Calc 3 Lecture Notes
Section 14.2
Page 1 of 7
Section 14.2: Line Integrals
Big idea: Line integrals are used to compute how a given quantity “adds up” along a curved
path.
Big skill: You should be able to compute line integrals (using the evaluation theorem).
Introductory examples:
 x
 kg
1. Compute the mass of a baseball bat with linear density   x     0.5 
for 0  x 
m
m
1 m. (Note that in this example, we are adding up a quantity over a curve, which is
simply a straight segment of length 1 m.)
2. Compute the mass of a helix of radius 1 m and that has 5 turns along 1 m of its axis, and
a linear mass density that increases linearly from 1 g/m to 26 g/m over its length. Notice
that the total arc length of this helix is:
5
 2 1m   0.2m
2
2
 10  2  0.01m  31.432m , so we can write the linear mass
25s  g

density as a function of arc length s as:   s   1 
 for s over the interval
 31.432m  m
0  s  31.432 m.
z
x
y
Note that in this example, we had to make an effort to relate the density to the arc length. This is
not usually how things are done. Instead, we parameterize the curve and leave the integrand
alone…
Calc 3 Lecture Notes
Section 14.2
Page 2 of 7
Definition 2.1: Line Integral with Respect to Arc Length
The line integral of f(x, y, z) with respect to arc length along the oriented curve C in threedimensional space, written as  f  x, y, z  ds is defined by
C
n
 f  x, y, z  ds  lim  f  x *, y *, z * s ,
P 0
C
i 1
i
i
i
i
provided the limit exists and is the same for all choices of evaluation points.
Theorem 2.1: Evaluation Theorem (for Line Integrals)
If f(x, y, z) is continuous in a region D containing a curve C, C can be described parametrically
by (x(t), y(t), z(t)) for a  t  b, and x(t), y(t), and z(t) have continuous first derivatives, then

C
b
f  x, y, z  ds   f  x  t  , y  t  , z  t  
 x  t     y t     z t   dt .
2
2
2
a
Likewise, if f(x, y) is continuous in a region D containing a curve C, C can be described
parametrically by (x(t), y(t)) for a  t  b, and x(t) and y(t) have continuous first derivatives, then

C
Proof:
b
f  x, y  ds   f  x  t  , y  t  
a
 x t     y t   dt .
2
2
Calc 3 Lecture Notes
Section 14.2
Page 3 of 7
Practice:
 25 z  g
1. Compute the mass of the helix from page 1 given that   x, y, z   1 
 .
m m

2. Evaluate
  x  y  z  ds , given a curve C specified by x = cos(t), y = sin(t), and z = cos(t)
C
for 0  t  2.
3. Evaluate
 xyds , given a curve C specified by the quarter ellipse in the first quadrant of
C
the x-y plane with a semi-major axis of length 2 aligned along the x-axis and semi-minor
axis of 1.
Calc 3 Lecture Notes
Section 14.2
Page 4 of 7
A curve C is called smooth if it can be described parametrically by (x(t), y(t), z(t)) for a  t  b,
x(t), y(t), and z(t) have continuous first derivatives, and  x  t     y  t     z  t    0 on the
interval [a, b].
2
2
2
For example, the curve C = (t2, t3, t2) is not smooth on the interval [-1, 1] because
2
2
2
 2t    3t    2t   0 when t = 0:
z
x
y
This means that the line integral

f  x, y, z  ds 
1
 f  x  t  , y  t  , z t    x t     y t     z t   dt is undefined, but if we
2
2
2
1
C
break up the curve into two subpieces C = C1  C2 over the intervals [-1,0] and [0, 1], then we
can break up the integral into two subpieces as well. A curve that can be subdivided into smooth
subpieces is called piecewise smooth.
Theorem 2.2: Line Integrals on a Piecewise Smooth Curve
If f(x, y, z) is continuous in a region D containing an oriented curve C, and C is piecewisesmooth with C = C1  C2  …  Cn and all the Ci are smooth and the terminal point of Ci is the
initial point of Ci+1 for all i = 1, 2, … n-1, THEN
 f  x, y, z  ds   f  x, y, z  ds
C
AND
C
 f  x, y, z  ds   f  x, y, z  ds   f  x, y, z  ds 
C
Practice:
4. Evaluate
C1
C2

