北邮高等数学英文课件Lecture 12-1

(C )
fds
( C1 )
fds
( C2 )
fds
( Cn )
fds .
Properties (Page 299
5
Computation of line integrals with respect to arc length (Page 300)
Proposition If the parametric equation of simple smooth space curve is
the line segment (C) is
x t, y t , z t , 0 t 1,
then
2 (t ) y 2 (t ) z 2 ( t ) 1 1 1 3. x
Therefore,
(C )

f ( x , y , z )ds f ( t , t , t ) 3dt ( t 3t 2 t ) 3dt 0.
(C )

f ( x , y , z )ds lim f ( k , k , k )sk ,
d 0 k 1
n f ( x , y )ds lim f ( k , k )sk . d 0 k 1 ( C )
Element of arc
lim
d 0 k 1
n
n f ( k , k , k )sk , lim f ( k , k )sk d 0 k 1
exists, then f is integrable over the curve C, and the limit is called the line integral of f along C with respect to arc length, which is denoted by n
2 ( k ) y 2 ( k ) z 2 ( k )t k lim f x ( k ), y( k ), z ( k ) x d 0
k 1
2 (t ) y 2 (t ) z 2 ( t )dt . f x ( t ), y( t ), z ( t ) x
I
9 x2 9 , 5 x 5
(C )

y ds
(C ) : { y
5}
o y ds x
11
{( 5 cos t , 3 sin t ), 0 t }

2 2
I 3sin t 5sin t 9cos tdt
0
Work Done by a Force Over a Curve in Space
A(t=a) F W= ? B(t=b)
12
Work Done by a Force Over a Curve in Space
If the coordinate of M k 1 and M k are ( xk 1 , yk 1 , zk 1 ) and ( xk , yk , zk ),
x x ( t ), y y( t ), z z ( t ) ( t )

and the function f is continuous on the curve (C ) , then
(C )

2 (t ) y 2 (t ) z 2 ( t )dt . f ( x , y , z )ds f x ( t ), y( t ), z ( t ) x
2 2
equation of the path
2 dx 2 y 1 y dy 0. I y ds y 1 dy 2 2 dy (C )
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Finish.
10
Evaluating a Line Integral
The lateral area of a cylinder Example Find the lateral area A of the part of the elliptic cylinder x2 y2 1 cut by the planes z = y and z = 0 located in the first and 5 9 z second octants. Hints

Proof
Partition the interval , into subintervals:
t 0 t1 t 2 t n .
The curves(C ) is divided into n segment arcs. Let the segmental arc ( sk ) correspond to
Since (C ) is smooth and the integrand
2 y 2 z 2 x
is continuous, then by the mean
2 ( k ) y 2 ( k ) z 2 ( k )t k , t k 1 k t k . x
d 0 k 1
n
This type of integral is called a line integral with respect to arc or line integral of the first type. If (C ) is a simple closed curve, the line integral is often denoted by
0
1
1
0
Finish.
9
Evaluating a Line Integral
Example Find I
y 2 2 x between the two points (2, 2) and (2,2).
y ds,where (C) is the segment of the parabola (C )
(C )
f ( x , y )ds or f ( x , y , z )ds.
(C )
4
Additivity of Line Integral with respect to Arc
Line integrals with respect to arc have the useful property that if a curve (C ) is made by joining a finite number of curves (C1 ),(C 2 ), ,(C n ) end to end, then the integral of a function over C is the sum of the integrals over the curves that make it up:
Suppose that the vector field F P ( x , y , z )i Q( x , y , z )j R( x , y , z )k represents a force throughout a region in space (it might be the force of gravity or an electromagnetic force of some kind) and that
3
Line integrals with respect to arc length
or
(C )

f ( x , y )ds lim f ( k ,k )sk ,
d 0 k 1
n
(C )

f ( x , y , z )ds lim f ( k ,k , k )sk .
There is a segment of a curve (C )
z
Suppose f is a function of three (or two)
variables defined on (C ), sk subarc. If the limit
O
y C
sk
x
is the arc length of the subarc, sk ,and M k is any point k ,k , k on the
Section 12.1
Line Integrals
1
Line integrals with respect to arc length
Mass of a plane curve body
B
Given density function f(x, y), how does one find the mass of the curved body?
7
Computation of line integrals with respect to arc length
Form the sum
sk x 2 ( k ) y 2 ( k ) z 2 ( k )tk
n k 1

k 1
n
2 2 2 f ( k , k , k )sk f x ( ), y ( ), z ( ) x ( ) y ( ) z ( k )t k . k k k k k
(C ) : r( t ) g( t )i h( t )j k ( t )k, a t b,
is a smooth curve in the region. Then the work done by F over the curve from A to B can be also obtain by the integral.
value theorem for the integral, we have
sk
Let
x ( k ) k ,
y( k ) k , z ( k ) k .
Obviously, the point should lie on the segmental arc ( sk ).
Since the integrand f is continuous on the curve (C ) , the line integral
(C )

f ( x , y , z )ds
exists. Then,
n d 0 k 1 n
(C )

f ( x , y , z )ds lim f ( k , k , k )sk
[ t k 1 , t k ]
and the arc length is sk .
6
Computation of line integrals with respect to arc length
It is easy to see that
sk
tk t k 1
2 (t ) y 2 (t ) z 2 ( t )dt . x
mk f ( k ,k )sk , k 1, 2, , n
Mk
A
Mk-1
f ( k , k )
n
m mk f ( k ,k )sk
k 1 k 1
n
n
m lim f ( k ,k )sk
d 0 k 1
2
Line integrals with respect to arc length


8
Evaluating a Line Integral
Example Integrate f ( x , y, z ) x 3 y 2 zover the line segment (C) joining the origin and the point (1,1,1).
Solution The parametric equation of
Solution
parameter y: Hence,
Choose y as the variable of integration and regard the
y 2 2 x as a parametric equation with the
y2 x , y y( 2 y 2). 2