Transcript No Slide Title

In addition to the multiple integral of a function f:RnR over a region
in Rn, there are many different types of integrals which can be defined,
each of which has its own interpretation and applications. These
include
the integral of a function f:RnR over a (parametrized) path (a path
integral),
the integral of a vector field F:RnRn over a (parametrized) path (a line
integral),
the integral of a function f:RnR over a (parametrized) surface, and
the integral of a vector field F:RnRn over a (parametrized) surface (a
surface integral).
We shall now consider integral of a vector field F:RnRn over a
(parametrized) path (a line integral).
c(t)
c  (t)
Suppose c(t) for a  t  b describes a smooth
(or “piecewise smooth”) path, the tangent
velocity vector c  (t)  0 for any t, and F is a
vector field.
b
The line integral of F over c is defined to be F(c(t)) • c  (t) dt denoted
(One possible interpretation is the
a
work done to move a particle along
F • ds .
the path through the vector field.)
c
In R3 this integral is sometimes written
(F1i+F2j+F3k) • (dxi+dyj+dzk) or
c
(F1dx+F2dy+F3dz) or
c
b
dx
dy
dz
( F1 — + F2 — + F3 — ) dt .
dt
dt
dt
a
Suppose F = (x + y)i + (x + z)j – (y + z)k ,
Example
c(t) = (t , t2 , 4t + 3) for 1  t  5 , and
b(u) = (u2 + 1 , u4 + 2u2 + 1 , 4u2 + 7) for 0  u  2 .
(a) Find
F • ds .
c
c  (t) = (1 , 2t , 4)
F(c(t)) • c  (t) = (t + t2 , 5t + 3 , – t2 – 4t – 3) • (1 , 2t , 4) =
t + t2 + 10t2 + 6t – 4t2 – 16t – 12 = 7t2 – 9t – 12
5
5
(7t2 – 9t – 12) dt
1
=
7t3/3 – 9t2/2 – 12t
t=1
=
800 / 6 = 400 / 3
(b) Find
F • ds .
b
b  (u) = (2u , 4u3 + 4u , 8u)
F(b(u)) • b  (u) =
(u4 + 3u2 + 2 , 5u2 + 8 , – u4 – 6u2 – 8) • (2u , 4u3 + 4u , 8u) =
14u5 + 10u3 – 28u
2
(14u5 + 10u3 – 28u) du =
0
2
7u6/3 + 5u4/2 – 14u2 = 800 / 6 = 400 / 3
u=0
(c) Do c(t) and b(u) describe the same path?
Yes, since c(u2+1) = b(u).
In general, suppose c(t) for a  t  b and b(u) = c(h(u)) for c  u  d
describe the same path. By making a change of variables in the line
integral of F over c, we can prove that the line integral of F over a path c
is the same no matter how the path is parametrized, that is, the line
integral is independent of parametrization of the path. (See Theorem 1
on page 437.)
Suppose F = f , that is, F a gradient vector field, and consider the line
integral of F over a path c(t) for a  t  b. From the chain rule, we know
that d
— f(c(t)) = f (c(t)) • c  (t) . We then observe that
dt
b
b
b
F(c(t)) • c  (t) dt =
F • ds =
c
a
f(c(b)) – f(c(a)) .
f(c(t)) • c  (t) dt =
a
(See Theorem 3 on page 440.)
f(c(t)) =
t=a
We see then that the line integral of a gradient field F = f over a path
c(t) for a  t  b depends only on the starting point c(a) and the ending
point c(b) of the path.
In other words, the line integral of a gradient field F = f over a path
from (x1 , y1 , z1) to (x2 , y2 , z2) will be equal to
f(x2 , y2 , z2) – f(x1 , y1 , z1)
no matter what path is chosen.
(x2 , y2 , z2)
(x1 , y1 , z1)
Example Let F = (x+y)i + (x+z)j – (y+z)k ,
V = xi + yj + zk ,
c(t) = (t , t2 , t3) for 1  t  4 ,
b(t) = (3t + 1 , 15t + 1 , 63t + 1) for 0  t  1 .
(a) Is F a gradient vector field? No, since curl F = – 2i  0
(b) Is V a gradient vector field? Yes, since curl V = 0.
x2 + y2 + z2
Also, V = f where f(x,y,z) = —————
2
(c) Do c(t) and b(t) begin at the same point and end at the same point?
Both paths begin at the point (1 , 1 , 1) and end at the point (4 , 16 , 64).
(d) Do c(t) and b(u) describe the same path?
No, they are two different paths from (1 , 1 , 1) to (4 , 16 , 64).
(e) Find
F • ds and
c
c  (t) = (1, 2t , 3t2)
F • ds .
b
b  (t) = (3 , 15 , 63)
F(c(t)) • c  (t) = (t + t2 , t + t3 , – t2 – t3) • (1 , 2t , 3t2) =
t + t2 + 2t2 + 2t4 – 3t4 – 3t5 = t + 3t2 – t4 – 3t5
F(b(t)) • b  (t) = (18t + 2 , 66t + 2 , – 78t – 2) • (3 , 15 , 63) =
54t + 6 + 990t + 30 – 4914t – 126 = – 3870t – 90
4
10913
F • ds = (t + 3t2 – t4 – 3t5) dt = – ———
5
c
1
1
F • ds =
(– 3870t – 90) dt = – 2025
b
0
(f) Find
V • ds and
c
V • ds .
b
V • ds =
c
V • ds =
b
(f) Find
V • ds and
c
V • ds =
c
V • ds .
b
V • ds = f(4 , 16 , 64) – f(1 , 1 , 1) =
b
(4)2 + (16)2 + (64)2
(1)2 + (1)2 + (1)2
———————— – ——————— =
2
2
4365
——
2
Example
Let c(t) be the counterclockwise path in the xy plane along the circle of
radius 3 centered at the origin starting and ending at (3,0).
Let b(t) be the path in R2 along the rectangle from (3,0) to (0,3) to (–3,0)
to (0,–3) to (3,0) .
(a) How can we parametrize the path c(t)?
c(t) = (3 cos t , 3 sin t) for 0  t  2
(b) How can we parametrize the path b(t)?
We can first define each of the line segments of the path separately:
b1(t) = ( 3 – 3t , 3t ) for 0  t  1 ,
b2(t) = (– 3t , 3 – 3t ) for 0  t  1 ,
b3(t) = ( – 3 + 3t , – 3t ) for 0  t  1 ,
b4(t) = ( 3t , – 3 + 3t ) for 0  t  1 .
We then say that b = b1  b2  b3  b4 .
(c) Suppose F(x,y) is a gradient vector field.
F • ds and
Find
F • ds .
b
c
Since F = f for some f(x,y), then we must have
F • ds =
c
F • ds =
b
f(3,0) – f(3,0) = 0
Look at the first homework problem in Section 7.2 (#2a, page 447):
x dy – y dx =
c
Note that the line integral is written in the following form:
(F1dx+F2dy)
c
This implies that the vector field F = ( – y , x , 0 ) is integrated over the given path