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Line Integrals

In this section we are now going to introduce a new kind of integral
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You should have seen some of this in your Calculus II course
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Here are some of the more basic curves that we’ll need to know how to do as well as limits on the
parameter if they are required
...
In fact, we will
be using the two-dimensional version of this in this section
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As we’ll eventually see the direction that the curve is
traced out can, on occasion, change the answer
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Now let’s move on to line integrals
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In this case we were thinking of x as taking all the values in this
interval starting at a and ending at b
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Note that this is different from the double
integrals that we were working with in the previous chapter where the points came out of some
two-dimensional region
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We will assume that the curve is smooth
(defined shortly) and is given by the parametric equations,

x = h (t )

y = g (t )

a≤t ≤b

We will often want to write the parameterization of the curve as a vector function
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The line integral of f ( x, y ) along C is denoted by,

∫ f ( x, y ) ds

C

We use a ds here to acknowledge the fact that we are moving along the curve, C, instead of the xaxis (denoted by dx) or the y-axis (denoted by dy)
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We’ve seen the notation ds before
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The ds is the same for both the arc
length integral and the notation for the line integral
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The
line integral is then,
b

∫

C

2
2
⌠
⎛ dx ⎞ ⎛ dy ⎞
f ( x, y ) ds = ⎮ f ( h ( t ) , g ( t ) ) ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
⌡a

Don’t forget to plug the parametric equations into the function as well
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Using this notation the line integral becomes,

∫ f ( x, y ) ds = ∫ a f ( h ( t ) , g ( t ) )

r′ ( t ) dt

b

C

Note that as long as the parameterization of the curve C is traced out exactly once at t increases
from a to b the value of the line integral will be independent of the parameterization of the curve
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Example 1 Evaluate

2
2
4
xy
ds
where
C
is
the
right
half
of
the
circle,
x
+
y
= 16 rotated in the
∫

C

counter clockwise direction
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This is given by,

x = 4 cos t

y = 4sin t

We now need a range of t’s that will give the right half of the circle
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−

π

2

≤t ≤

π

2

Now, we need the derivatives of the parametric equations and let’s compute ds
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A piecewise smooth
curve is any curve that can be written as the union of a finite number of smooth curves, C1 ,…, Cn

where the end point of Ci is the starting point of Ci +1
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Evaluation of line integrals over piecewise smooth curves is a relatively simple thing to do
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The line integral
for some function over the above piecewise curve would be,

∫ f ( x, y ) ds = ∫ f ( x, y ) ds + ∫ f ( x, y ) ds + ∫ f ( x, y ) ds + ∫ f ( x, y ) ds

C

C1

C2

C3

Let’s see an example of this
...

C

Solution
So, first we need to parameterize each of the curves
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∫ 4x

3

0

ds = ∫ 4t

(1) + ( 0 )
2

3

−2

C1

1

3
⌠
⎮
4
x
ds
=
4
t
∫C
⌡0
3

(1)

2

0

dt = ∫ 4t dt = t

2

3

4 0

−2

+ ( 3t

)

2 2

−2

= −16

dt

2

1

= ∫ 4t 3 1 + 9t 4 dt
0

1⎛2⎞
= ⎜ ⎟ (1 + 9t
9⎝3⎠

∫ 4x

C3

3

ds = ∫ 4 (1)
2

3

0

3 1
4 2

)

0

2 ⎛ 32 ⎞
= ⎜ 10 − 1⎟ = 2
...
268 + 8
= −5
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The direction of motion
along a curve may change the value of the line integral as we will see in the next section
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Let’s first see what happens to the line integral if we change the path between these two

points
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C

Solution
From the parameterization formulas at the start of this section we know that the line segment start
at ( −2, −1) and ending at (1, 2 ) is given by,

r ( t ) = (1 − t ) −2, −1 + t 1, 2
= −2 + 3t , −1 + 3t
for 0 ≤ t ≤ 1
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When doing these integrals don’t forget simple Calc I substitutions to avoid having to do things
like cubing out that term
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So, the previous two examples seem to suggest that if we change the path between two points
then the value of the line integral (with respect to arc length) will change
...
In a later section we will investigate
this idea in more detail
Next, let’s see what happens if we change the direction of a path
...