 f  x, y, z  ds
Cn
 xyds , given a curve C specified by a sector of outer radius 2 and inner radius
C
1 and angular extent of /2 symmetric to the x-axis.
Calc 3 Lecture Notes
Section 14.2
Page 5 of 7
Geometric interpretation of the line integral:
b
Just as
 f  x  dx is the area bounded by x = a, x = b, y = 0 and y = f(x),
a

C
b
f  x, y  ds   f  x  t  , y  t   ds is the area of the “vertical” surface bounded by the “vertical”
a
line through (x(a), y(a)), the vertical line through (x(b), y(b)), the parametric curve (x(t), y(t), 0),
and the parametric curve ( x(t), y(t), f(x(t), y(t)) ).
Theorem 2.3: Arc Length of a Curve from a Line Integral
For any piecewise-smooth curve C,  1ds gives the arc length of the curve C.
C
Next, we worry about evaluating a line integral where the integrand is the dot product of a vector
field and the differential tangent vector to the curve, as arises in the computation of work:
Calc 3 Lecture Notes
Section 14.2
Page 6 of 7
Definition 2.2: Component-Wise Line Integrals
The line integral of f(x, y, z) with respect to x along the oriented curve C in three-dimensional
space is written as  f  x, y, z  dx and is defined by:
C

n
f  x, y, z  dx  lim  f  xi *, yi *, zi * xi ,
P 0
C
i 1
provided the limit exists and is the same for all choices of evaluation points.
The line integral of f(x, y, z) with respect to y along the oriented curve C in three-dimensional
space is written as  f  x, y, z  dy and is defined by:
C

n
f  x, y, z  dy  lim  f  xi *, yi *, zi * yi ,
P 0
C
i 1
provided the limit exists and is the same for all choices of evaluation points.
The line integral of f(x, y, z) with respect to z along the oriented curve C in three-dimensional
space is written as  f  x, y, z  dz and is defined by:
C

C
n
f  x, y, z  dz  lim  f  xi *, yi *, zi * zi ,
P 0
i 1
provided the limit exists and is the same for all choices of evaluation points.
Theorem 2.4: Evaluation Theorem (for Component-Wise Line Integrals)
If f(x, y, z) is continuous in a region D containing a curve C, C can be described parametrically
by (x(t), y(t), z(t)) for a  t  b, and x(t), y(t), and z(t) have continuous first derivatives, then

C
b
f  x, y, z  dx   f  x  t  , y  t  , z  t   x  t  dt
a
b
 f  x, y, z  dy   f  x  t  , y  t  , z  t   y t  dt .
C
a
b
 f  x, y, z  dz   f  x  t  , y  t  , z t   z t  dt
C
a
Calc 3 Lecture Notes
Section 14.2
Page 7 of 7
Theorem 2.5: Evaluation Theorem (for Component-Wise Line Integrals)
If f(x, y, z) is continuous in a region D containing an oriented curve C, then:
If C is piecewise-smooth:
 f  x, y, z  dx   f  x, y, z  dx
C
C
 f  x, y, z  dy   f  x, y, z  dy
C
C
 f  x, y, z  dz    f  x, y, z  dz
C
C
If C is piecewise-smooth and forms a closed loop:
 f  x, y, z  dx   f  x, y, z  dx   f  x, y, z  dx 

 f  x, y, z  dy   f  x, y, z  dy   f  x, y, z  dy 

 f  x, y, z  dz   f  x, y, z  dz   f  x, y, z  dz 

C
C1
C
C1
C
C1
Consequence:
W   F  x, y, z   dr
C
C1
C1
C1
 f  x, y, z  dx
Cn
 f  x, y, z  dy
Cn
 f  x, y, z  dz
Cn