Chap
.
10
Vector Integral Calculus
.
Integral Theorems
Sec
.
10
.
1
Line Integrals
Problem Set 10
.
1
.
Page 425
1
.
Line integral in the plane
.
This generalizes a definite integral of calculus. Instead of integrating along
the
x
axis you now integrate over a curve
C
, a portion of a parabola, in the
xy
plane.
Integrals of the form (3) have various applications, for instance, in connection with work done by
a force in a displacement. The right side of (3) shows how such an integral is converted to a definite
integral with
t
as the variable of integration. The conversion is done by using the representation of the
path of integration
C
.
The problem parallels Example 1 in the text. The parabola
C
can be represented by
r
ue0a2
t
ue0a3
u003d
ue0a4
t
,
5
t
2
ue0a5
,
in components,
x
u003d
t
,
y
u003d
5
t
2
,
(I)
where
t
varies from
t
u003d
0 (the initial point of
C
on the
x
axis) to
t
u003d
2 (the terminal point of
C
).
The given function is a vector function
F
u003d
ue0a4
y
3
,
x
3
ue0a5
.
(II)
F
defines a vector field in the
xy
plane. At each point
ue0a2
x
,
y
ue0a3
it gives a certain vector, which you could
draw as a little arrow. In particular, at each point of
C
the vector function
F
gives a vector. You can
obtain these vectors simply by substituting
x
and
y
from (I) into (II). This gives
F
ue0a2
r
ue0a2
t
ue0a3ue0a3
u003d
ue0a4
125
t
6
,
t
3
ue0a5
.
(III)
This is now a vector function of
t
defined on the parabola
C
.
Now comes an important point to observe. You do not integrate
F
itself, but you integrate the dot
product of
F
in (III) and the tangent vector
r
u2032
ue0a2
t
ue0a3
of
C
. This dot product
F
u2219
r
u2032
can be “visualized”
because it is the component of
F
in the direction of the tangent of
C
(times the factor 
r
u2032
ue0a2
t
ue0a3
), as you can
see from (11) in Sec. 9.2 with
F
playing the role of
a
and
r
u2032
playing the role of
b
.
(Note that if
t
is the arc
length
s
of
C
, then
r
u2032
is a unit vector, so that that factor equals 1 and you get exactly that tangential
projection.) Think this over before you go on calculating.
Differentiation with respect to
t
gives the tangent vector
r
u2032
ue0a2
t
ue0a3
u003d
ue0a4
1,
10
t
ue0a5
.
Hence the dot product is
F
ue0a2
r
ue0a2
t
ue0a3ue0a3
u2219
r
u2032
ue0a2
t
ue0a3
u003d
125
t
6
u2219
1
u002b
t
3
u2219
10
t
u003d
125
t
6
u002b
10
t
4
.
Integration with respect to
t
(the parameter in the parametric representation of the path of integration
C
)
from
t
u003d
0 to
t
u003d
2 gives
u222b
0
2
F
ue0a2
r
ue0a2
t
ue0a3ue0a3
u2219
r
u2032
ue0a2
t
ue0a3
dt
u003d
u222b
0
2
ue0a2
125
t
6
u002b
10
t
4
ue0a3
dt
u003d
125
t
7
7
u002b
10
t
5
5
0
2
u003d
125
7
u2219
2
7
u002b
2
u2219
2
5
u003d
2349.71
rounded to 2350 on p. A27 in Appendix 2 of the book.
7
.
Line integral
(
3
)
in space
.
Line integrals in space are handled by the same method as line integrals in
the plane. Quite generally, it is a great advantage of vector methods that methods in space and in the
plane are very similar in most cases. Integrals (3), as just discussed, are suggested by work.
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98
Linear Algebra, Vector Calculus
Part B
In the present problem, the path of integration
C
is a portion of a helix (see Example 4 on p. 391)
C
:
r
ue0a2
t
ue0a3
u003d
ue0a4
cos
t
, sin
t
,
t
ue0a5
with
t
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 Spring '12
 M.lee
 Vector Calculus, Sin, Cos

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