C

Solution
This one isn’t much different, work wise, from the previous example
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r ( t ) = (1 − t ) 1, 2 + t −2, −1
= 1 − 3t , 2 − 3t

for 0 ≤ t ≤ 1
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Here is the line integral
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213
So, it looks like when we switch the direction of the curve the line integral (with respect to arc
length) will not change
...
However, there
are other kinds of line integrals in which this won’t be the case
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Before working another example let’s formalize this idea up somewhat
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Let’s also suppose that the initial point
on the curve is A and the final point on the curve is B
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Finally, let −C be the curve with the same points as C, however in
this case the curve has B as the initial point and A as the final point, again t is increasing as we
traverse this curve
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We then have the following fact about line integrals with respect to arc length
...
This is a useful fact to remember as some line integrals will be
easier in one direction than the other
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C

(a) C1 : y = x , −1 ≤ x ≤ 1 [Solution]
2

(b) C2 : The line segment from ( −1,1) to (1,1)
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[Solution]

Solution
Before working any of these line integrals let’s notice that all of these curves are paths that
connect the points ( −1,1) and (1,1)
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Here is a sketch of the three curves and note that the curves illustrating C2 and C3 have been
separated a little to show that they are separate curves in some way even thought they are the
same line
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C1 : x = t, y = t 2 , −1 ≤ t ≤ 1
Here is the line integral
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There are two parameterizations that we could use here for this curve
...
That parameterization is,

C2 : r ( t ) = (1 − t ) −1,1 + t 1,1
= 2t − 1,1

for 0 ≤ t ≤ 1
...
However, in this case there is a
second (probably) easier parameterization
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Using this the parameterization is,

C2 : x = t, y = 1, −1 ≤ t ≤ 1
This will be a much easier parameterization to use so we will use this
...

1

1 2
∫C x ds = ∫ −1 t 1 + 0 dt = 2 t −1 = 0
2
1

Note that this time, unlike the line integral we worked with in Examples 2, 3, and 4 we got the

same value for the integral despite the fact that the path is different
...
We should also not expect this integral to be the same for all paths between these two
points
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It is completely possible that there is another path between these two points that will give
a different value for the line integral
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Now, according to our fact above we really don’t need to do anything here since we know that
C3 = −C2
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e
...
However, let’s verify that, plus there is a point we need to make here about the
parameterization
...


C3 : r ( t ) = (1 − t ) 1,1 + t −1,1
= 1 − 2t ,1

for 0 ≤ t ≤ 1
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Here is the line integral for this curve
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[Return to Problems]

To this point in this section we’ve only looked at line integrals over a two-dimensional curve
...
We can do line integrals over threedimensional curves as well
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r (t ) = x (t ) , y (t ) , z (t )

Notice that we changed up the notation for the parameterization a little
...

Also notice that, as with two-dimensional curves, we have,
2

2

2

⎛ dx ⎞ ⎛ dy ⎞ ⎛ dz ⎞
⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = r′ (t )
⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠
and the line integral can again be written as,

∫ f ( x, y, z ) ds = ∫ a f ( x ( t ) , y ( t ) , z ( t ) ) r′ ( t ) dt
b

C

So, outside of the addition of a third parametric equation line integrals in three-dimensional space
work the same as those in two-dimensional space
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Example 6 Evaluate
0 ≤ t ≤ 4π
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Here
is a quick sketch of the helix
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∫ xyz ds = ∫

4π
0

3t cos ( t ) sin ( t ) sin 2 t + cos 2 t + 9 dt

C

4π

⎛1
⎞
⌠
= ⎮ 3t ⎜ sin ( 2t ) ⎟ 1 + 9 dt
⎝2
⎠
⌡0
3 10 4π
=
t sin ( 2t ) dt
∫
2 0
4π

3 10 ⎛ 1
t
⎞
=
sin
2
t
−
cos
2
t
( )
( )⎟
⎜
2 ⎝4
2
⎠0
= −3 10 π

You were able to do that integral right? It required integration by parts
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In the previous section we looked at line integrals with respect to arc length
...

