you have seen,
The line integral of f over C with respect to arc length s is defined by setting
( ) ( ) ( )
2( )
2( )
2s u′ = r′ u = x u′ + y u′ + z u′
Line Integrals with Respect to Arc Length
Suppose now that C represents a thin wire (a material curve) of varying mass density λ = λ(r). (Here mass density is mass per unit length.) The length of the wire can be written
The mass of the wire is given by
and the center of mass rM can be obtained from the vector equation
The moment of inertia about an axis is given by the formula
Green’s Theorem
Green’s Theorem
Green’s Theorem
Green’s Theorem
Regions Bounded by Two or More Jordan Curves
Parametrized Surfaces; Surface Area
Parametrized Surfaces; Surface Area
Parametrized Surfaces; Surface Area
is called the fundamental vector product of the surface.
Parametrized Surfaces; Surface Area
The Area of a Parametrized Surface
Parametrized Surfaces; Surface Area
More generally, let’s suppose that we have a surface S parametrized by a continuously differentiable function
r = r (u, v), (u, v) ∈ Ω.
We assume that Ω is a basic region in the uv-plane and that r is one-to-one on the interior of Ω. (We don’t want r to cover parts of S more than once.) Also we assume that the fundamental vector product N = r´u × r´v is never zero on the interior of Ω. (We can then use it as a normal.) Under these conditions we call S a smooth surface and define
Parametrized Surfaces; Surface Area
The Area of a Surface z = f (x, y)
Above each point (x, y) of Ω there is one and only one point of S. The surface S is then the graph of a function
z = f (x, y), (x, y) ∈ Ω .
As we show, if f is continuously differentiable, then
Surface Integrals
The Mass of a Material Surface
Imagine a thin distribution of matter spread out over a surface S. We call this a material surface.
Surface Integrals
Surface Integrals
The double integral in (18.7.1) can be calculated not only for a mass density function λ but for any scalar field H continuous over S. We call this integral the surface integral of H over S and write
Note that, if H(x, y, z) is identically 1, then the right-hand side of (18.7.2) gives the area of S. Thus
Surface Integrals
Like the other integrals you have studied, the surface integral satisfies a mean- value condition; namely, if the scalar field H is continuous, then there is a point (x0, y0, z0) on S for which
We call H(x0, y0, z0) the average value of H on S. Thus we can write
We can also take weighted averages: if H and G are continuous on S and G is nonnegative on S, then there is a point (x0, y0, z0) on S for which
As you would expect, we call H(x , y , z ) the G-weighted average of H on S.
(
, ,) (
0, 0, 0)(
area of)
S
H x y z dσ = H x y z S
∫∫
Surface Integrals
Suppose that a material surface S rotates about an axis. The moment of inertia of the surface about that axis is given by the formula
where λ = λ(x, y, z) is the mass density function and R(x, y, z) is the distance
from the axis to the point (x, y, z). (As usual, the moments of inertia about the x, y, z axes are denoted by Ix , Iy , Iz .)
Surface Integrals
The Flux of a Vector Field Suppose that
S : r = r (u, v), (u, v) ∈ Ω
is a smooth surface with a unit normal n = n(x, y, z) that is continuous on all of S.
Such a surface is called an oriented surface. Note that an oriented surface has two sides: the side with normal n and the side with normal −n. If v = v(x, y, z) is a vector field continuous on S, then we can form the surface integral
This surface integral is called the flux of v across S in the direction of n.
The Vector Differential Operator
The vector differential operator is defined formally by setting
The Vector Differential Operator
If v = v1 i + v2 j + v3 k is a differentiable vector field, then, by definition,
The “product,” · v, defined in imitation of the ordinary dot product, is called the divergence of v:
· v = div v.
The Vector Differential Operator
The “product,” × v, defined in imitation of the ordinary cross product, is called the curl of v:
× v = curl v.
The Vector Differential Operator
The Vector Differential Operator
The next two identities are product rules. Here f is a scalar field and v is a vector field.
As usual, set r = x i + y j + z k and r = We know that · r = 3 and × r = 0 at all points of space. Now we can show that if n is an integer, then, for all r r . ≠ 0,
The Vector Differential Operator
The Laplacian
From the operator we can construct other operators, the most important of which is the Laplacian The Laplacian (named after the French mathematician Pierre-Simon Laplace) operates on scalar fields according to the following rule:
∇
2 .
∇ = ∇ ⋅∇
The Divergence Theorem
The Divergence Theorem
Divergence as Outward Flux per Unit Volume
Choose a point P and surround it by a closed ball N∈ of radius ∈. According to the divergence theorem,
Think of v as the velocity of a fluid. As suggested in Section 18.8, negative divergence at P signals an accumulation of fluid near P:
Positive divergence at P signals a flow of liquid away from P:
Points at which the divergence is negative are called sinks; points at which the
divergence is positive are called sources. If the divergence of v is 0 throughout, then the flow has no sinks and no sources and v is called solenoidal.
( )
flux of out ofN
dx dy dz N
∈
∇ ⋅ = ∈
∫∫∫
v v0 at P flux out of N∈ 0 net flow into N∈.
∇ ⋅ <v ⇒ < ⇒
0 at P flux out of N∈ 0 net flow out of N∈.
∇ ⋅v > ⇒ > ⇒
The Divergence Theorem
Solids Bounded by Two or More Closed Surfaces
( ) ( )
1 bdry of 1
T T
dx dy dz dσ
∇ ⋅ = ⋅
∫∫∫
v∫∫
v n( ) ( )
2 bdry of 2
T T
dx dy dz dσ
∇ ⋅ = ⋅
∫∫∫
v∫∫
v n(
∇ ⋅)
=(
⋅) σ
∫∫∫ ∫∫
Stokes’s Theorem
Stokes’s Theorem
Stokes’s Theorem
The Normal Component of × v as Circulation per Unit Area; Irrotational Flow Take a point P within the flow and choose a unit vector n. Let D∈ be the ∈-disk
that is centered at P and is perpendicular to n. Let C∈ be the circular boundary of D∈ directed in the positive sense with respect to n. By Stokes’s theorem,
( ) ( )
C D
n d
σ
∈
∈
∇ × ⋅ = ⋅
∫∫
n ∫
v r drAt each point P the component of × v in any direction n is the circulation of v per unit area in the plane normal to n. If × v = 0 identically, the fluid has no rotational tendency, and the flow is called irrotational.
