Suppose that a continuous force field
accelerates a mass
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
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
is a conservative force field:
In our derivation of work- energy formula we showed that
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
(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.)
the line integral
can be written
The notation arises as follows. With
the line integral
Writing the sum of these integrals as
lies in the
drop out and the line integral reduces to
The notation is easy to use; for example, with as above,
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
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
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
be a Jordan region with a piecewise-smooth boundary
are scalar fields continuously differentiable on an open set that contains
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.
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.
is a Jordan region that can be broken up into two elementary regions
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
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
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
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
be the region enclosed by
In the first integral
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
An annular region
is not a Jordan region: the boundary consists of two Jordan curves
We cannot apply Green's theorem to
directly, but we can break up
into two Jordan regions
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
Solution. In this case
If does not enclose the origin, then throughout the region enclosed by , and, by Green's theorem, the line integral is .
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
For a region bounded by three Jordan curves:
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.
10. where is the rectangle with vertices
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