As with the last section we will start with a two-dimensional curve C with parameterization,

x = x (t )

y = y (t )

a≤t ≤b

The line integral of f with respect to x is,

∫ f ( x, y ) dx = ∫ a f ( x ( t ) , y ( t ) ) x′ ( t ) dt
b

C

The line integral of f with respect to y is,

∫ f ( x, y ) dy = ∫ a f ( x ( t ) , y ( t ) ) y′ ( t ) dt
b

C

Note that the only notational difference between these two and the line integral with respect to arc
length (from the previous section) is the differential
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So when evaluating line integrals be careful to first note
which differential you’ve got so you don’t work the wrong kind of line integral
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∫ Pdx + Q dy = ∫ P ( x, y ) dx + ∫ Q ( x, y ) dy

C

C

C

Let’s take a quick look at an example of this kind of line integral
...

C

Solution
Here is the parameterization of the curve
...
Let’s see what happens with line
integrals with respect to x and/or y
...

2

C

Solution
So, we simply changed the direction of the curve
...


r ( t ) = (1 − t ) 1, 4 + t 0, 2 = 1 − t , 4 − 2t

0 ≤ t ≤1

The line integral in this case is,
2
2
sin
π
y
dy
+
yx
dx
=
sin
π
y
dy
+
yx
(
)
(
)
∫
∫
∫ dx

C

C

C

= ∫ sin (π ( 4 − 2t ) ) ( −2 ) dt + ∫ ( 4 − 2t )(1 − t ) ( −1) dt
1

1

0

0

2

1

1

⎛ 1 4 8 3
⎞
2
= cos ( 4π − 2π t ) − ⎜ − t + t − 5t + 4t ⎟
π
3
⎝ 2
⎠0
0
7
=−
6
1

So, switching the direction of the curve got us a different value or at least the opposite sign of the
value from the first example
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Fact
If C is any curve then,

∫ f ( x, y ) dx = −∫ f ( x, y ) dx

−C

∫ f ( x, y ) dy = − ∫ f ( x, y ) dy

and

−C

C

C

With the combined form of these two integrals we get,

∫ Pdx + Q dy = −∫ Pdx + Q dy

−C

C

We can also do these integrals over three-dimensional curves as well
...


∫ f ( x, y, z ) dx = ∫ a f ( x ( t ) , y ( t ) , z ( t ) ) x′ ( t ) dt
b

C

∫ f ( x, y, z ) dy = ∫ a f ( x ( t ) , y ( t ) , z ( t ) ) y′ ( t ) dt
b

C

∫ f ( x, y, z ) dz = ∫ a f ( x ( t ) , y ( t ) , z ( t ) ) z′ ( t ) dt
b

C

where the curve C is parameterized by

x = x (t )

y = y (t )

z = z (t )

a≤t ≤b

As with the two-dimensional version these three will often occur together so the shorthand we’ll
be using here is,

∫ Pdx + Q dy + R dz = ∫ P ( x, y, z ) dx + ∫ Q ( x, y, z ) dy + ∫ R ( x, y, z ) dz

C

C

C

C

Let’s work an example
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2
y
dx
+
x
dy
+
z
dz
where
C
is
given
by
x
=
cos
t
,
y
=
sin
t
,
z
=
t
,
∫

C

Solution
So, we already have the curve parameterized so there really isn’t much to do other than evaluate
the integral

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