# JACOBIANS; CHANGING VARIABLES IN MULTIPLE INTEGRATION

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 Varying Force over a Curved Path

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:

# Line Integrals

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

# THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS

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:

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