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 $\Gamma $ be a basic region in a plane that we are calling the $uv$-plane. (In this plane we denote the abscissa of a point by $u$ and the ordinate by $v$.) Suppose that
MATH
are functions continuously differentiable on $\Gamma $. As $(u,v)$ ranges over $\Gamma $, the point MATH generates a region $\Omega $ in the $xy$-plane. If the mapping
MATH
is one-to-one on the interior of $\Gamma $ and the Jacobian
MATH
is never zero on the interior of $\Gamma $, then
MATH

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 $f=f(x,y)$ over $\Omega .$ If this proves difficult to do directly, then we can change variables to $u,v$ and try to integrate over $\Gamma $ instead:
MATH

The derivation of this formula from the area formula follows the usual lines. Break up $\Gamma $ into $N$ little basic regions MATH These induce a decomposition of MATH into $N$ little basic regions MATH. We can then write
MATH
This last expression is a Riemann sum for
MATH
and tends to that integral as the maximum diameter of the $\Omega _{i}$ tends to zero. This we can ensure by letting the maximum diameter of the $\Gamma _{i}$ tend to zero.

Problem 1. Evaluate
MATH
where $\Omega $ is the parallelogram bounded by the lines
MATH

Solution. The boundaries suggest that we set
MATH
We want $x$ and $y$ in terms of $u$ and $v$. Since

MATH and MATH

we have
MATH
The Jacobian is given by
MATH
Therefore
MATH

Problem 2. Evaluate
MATH
where MATH is the first-quadrant region bounded by the curves
MATH

Solution. The boundaries suggest that we set
MATH
We want $x$ and $y$ in terms of $u$ and $v$. Since

$u+v=2x^2$ and $u-v=2y^2,$

we have
MATH
The transformation has Jacobian
MATH
Therefore
MATH

Earlier we saw the formula for changing variables from rectangular coordinates $(x,y)$ to polar coordinates $(r,\theta ).$ The formula reads
MATH
The factor $r$ in the double integral over $\Gamma $ is the Jacobian of the transformation $x=r\cos \,\theta $,$y=r\sin \,\theta $:
MATH
When changing variables in a triple integral we make three coordinate changes:
MATH
If these functions carry a basic solid $\Gamma $ onto a solid $T$, then, under conditions analogous to the two-dimensional case,


MATH
where now the Jacobian is a three-by-three determinant:
MATH
In this case the change of variables formula reads
MATH

MATH

LINE INTEGRALS AND SURFACE INTEGRALS

The Work Done by a Varying Force over a Curved Path

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

$W=($comp$_d$ $F$) MATH.

We can write this more briefly as a dot product:
MATH
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
MATH
subject to a continuous force $F$. (The vector field $F$ 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 $r^{\prime }$ is continuous and never zero. What we want to do here is define the total work done by $F$ along the curve $C$.

To decide how to do this, we begin by focusing our attention on what happens over a short parameter interval $[u,u+h]$. As an estimate for the work done over this interval we can use the dot product
MATH
In making this estimate we are using the force vector at $r(u)$ and we are replacing the curved path from $r(u)$ to $r(u+h)$ by the line segment from $r(u)$ to $r(u+h)$. If we set

$W(u)=$ total work done by $F$ from $r(a)$ to $r(u)$ and

$W(u+h)=$ total work done by $F$ from $r(a)$ to $r(u+h),$

then the work done by $F$ from $r(u)$ to $r(u+h)$ must be the difference

$W(u+h)-W(u).$

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

MATH

which, upon division by $h$, becomes
MATH
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 $h$ tends to zero; in other words, by requiring that
MATH
The rest is now determined. Since

$W(a)=0$ and $W(b)=$ total work done on $C$,

we have

total work done on MATH

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

Line Integrals

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

DEFINITION LINE INTEGRAL

Let $h$ be a vector field continuous on a smooth curve
MATH
The line integral of $h$ over $C$ is the number
MATH
Note that, although we speak of integrating over $C$, we actually carry out the calculations over the parameter set $[a,b]$. If our definition of line integral is to make sense, the line integral as defined must be independent of the particular parametrization chosen for $C.$ Within the limitations spelled out below this is indeed the case:

the line integral
MATH
is left invariant by every sense-preserving change of parameter.

