Suppose that a continuous force field
accelerates a mass
from
to
along some smooth curve
.
The object undergoes a change in kinetic energy:

The force does a certain amount of work
.
How are these quantities related? They are equal:

This relation is called the **Work-Energy Formula.**

We parametrize the path of the motion by the time parameter

Note that
and
The work done by
is given by the formula

From Newton's second law of motion, we know that at time

It follows that

Substituting this last expression into the work integral, we see that

as asserted.

In general, if an object moves from one point to another, the work done (and
hence the change in kinetic energy) depends on the path of the motion. There
is, however, an important exception: if the force field is a gradient field,

then the work done (and hence the change in kinetic energy) depends only on
the endpoints of the path and not on the path itself. (This follows directly
from The Fundamental Theorem for Line Integrals.) A force field that is a
gradient field is called a **conservative field.**

Since the line integral over a closed path is zero, the work done by a conservative field over a closed path is always zero. An object that passes through a given point with a certain kinetic energy returns to that same point with exactly the same kinetic energy.

Suppose that
is a conservative force field. It is then a gradient field. Then
is a gradient field. The functions
for which
are called **potential energy functions** for

Suppose that
is a conservative force field:
In our derivation of work- energy formula we showed that

Since

we have

and therefore

An an object moves in a conservative force field, its kinetic energy can vary
and its potential energy can vary, but the sum of these two quantities remains
constant. We call this constant the **total mechanical energy.**

The total mechanical energy is usually denoted by the letter
The law of conservation of mechanical energy can then be written

The conservation of energy is one of the cornerstones of physics. Here we have
been talking about mechanical energy. There are other forms of energy and
other energy conservation laws.

Potential energy at a particular point has no physical significance. Only
differences in potential energy are significant:

is the work required to move from
to
against the force field

**Problem 1.** A planet moves in the gravitational field of the
sun:

(m is the mass of the planet) Show that the force field is conservative, find
a potential energy function, and determine the total energy of the planet. How
does the planet's speed vary with the planet's distance from the sun?

*Solution.* The field is conservative since

As a potential energy function we can use

The total energy of the planet is the constant

Solving the energy equation for
we have

As decreases, increases, and increases; as increases, decreases, and decreases. Thus every planet speeds up as it comes near the sun and slows down as it moves away. (The same holds true for Halley's comet. The fact that it slows down as it gets farther away helps explain why it comes back. The simplicity of all this is a testimony to the power of the principle of energy conservation.)

If
the line integral
can be written

The notation arises as follows. With

the line integral

expands to

Now set

Writing the sum of these integrals as

we have

If
lies in the
-plane
and
then the
's
drop out and the line integral reduces to

The notation is easy to use; for example, with
as above,

Suppose that
is a scalar field continuous on a piecewise-smooth curve

If
is the length of the curve from the tip of
to the tip of
,
then

The line integral of
over
**with respect to arc length**
is defined by setting

In the
-notation
we have

which, in the two-dimensional case, becomes

Suppose now that
represents a thin wire (a material curve) of varying mass density
.
[Here mass density is mass per unit length.] The length of the wire can be
written

The *mass* of the wire is given by

and the *center of mass*
can be obtained from the vector equation

The equivalent scalar equations read

Finally, the *moment of inertia* about an axis is given by the formula

where
is the distance from the axis to the tip of

**Problem 1.** The mass density of a semicircular wire of radius
varies directly as the distance from the diameter that joins the two endpoints
of the wire. (a) Find the mass of the wire. (b) Locate the center of mass. (c)
Determine the moment of inertia of the wire about the diameter.

*Solution.* The wire can be parametrized by

and the mass density function can be written
.
Since
we have

Therefore

By the symmetry of the configuration,
To find
we have to integrate:

Since
,
we have
.
The center of mass lies on the perpendicular bisector of the wire at a
distance
from the diameter.

Now let's find the moment of inertia about the diameter:

It is customary to express
in terms of
.
With
,
we have

Recall that a Jordan curve is a plane curve that is both closed and simple. Thus circles, ellipses, and triangles are Jordan curves; figure eights are not.

A closed region the total boundary of which is a Jordan curve is called a
**Jordan region.**

We know how to integrate over a Jordan region , and, if its boundary is piecewise smooth, we know how to integrate over . Green's theorem expresses a double integral over as a line integral over

**THEOREM** GREEN'S THEOREM

Let
be a Jordan region with a piecewise-smooth boundary
If
and
are scalar fields continuously differentiable on an open set that contains
then

where the integral on the right is the line integral over
taken in the counterclockwise direction.

