You start by finding
dy
directly
from the given path:
2
4
4
2
y
x
x
dy
x dx
And then transform the line integral so that everything is with respect to
x
and
dx
:
4
2
2
1
2
4
2
3
1
4
4
2
69
4
3
2
2
C
y dx
x dy
d
x
x
x dx
x
x
x
dx
x
x
This example illustrates a convenient feature of the line integral in differential form:
It allows you to work with the
curve
C
directly without having to find the parametric equation for it
. Because you convert the line integral to