3
3
0
4.1 Line Integrals
141
So in both cases, if the vector field f( x, y) = ( x 2 + y 2)i+2 xyj represents the force moving an object from (0, 0) to (1, 2) along the given curve C, then the work done is 13 . This may lead
3
you to think that work (and more generally, the line integral of a vector field) is independent
of the path taken. However, as we will see in the next section, this is not always the case.
Although we defined line integrals over a single smooth curve, if C is a piecewise smooth
curve, that is
C = C 1 ∪ C 2 ∪ ... ∪ Cn
is the union of smooth curves C , . . . , C , then we can define
1
n
f · dr =
f · dr
f
f
1 +
· dr2 + ... +
· dr n
C
C 1
C 2
Cn
where each r is the position vector of the curve C .
i
i
Example 4.3. Evaluate
( x 2
C
+ y 2) dx + 2 xy d y, where C is the polygonal path from (0,0) to
(0, 2) to (1, 2).
y
Solution: Write C = C
, where C is the curve given by x
1 ∪ C 2
1
= 0, y = t,
2
(1, 2)
0 ≤ t ≤ 2 and C is the curve given by x
2
= t, y = 2, 0 ≤ t ≤ 1 (see Figure
C 2
4.1.5). Then
C 1
( x 2 + y 2) dx + 2 xy d y =
( x 2 + y 2) dx + 2 xy d y
x
C
C 1
0
1
+
( x 2 + y 2) dx + 2 xy d y
C
Figure 4.1.5
2
2
1
=
(02 + t 2)(0) + 2(0) t(1) dt +
( t 2 + 4)(1) + 2 t(2)(0) dt
0
0
2
1
=
0 dt +
( t 2 + 4) dt
0
0
t 3
1
1
13
=
+ 4 t
=
+ 4 =
3
3
3
0
Line integral notation varies quite a bit. For example, in physics it is common to see the
notation
b f
a
· dl, where it is understood that the limits of integration a and b are for the
underlying parameter t of the curve, and the letter l signifies length. Also, the formulation
f
C
· T ds from Theorem 4.1 is often preferred in physics since it emphasizes the idea of
integrating the tangential component f· T of f in the direction of T (i.e. in the direction of C),
which is a useful physical interpretation of line integrals.
142
CHAPTER 4. LINE AND SURFACE INTEGRALS
Exercises
A
For Exercises 1-4, calculate
f ( x, y) ds for the given function f ( x, y) and curve C.
C
1. f ( x, y) = xy;
C : x = cos t, y = sin t, 0 ≤ t ≤ π/2
x
2. f ( x, y) =
;
C : x = t, y = 0, 0 ≤ t ≤ 1
x 2 + 1
3. f ( x, y) = 2 x + y;
C: polygonal path from (0, 0) to (3, 0) to (3, 2)
4. f ( x, y) = x + y 2;
C: path from (2, 0) counterclockwise along the circle x 2 + y 2 = 4 to the
point (−2,0) and then back to (2,0) along the x-axis
5. Use a line integral to find the lateral surface area of the part of the cylinder
x 2 + y 2 = 4 below the plane x + 2 y + z = 6 and above the xy-plane.
For Exercises 6-11, calculate
f
C · dr for the given vector field f( x, y) and curve C.
6. f( x, y) = i − j;
C : x = 3 t, y = 2 t, 0 ≤ t ≤ 1
7. f( x, y) = yi − x j;
C : x = cos t, y = sin t, 0 ≤ t ≤ 2 π
8. f( x, y) = x i + yj;
C : x = cos t, y = sin t, 0 ≤ t ≤ 2 π
9. f( x, y) = ( x 2 − y)i + ( x − y 2)j;
C : x = cos t, y = sin t, 0 ≤ t ≤ 2 π
10. f( x, y) = xy 2 i + xy 3 j;
C : the polygonal path from (0, 0) to (1, 0) to (0, 1) to (0, 0)
11. f( x, y) = ( x 2 + y 2)i;
C : x = 2 + cos t, y = sin t, 0 ≤ t ≤ 2 π
B
12. Verify that the value of the line integral in Example 4.1 is unchanged when using the
parametrization of the circle C given in formulas (4.8).
13. Show that if f ⊥ r′( t) at each point r( t) along a smooth curve C, then
f
C · dr = 0.
14. Show that if f points in the same direction as r ′( t) at each point r( t) along a smooth
curve C, then
f
f ds.