We will prove the theorem only for special cases. First of all let's assume
that
is an *elementary region*, a region that is both of type I and type II.

Since
is of type I, we are to show that

In the first place

The graph of
parametrized from left to right is the curve

the graph of
,
also parametrized from left to right, is the curve

Since
traversed counterclockwise consists of
followed by
, we can see that

Since
is a dummy variable, it can be replaced by
.
This proves the claim. Similarly one can verify that

by using the fact that
is of type II. This will complete the proof of the theorem for regions that
are both of type I and type II.

A slight modification of this argument applies to elementary regions which are bordered entirely or in part by line segments parallel to the coordinate axes.

Suppose that
is a Jordan region that can be broken up into two elementary regions
and
Green's theorem applied to the elementary parts tells us that

We now add these equations. The sum of the double integrals is, by additivity,
the double integral over
.
The sum of the line integrals is the integral over
plus the integrals over the crosscut. Since the crosscut is traversed twice
and in opposite directions, the total contribution of the crosscut is zero and
therefore

This same argument can be extended to a Jordan region that breaks up into elementary regions The double integrals over the add up to the double integral over , and, since the line integrals over the crosscuts cancel, the line integrals over the boundaries of the add up to the line integral over . (This is as far as we will carry the proof of Green's theorem. It is far enough to cover all the Jordan regions encountered in practice.)

**Problem 1.** Use Green's theorem to evaluate

where
is the circle
.

*Solution.* Let
be the closed disc
With

and

we have

By Green's theorem

**Problem 2.** Use Green's theorem to evaluate

where
is the square with vertices
.

*Solution.* Let
be the square region enclosed by
.
With

and ,

we have

By Green's theorem the line integral equals

where
is the
-coordinate
of the centroid of the square region. Obviously
and the integral equals

**Problem 3.** Use Green's theorem to evaluate

where
is the circle

*Solution.* The curve bounds the closed disc
of radius
centered at
.
Here

Therefore the line integral over
equals

With Green's theorem we can show that, if is a piecewise-smooth Jordan curve, then

the area enclosed by

Proof. Let
be the region enclosed by
.
In the first integral

Therefore

Thus by Green's theorem

That the second integral also gives the area of
can be verified in a similar manner.

**Problem 4.** Let
be a Jordan region of area
with a piecewise-smooth boundary
.
Show that the coordinates of the centroid of
are given by

*Solution*.

An annular region
is not a Jordan region: the boundary consists of two Jordan curves
and
We cannot apply Green's theorem to
directly, but we can break up
into two Jordan regions
and
and then apply Green's theorem to each piece.

When we add the double integrals, we get the double integral over
.
When we add the line integrals, the integrals over the crosscuts cancel and we
are left with the counterclockwise integral over
and the clockwise integral over
Thus, for the annular region,

As a corollary to this we see that, if
throughout
,
then the double integral on the left is
,
the sum of the integrals on the right is
,
and therefore

**Problem 5.** Let
be a Jordan curve that does not pass through the origin
.
Show that

*Solution.* In this case

Thus

If
does not enclose the origin, then
throughout the region enclosed by
,
and, by Green's theorem, the line integral is
.

If
does enclose the origin, we draw within the inner region of
a small circle centered at the origin

Since
on the annular region bounded by
and
,
we conclude that the line integral over
equals the line integral over
.
All we have to show now is that the line integral over
is
This is straightforward. Parametrizing the circle by

we have

For a region bounded by three Jordan curves:
and
both exterior to one another, both within
Green's theorem gives

To see this, break up
into two regions by making the crosscuts. The general formula for
configurations of this type reads

Evaluate the line integral using Green's Theorem.

1. , where is the square with vertices and oriented counterclockwise.

2. where is the rectangle bounded by and

3. where is the square with vertices and

4. where is the circle

5. where is the triangle

with vertices and

6. where is the square with vertices and

7. where is the boundary of the region enclosed by and

8. Use the formula to find the area of the region swept out by the line from the origin to the ellipse if varies from to

Evaluate by Green's theorem.

9.

10. where is the rectangle with vertices

11.

12.

13.

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