In addition to the multiple integral of a function f:RnR over a region in Rn, there are many differe

1
In addition to the multiple integral of a
function fRn?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 fRn?R over a
(parametrized) path (a path integral),
the integral of a vector field FRn?Rn over a
(parametrized) path (a line integral),
the integral of a function fRn?R over a
(parametrized) surface, and
the integral of a vector field FRn?Rn over a
(parametrized) surface (a surface integral).
We shall now consider integral of a vector field
FRn?Rn over a (parametrized) path (a line
integral).
2
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.
c ? (t)
c(t)
b
The line integral of F over c is defined to be
F(c(t)) c ? (t) dt denoted
(One possible interpretation the work done to
move a particle along the path through the vector
field.)
a
F ds .
c
In R3 this integral is sometimes written
(F1dxF2dyF3dz) or
(F1iF2jF3k) (dxidyjdzk) or
c
c
b
dx dy dz ( F1 F2
F3 ) dt . dt dt dt
a
3
Example Suppose F (x y)i (x z)j (y
z)k , 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)
(t t2 , 5t 3 , t2 4t 3) (1 , 2t , 4)
F(c(t)) c ? (t)
t t2 10t2 6t 4t2 16t 12 7t2 9t
12
5
5
(7t2 9t 12) dt
7t3/3 9t2/2 12t
800 / 6 400 / 3
1
t 1
4
(b) Find
F ds .
b
(2u , 4u3 4u , 8u)
b ? (u)
F(b(u)) b ? (u)
(u4 3u2 2 , 5u2 8 , u4 6u2 8) (2u ,
4u3 4u , 8u)
14u5 10u3 28u
2
2
(14u5 10u3 28u) du
7u6/3 5u4/2 14u2
800 / 6 400 / 3
u 0
0
(c) Do c(t) and b(u) describe the same path?
Yes, since c(u21) b(u).
5
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
F(c(t)) c ? (t) dt
f(c(t))
c
a
a
t a
(See Theorem 3 on page 440.)
f(c(b)) f(c(a)) .
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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 no matter what
path is chosen.
f(x2 , y2 , z2) f(x1 , y1 , z1)
(x2 , y2 , z2)
(x1 , y1 , z1)
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Example Let F (xy)i (xz)j (yz)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. Also, V ?f where
f(x,y,z)
x2 y2 z2 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).
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c ? (t)
(1, 2t , 3t2)
(e) Find
F ds and F ds .
b ? (t)
(3 , 15 , 63)
c
b
(t t2 , t t3 , t2 t3) (1 , 2t , 3t2)
F(c(t)) c ? (t)
t t2 2t2 2t4 3t4 3t5 t 3t2 t4
3t5
(18t 2 , 66t 2 , 78t 2) (3 , 15 , 63)
F(b(t)) b ? (t)
54t 6 990t 30 4914t 126 3870t 90
4
10913 5
F ds
(t 3t2 t4 3t5) dt
c
1
1
F ds
( 3870t 90) dt
2025
b
0
(f) Find
V ds and V ds .
V ds V ds
c
b
c
b
9
(f) Find
V ds and V ds .
c
b
V ds V ds
f(4 , 16 , 64) f(1 , 1 , 1)
c
b
(4)2 (16)2 (64)2 (1)2 (1)2
(1)2 2 2
4365 2
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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 .
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(c) Suppose F(x,y) is a gradient vector
field. Find
F ds and F ds .
c
b
Since F ?f for some f(x,y), then we must have
F ds F ds
f(3,0) f(3,0) 0
c
b
12
Look at the first homework problem in Section 7.2
(2a, page 447)
x dy y dx
c
(F1dxF2dy)
Note that the line integral is written in the
following form
c
This implies that the vector field F (
) is integrated over the given path
y , x , 0