C · dr = C
C
15. Prove that
f ( x, y) ds
f ( x, y) ds. ( Hint: Use formulas (4.9) . )
C
= − C
16. Let C be a smooth curve with arc length L, and suppose that f( x, y) = P( x, y)i + Q( x, y)j is a vector field such that f( x, y) ≤ M for all ( x, y) on C. Show that
f
g( x) dx
C · dr ≤ M L. ( Hint: Recall that
b
a
≤ b
a | g( x)| d x for Riemann integrals. )
17. Prove that the Riemann integral
b f ( x) dx is a special case of a line integral.
a
4.2 Properties of Line Integrals
143
4.2 Properties of Line Integrals
We know from the previous section that for line integrals of real-valued functions (scalar
fields), reversing the direction in which the integral is taken along a curve does not change
the value of the line integral:
f ( x, y) ds =
f ( x, y) ds
(4.17)
C
− C
For line integrals of vector fields, however, the value does change. To see this, let f( x, y) =
P( x, y) i + Q( x, y)j be a vector field, with P and Q continuously differentiable functions. Let C be a smooth curve parametrized by x = x( t), y = y( t), a ≤ t ≤ b, with position vector r( t) =
x( t) i + y( t)j (we will usually abbreviate this by saying that C : r( t) = x( t)i + y( t)j is a smooth curve). We know that the curve − C traversed in the opposite direction is parametrized by
x = x( a + b − t), y = y( a + b − t), a ≤ t ≤ b. Then b
d
P( x, y) dx =
P( x( a + b − t), y( a + b − t))
( x( a + b − t)) dt
− C
a
dt
b
=
P( x( a + b − t), y( a + b − t))(− x ′( a + b − t)) dt (by the Chain Rule) a
a
=
P( x( u), y( u)) (− x ′( u))(− du) (by letting u = a + b − t)
b
a
=
P( x( u), y( u)) x ′( u) du
b
b
a
b
= −
P( x( u), y( u)) x ′( u) du , since
= −
, so
a
b
a
P( x, y) dx = −
P( x, y) dx
− C
C
since we are just using a different letter ( u) for the line integral along C. A similar argument
shows that
Q( x, y) d y = −
Q( x, y) d y ,
− C
C
and hence
f · dr =
P( x, y) dx +
Q( x, y) d y
− C
− C
− C
= −
P( x, y) dx + −
Q( x, y) d y
C
C
= −
P( x, y) dx +
Q( x, y) d y
C
C
f · dr = −
f · dr .
(4.18)
− C
C
144
CHAPTER 4. LINE AND SURFACE INTEGRALS
The above formula can be interpreted in terms of the work done by a force f( x, y) (treated
as a vector) moving an object along a curve C: the total work performed moving the object
along C from its initial point to its terminal point, and then back to the initial point moving
backwards along the same path, is zero. This is because when force is considered as a vector,
direction is accounted for.
The preceding discussion shows the importance of always taking the direction of the curve
into account when using line integrals of vector fields. For this reason, the curves in line
integrals are sometimes referred to as directed curves or oriented curves.
Recall that our definition of a line integral required that we have a parametrization x =
x( t), y = y( t), a ≤ t ≤ b for the curve C. But as we know, any curve has infinitely many parametrizations. So could we get a different value for a line integral using some other
parametrization of C, say, x = ˜ x( u), y = ˜ y( u), c ≤ u ≤ d ? If so, this would mean that our definition is not well-defined. Luckily, it turns out that the value of a line integral of a
vector field is unchanged as long as the direction of the curve C is preserved by whatever
parametrization is chosen:
Theorem 4.2. Let f( x, y) = P( x, y)i + Q( x, y)j be a vector field, and let C be a smooth curve parametrized by x = x( t), y = y( t), a ≤ t ≤ b. Suppose that t = α( u) for c ≤ u ≤ d, such that a = α( c), b = α( d), and α′( u) > 0 on the open interval ( c, d) (i.e. α( u) is strictly increasing on
[ c, d]). Then
f
C
· dr has the same value for the parametrizations x = x( t), y = y( t), a ≤ t ≤ b and x = ˜ x( u) = x( α( u)), y = ˜ y( u) = y( α( u)), c ≤ u ≤ d.
Proof: Since α( u) is strictly increasing and maps [ c, d] onto [ a, b], then we know that t =
α( u) has an inverse function u = α−1( t) defined on [ a, b] such that c = α−1( a), d = α−1( b), and du
. Also, dt
dt =
1
α ′( u)
= α′( u) du, and by the Chain Rule
d ˜ x
d
dx dt
˜ x ′( u)
˜ x ′( u) =
=
( x( α( u))) =
= x ′( t) α′( u)
⇒
x ′( t) =
du
du
dt du
α ′( u)
so making the susbstitution t = α( u) gives
b
α−1( b)
˜ x ′( u)
P( x( t), y( t)) x ′( t) dt =
P( x( α( u)), y( α( u)))
( α ′( u) du)
a