section 17.2 line integrals. let c be a smooth plane curve given by x = x(t), y = y(t), a ≤ t ≤...
TRANSCRIPT
Section 17.2
Line Integrals
Let C be a smooth plane curve given by
x = x(t), y = y(t), a ≤ t ≤ b.
We divide the parameter interval [a, b] into n subintervals [ti − 1, ti]of equal width, and we let xi = x(ti) and yi = y(ti). Then the corresponding points Pi(xi, yi) divide C into n subarcs with lengths Δsi. Let be a point on the subarc Ci. If f is defined on a smooth curve C, then the line integral of f along C
if the limit exists.
LINE INTEGRALS
),( ***ii yxP
i
C
n
iiii
nsyxfdsyxf
1
** ),(lim),(
Recall from Section 11.2 that the arc length of C is
If f is a continuous function, the limit on the previous slide always exists. The following formula can be used to evaluate the line integral.
EVALUATING LINE INTEGRALS
b
a
dtdt
dy
dt
dxL
22
C
b
a
dtdt
dy
dt
dxtytxfdsyxf
22
)(),(),(
INTERPRETATION OF THELINE INTEGRAL
If z = f (x, y) ≥ 0 and C is a curve in the plane, then line integral
gives the area of the curved “curtain” below the surface and above C. See Figure 2 on page 1099.
Cdsyxf ),(
EXAMPLE
Evaluate the following line integral where
C is the line segment joining (1, 2) to (4, 7)
C
x dsey
PIECEWISE-SMOOTH CURVES AND LINE INTEGRALS
If C is a piecewise-smooth curve then C can be written as a finite union of smooth curves; that is,
C = C1 U C2 . . . U Cn
The line integral of f along C is defined as the sum of the line integrals of f along each of the smooth pieces of C; that is,
nC
CCC
dsyxf
dsyxfdsyxfdsyxf
),(
),(),(),(21
Evaluate where C is the piecewise-
smooth curve formed by the boundary region bounded by y = x and y = x2.
Cdsx
EXAMPLE
AN INTERPRETATION OF THE LINE INTEGRAL
Suppose that ρ(x, y) represents the density of a thin wire that is shaped like the plane curve C. The mass of the wire is given by
The center of mass of the wire is given by
CC
C
dsyxym
ydsyxxm
x
dsyxm
),(1
),(1
),(
EXAMPLE
A thin wire is bent in the shape of the semicircle
x = cos t, y = sin t, 0 ≤ t ≤ π
If the density of the wire at a point is proportional to its distance from the x-axis, find the mass and center of mass of the wire.
LINE INTEGRALS WITH RESPECT TO x AND y
Two other line integrals can be obtained by replacing Δsi by either Δxi = xi − xi − 1 or Δyi = yi − yi − 1. They are called the line integrals of f along C with respect to x and y.
n
iiii
nC
n
iiii
nC
yyxfdyyxf
xyxfdxyxf
1
**
1
**
),(lim),(
),(lim),(
DISTINGUISHING FROM THE ORIGINAL LINE INTEGRAL
To distinguish the line integral with respect to x and y from the original line integral ∫C f (x, y) ds, we call ∫C f (x, y) ds the line integral with respect to arc length.
EVALUATING LINE INTEGRALS WITH RESPECT TO x AND y
b
aC
b
aC
dttytytxfdyyxf
dttxtytxfdxyxf
)()(),(),(
)()(),(),(
A SPECIAL NOTATIONThe line integrals with respect to x and y frequently occur together. We write this as follows.
CCC
dyyxQdxyxPdyyxQdxyxP ),(),(),(),(
ORIENTATION AND LINE INTEGRALS
Recall that a given parametrization x = x(t), y = y(t), a ≤ t ≤ b, determines an orientation of a curve C. If we let −C denote the curve consisting of the same points as C but with opposite orientation, then we have:
CC
CCCC
dsyxfdsyxf
dyyxfdyyxfdxyxfdxyxf
),(),(
),(),(),(),(
NOTE: The line integral with respect to arc length DOES NOT change sign.
LINE INTEGRALS IN SPACE
Suppose that C is a smooth space curve given by
x = x(t), y = y(t), z = z(t), a ≤ t ≤ b.
Suppose that f is function of three variables that is continuous on some region containing C, then the line integral of f along C is defined in a similar manner as for plane curves:
n
iiiii
nCszyxfdszyxf
1
*** ),,(lim),,(
EVALUATING LINE INTEGRALS IN SPACE
b
aC
dtdt
dz
dt
dy
dt
dxzyxfdszyxf
222
),,(),,(
VECTOR NOTATION FOR LINE INTEGRALS
If r(t) is the vector form of either a plane curve or a space curve, then the formula for evaluating a line integral with respect to arc length can be written compactly as
dtttfb
a|)(|)( rr
Suppose that F = P i + Q j + R k is a continuous force field in three dimensions, such as a gravitational field. To compute the work done by this force in moving a particle along the smooth curve C, we divide C into subarcs Pi−1Pi with lengths Δsi by dividing the parameter interval [a, b] into subintervals of equal width. Choose a point on the ith subarc corresponding to the parameter . If Δsi is small, then as the particle moves from Pi−1 to Pi along the curve, it proceeds approximately in the direction of , the unit tangent vector at . The work done by the force F in moving to particle from Pi−1 to Pi is approximately
The total work done in moving the particle along C is approximately
WORK AND LINE INTEGRALS
),,( ****iiii zyxP
*it
)( *itT
*iP
iiiiiiiiii stzyxtszyx )(),,()(),,( ******** TFTF
n
iiiiii stzyx
1
**** )(),,( TF
WORK (CONCLUDED)
Based on the derivation on the previous slide, we define the work W done by the force field F in moving a particle along C as the limit of the Riemann sums, namely,
CC
dsdszyxzyxW TFTF ),,(),,(
EVALUATING A LINE INTEGRAL FOR WORK
If the curve C is given by the vector equation
r(t) = x(t)i + y(t)j + z(t)k
then T(t) = r′(t)/|r′(t)|. So, the line integral for work can be rewritten as
dtttdttt
ttW
b
a
b
a
)()(|)(||)(|
)()( rrFr
r
rrF
EXAMPLE
Find the work done by the force field
on a particle as it moves along the helix given by
r(t) = cos ti + sin tj + tk
from the point (1, 0, 0) to (−1, 0, 3π)
kjiF4
1
2
1
2
1),,( yxzyx
LINE INTEGRAL OF A VECTOR FIELD
Definition: Let F be a continuous vector field defined on a smooth curve C given by a vector function r(t), a ≤ t ≤ b. Then the line integral of F along C is
where F = Pi + Qj + Rk.
C
b
aCdsdtttd TFrrFrF )()(
NOTE 1
Even though ∫C F ∙ dr = ∫C F ∙ T ds and integrals with respect to arc length are unchanged when orientation is reversed, it is still true that
because the unit tangent vector T is replaced by its negative when C is replaced by −C.
CC
dd rFrF
NOTE 2
CC
dzRdyQdxPdrF
where F = Pi + Qj + Rk