Proof. Suppose that $\phi $ maps MATH onto MATH and that $\phi ^{\prime }$ is positive and continuous on MATH. We must show that the line integral over $C$ as parametrized by
MATH
equals the line integral over $C$ as parametrized by $r$. The argument is straightforward:
MATH

MATH

Problem 1. Calculate
MATH
given that

$h(x,y)=xyi+y^2j$ and MATH

Solution. Here
MATH
and
MATH
It follows that
MATH

Problem 2. Integrate the vector field MATH over the twisted cubic
MATH
from $(-1,1,-1)$ to $(1,1,1).$

Solution. The path of integration begins at $u=-1$ and ends at $u=1.$ In this case
MATH
Therefore
MATH

MATH
and
MATH

If a curve $C$ is not smooth but is made up of a finite number of adjoining smooth pieces MATH, then we define the integral over $C$ as the sum of the integrals over the $C_i$:
MATH
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 $a=(a_1,a_2)$ and ends at $b=(b_1,b_2)$ can be parametrized by setting
MATH

Problem 3. Evaluate the line integral
MATH
if MATH and $C$ is the triangle with vertices $(1,0),$ $(0,1),(-1,0)$ traversed counterclockwise.

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


MATH

MATH

MATH
We verify,
MATH

MATH

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

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 $-1.$ To be precise, let $C$ be a piecewise-smooth curve and let $-C$denote the same path traversed in the opposite direction. If $C$ is parametrized by a vector function $r$ defined on $[a,b]$, then $-C$ can be parametrized by setting
MATH
Our assertion is that
MATH
or more briefly that
MATH

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

Problem 4. An object, acted upon by various forces, moves along the parabola $y=3x^2$ from the origin to the point $(1,3).$ One of the forces acting on the object is $F(x,y)=x^3i+yj.$ Calculate the work done by $F.$

Solution. We can parametrize the path by setting
MATH
Here
MATH
and
MATH
It follows that


MATH

If an object of mass $m$ moves so that at time $t$ it has position $r(t)$, then, from Newton's second law, the total force acting on the object at time $t$ must be MATH

Problem 5. An object of mass $m$ moves from time $t=0$ to $t=1$ so that its position at time $t$ is given by the vector function
MATH
Find the total force acting on the object at time $t$ and calculate the total work done by this force.

Solution. Differentiation gives
MATH
The total force on the object at time $t$ is therefore
MATH
We can calculate the total work done by this force by integrating the force over the curve
MATH
Thus
MATH

THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS

In general, if we integrate a vector field $h$ 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,
MATH
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 MATH be a piecewise-smooth curve that begins at $a$ and ends at $b$. If the scalar field $f$ is continuously differentiable on an open set that contains the curve $C$, then
MATH

Proof. If $C$ is smooth,
MATH

If $C$ is not smooth but only piecewise smooth, then we break up $C$ into smooth pieces
MATH
With obvious notation,
MATH
The theorem we just proved has an important corollary:

if the curve $C$ is closed (that is, if $a=b$), then $f(b)=f(a)$ and
MATH

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

Solution. First we try to determine whether $h$ is a gradient. Note that $h(x,y)$ has the form $P(x,y)i+Q(x,y)j$ with

$P(x,y)=y^2$ and $Q(x,y)=2xy-e^y$. Since $P$ and $Q$ are continuously differentiable everywhere and
MATH
we can conclude that $h$ is a gradient. Since the integral depends then only on the endpoints of $C$ and not on $C$ itself, we can simplify the computations by integrating over the line segment $C^{\prime }$ that joins these same endpoints.

We parametrize $C^{\prime }$ by setting
MATH
We then have
MATH

Alternative Solution. Once we have recognized that MATH is a gradient $\nabla f(x,y),$ we can determine $f(x,y)$ by the methods as follows. Since
MATH
we have
MATH
The two expressions for MATH can be reconciled only if
MATH
This means that
MATH
Since the curve $C$ begins at $(1,0)$ and ends at $(0,1)$, we see that
MATH

Problem 2. Evaluate the line integral
MATH
where $C$ is the unit circle
MATH
and
MATH

Solution. Although MATH is not a gradient, part of it,
MATH
is a gradient. Since we are integrating over a closed curve, the contribution of the gradient part is $0$. The contribution of the remaining part is
MATH
This last integral is easy to evaluate:
MATH

This document created by Scientific WorkPlace 4.0.