During the course of the last few sections we have met several formulas for changing variables in multiple integration: to polar coordinates, to cylindrical coordinates, to spherical coordinates. The purpose of this section is to bring some unity into that material and provide a general prescription for other changes of variable.

We begin with a consideration of area. Let
be a basic region in a plane that we are calling the
-plane.
(In this plane we denote the abscissa of a point by
and the ordinate by
.)
Suppose that

are functions continuously differentiable on
.
As
ranges over
,
the point
generates a region
in the
-plane.
If the mapping

is one-to-one on the interior of
and the **Jacobian
**is never zero on the interior of
,
then

It is very difficult to prove this assertion without making additional
assumptions. At this point we will simply assume this area formula and go on
from there. Suppose now that we want to integrate some continuous function
over
If this proves difficult to do directly, then we can change variables to
and try to integrate over
instead:

The derivation of this formula from the area formula follows the usual lines.
Break up
into
little basic regions
These induce a decomposition of
into
little basic regions
.
We can then write

This last expression is a Riemann sum for

and tends to that integral as the maximum diameter of the
tends to zero. This we can ensure by letting the maximum diameter of the
tend to zero.

**Problem 1.** Evaluate

where
is the parallelogram bounded by the lines

*Solution.* The boundaries suggest that we set

We want
and
in terms of
and
.
Since

and

we have

The Jacobian is given by

Therefore

**Problem 2.** Evaluate

where
is the first-quadrant region bounded by the curves

*Solution.* The boundaries suggest that we set

We want
and
in terms of
and
.
Since

and

we have

The transformation has Jacobian

Therefore

Earlier we saw the formula for changing variables from rectangular coordinates
to polar coordinates
The formula reads

The factor
in the double integral over
is the Jacobian of the transformation
,:

When changing variables in a triple integral we make three coordinate changes:

If these functions carry a basic solid
onto a solid
,
then, under conditions analogous to the two-dimensional case,

where now the Jacobian is a three-by-three determinant:

In this case the change of variables formula reads

The work done by a constant force on an object that moves along a straight line is, by definition, the component of in the direction of the displacement multiplied by the length of the displacement vector :

comp ) .

We can write this more briefly as a dot product:

This elementary notion of work is useful, but it is not sufficient. Consider,
for example, an object that moves through a magnetic field or a gravitational
field. The path of the motion is then usually not a straight line but a curve,
and the force, rather than remain constant, tends to vary from point to point.
What we want now is a notion of work that applies to this more general
situation. Let's suppose that an object moves along a curve

subject to a continuous force
.
(The vector field
may vary from point to point, not only in magnitude but also in direction.) We
will suppose that the curve is smooth; namely, we will suppose that the
tangent vector
is continuous and never zero. What we want to do here is define the total work
done by
along the curve
.

To decide how to do this, we begin by focusing our attention on what happens
over a short parameter interval
.
As an estimate for the work done over this interval we can use the dot product

In making this estimate we are using the force vector at
and we are replacing the curved path from
to
by the line segment from
to
.
If we set

total work done by from to and

total work done by from to

then the work done by from to must be the difference

Bringing our estimate into play, we are led to the approximate equation

which, upon division by
,
becomes

The quotients here are average rates of change, and the equation is only an
approximate one. The notion of work is made precise by requiring that both
sides have exactly the same limit as
tends to zero; in other words, by requiring that

The rest is now determined. Since

and total work done on ,

we have

total work done on

In short, we have arrived at the following definition of work:

The integral on the right of the above equation can be calculated not only for a force function but for any vector field continuous on

**DEFINITION** LINE INTEGRAL

Let
be a vector field continuous on a smooth curve

The line integral of
over
is the number

Note that, although we speak of integrating over
,
we actually carry out the calculations over the parameter set
.
If our definition of line integral is to make sense, the line integral as
defined must be independent of the particular parametrization chosen for
Within the limitations spelled out below this is indeed the case:

the line integral

is left invariant by every sense-preserving change of parameter.

*Proof.* Suppose that
maps
onto
and that
is positive and continuous on
.
We must show that the line integral over
as parametrized by

equals the line integral over
as parametrized by
.
The argument is straightforward:

**Problem 1.** Calculate

given that

and

*Solution.* Here

and

It follows that

**Problem 2.** Integrate the vector field
over the twisted cubic

from
to

*Solution.* The path of integration begins at
and ends at
In this case

Therefore

and

If a curve
is not smooth but is made up of a finite number of adjoining smooth pieces
,
then we define the integral over
as the sum of the integrals over the
:

A curve of this type is said to be **piecewise smooth**.

All polygons are piecewise-smooth curves. In the next problem we integrate
over a triangle. We do this by integrating over each of the sides and then
adding up the results. Observe that the directed line segment that begins at
and ends at
can be parametrized by setting

**Problem 3.** Evaluate the line integral

if
and
is the triangle with vertices
traversed counterclockwise.

Solution. The path is made up of three line segments:

We verify,

The integral over the entire triangle is the sum of these integrals:

When we integrate over a parametrized curve, we integrate in the direction
determined by the parametrization. If we integrate in the opposite direction,
our answer is altered by a factor of
To be precise, let
be a piecewise-smooth curve and let
denote
the same path traversed in the opposite direction. If
is parametrized by a vector function
defined on
,
then
can be parametrized by setting

Our assertion is that

or more briefly that

We were led to the definition of line integral by the notion of work. It is
clear that, if a force
is continually applied to an object that moves over a piecewise-smooth curve
then the work done by
is the line integral of
over
:

**Problem 4.** An object, acted upon by various forces, moves
along the parabola
from the origin to the point
One of the forces acting on the object is
Calculate the work done by

*Solution.* We can parametrize the path by setting

Here

and

It follows that

If an object of mass moves so that at time it has position , then, from Newton's second law, the total force acting on the object at time must be

**Problem 5.** An object of mass
moves from time
to
so that its position at time
is given by the vector function

Find the total force acting on the object at time
and calculate the total work done by this force.

*Solution.* Differentiation gives

The total force on the object at time
is therefore

We can calculate the total work done by this force by integrating the force
over the curve

Thus

In general, if we integrate a vector field
from one point to another, the value of the line integral depends upon the
path chosen. There is, however, an important exception. If the vector field is
a *gradient field,*

then the value of the line integral depends only on the endpoints of the path
and not on the path itself. The details are spelled out in the following
theorem.

**THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS**

Let
be a piecewise-smooth curve that begins at
and ends at
.
If the scalar field
is continuously differentiable on an open set that contains the curve
,
then

Proof. If
is smooth,

If
is not smooth but only piecewise smooth, then we break up
into smooth pieces

With obvious notation,

The theorem we just proved has an important corollary:

if the curve
is closed (that is, if
),
then
and

**Problem 1.** Integrate the vector field
over the circular arc

*Solution.* First we try to determine whether
is a gradient. Note that
has the form
with

and
.
Since
and
are continuously differentiable everywhere and

we can conclude that
is a gradient. Since the integral depends then only on the endpoints of
and not on
itself, we can simplify the computations by integrating over the line segment
that joins these same endpoints.

We parametrize
by setting

We then have

*Alternative Solution.* Once we have recognized that
is a gradient
we can determine
by the methods as follows. Since

we have

The two expressions for
can be reconciled only if

This means that

Since the curve
begins at
and ends at
,
we see that

**Problem 2.** Evaluate the line integral

where
is the unit circle

and

*Solution.* Although
is not a gradient, part of it,

is a gradient. Since we are integrating over a closed curve, the contribution
of the gradient part is
.
The contribution of the remaining part is

This last integral is easy to evaluate: