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Consider the curve
defined by the parametrization
If the curve
passes through the domain
of a function
w=f(x,y,z), the values of f on the curve
are given by
w=f(x(t),y(t),z(t)).
If we integrate this function of the variable t with respect to the arc length for
t in the interval [a,b], we get the line integral of f on
;
it is denoted
We suppose that the curve
is smooth, i.e. for every value of t, the functions x, y and z are differentiable and nowhere all equal to 0.
Denote
.
Example 8.2
Let
f(
x,
y,
z)=
xy+
yz^{2}+
x and let
be the segment joining the origin to the point
A(1,2,1).
A parametrization of
is given by
.
Thus
and
.
We have:
Example 8.3
We define a function by
f(
x,
y,
z)=
z+
y
z and consider the curve
,
dispalyed on Fig.
1, with the following parametrization:
Figure 1:
A loop of an helix.

Then:
and
Remark 8.4
After replacing
ds by
,
the integral becomes a regular integral of a function of one real variable on an interval.
Example 8.6
Let
f(
x,
y,
z)=
x+
y+2
z; we compute the integral of
f on the broken line
displayed on Fig.
2. This curve is the union of two line segments:
 The segment OA, with parametrization:
.
 The segment AB, with parametrization:
.
Figure 2:
A ``broken'' path.

We have:
 On OA,
,
hence
.
Therefore:
 On AB,
,
hence
.
Therefore:
Finally, we have:
Next: Mass of a wire,
Up: Line integrals.
Previous: Line integrals.
Noah DanaPicard
20010530
ar