WORK-ENERGY FORMULA; CONSERVATION OF MECHANICAL ENERGY

Suppose that a continuous force field $F=F(r)$ accelerates a mass $m$ from $r=a$ to $r=b$ along some smooth curve $C$. The object undergoes a change in kinetic energy:
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The force does a certain amount of work $W$. How are these quantities related? They are equal:
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This relation is called the Work-Energy Formula.

Derivation of the Work- Energy Formula

We parametrize the path of the motion by the time parameter $t:$
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Note that $r(a)=a$ and $r(b)=b.$ The work done by $F$ is given by the formula
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From Newton's second law of motion, we know that at time $t,$
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It follows that
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Substituting this last expression into the work integral, we see that
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as asserted.

Conservative Force Fields

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,
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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.

Potential Energy Functions

Suppose that $F$ is a conservative force field. It is then a gradient field. Then $-F$ is a gradient field. The functions $U$ for which $\nabla U=-F$ are called potential energy functions for $F.$

The Conservation of Mechanical Energy

Suppose that $F$ is a conservative force field: $F=-\nabla U.$ In our derivation of work- energy formula we showed that
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Since
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we have
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and therefore
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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 $E.$ The law of conservation of mechanical energy can then be written
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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.

Differences in Potential Energy

Potential energy at a particular point has no physical significance. Only differences in potential energy are significant:
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is the work required to move from $r=a$ to $r=b$ against the force field $F.$

Problem 1. A planet moves in the gravitational field of the sun:
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(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
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As a potential energy function we can use
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The total energy of the planet is the constant
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Solving the energy equation for $v$ we have
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As $r$ decreases, $2\rho /r$ increases, and $v$ increases; as $r$ increases, $2\rho /r$ decreases, and $v$ 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.)

LINE INTEGRALS WITH RESPECT TO ARC LENGTH

If MATH the line integral $\int_Ch(r)\cdot dr$ can be written
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The notation arises as follows. With


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the line integral
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expands to
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Now set
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Writing the sum of these integrals as
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we have
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If $C$ lies in the $xy$-plane and MATH then the $z$'s drop out and the line integral reduces to
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The notation is easy to use; for example, with $C$ as above,
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Line Integrals with Respect to Arc Length

Suppose that $f$ is a scalar field continuous on a piecewise-smooth curve
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If $s(u)$ is the length of the curve from the tip of $r(a)$ to the tip of $r(u)$, then
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The line integral of $f$ over $C$ with respect to arc length $s$ is defined by setting
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In the $xyz$-notation we have
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which, in the two-dimensional case, becomes
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Suppose now that $C$ represents a thin wire (a material curve) of varying mass density MATH. [Here mass density is mass per unit length.] The length of the wire can be written
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The mass of the wire is given by
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and the center of mass $r_M$ can be obtained from the vector equation
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The equivalent scalar equations read
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Finally, the moment of inertia about an axis is given by the formula
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where $R(r)$ is the distance from the axis to the tip of $r.$

Problem 1. The mass density of a semicircular wire of radius $a$ 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
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and the mass density function can be written $\lambda (x,y)=ky$. Since MATH we have
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Therefore
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By the symmetry of the configuration, $x_M=0.$ To find $y_M$ we have to integrate:
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Since $M=2ka^2$, we have MATH. The center of mass lies on the perpendicular bisector of the wire at a distance $\frac{a\pi }4$ from the diameter.

Now let's find the moment of inertia about the diameter:
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It is customary to express $I$ in terms of $M$. With $M=2ka^{2}$, we have
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GREEN'S THEOREM

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 $\Omega $, and, if its boundary $C$ is piecewise smooth, we know how to integrate over $C$. Green's theorem expresses a double integral over $\Omega $ as a line integral over $C.$

THEOREM GREEN'S THEOREM

Let $\Omega $ be a Jordan region with a piecewise-smooth boundary $C.$ If $P$ and $Q$ are scalar fields continuously differentiable on an open set that contains $\Omega $ then
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where the integral on the right is the line integral over $C$ taken in the counterclockwise direction.

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

Since $\Omega $ is of type I, we are to show that
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In the first place
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The graph of $\phi _{1}$ parametrized from left to right is the curve
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the graph of $\phi _{2}$, also parametrized from left to right, is the curve
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Since $C$ traversed counterclockwise consists of $C_{\QTR{group}{1}}$ followed by $-C_{2}$ , we can see that
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Since $u$ is a dummy variable, it can be replaced by $x$. This proves the claim. Similarly one can verify that
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by using the fact that $\Omega $ 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 $\Omega $ is a Jordan region that can be broken up into two elementary regions $\Omega _1$ and $\Omega _2.$ Green's theorem applied to the elementary parts tells us that
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We now add these equations. The sum of the double integrals is, by additivity, the double integral over $\Omega $. The sum of the line integrals is the integral over $C$ 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
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This same argument can be extended to a Jordan region $\Omega $ that breaks up into $n$ elementary regions MATH The double integrals over the $\Omega _i$ add up to the double integral over $\Omega $, and, since the line integrals over the crosscuts cancel, the line integrals over the boundaries of the MATH add up to the line integral over $C$. (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
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where $C$ is the circle $x^{2}+y^{2}=a^{2}$.

Solution. Let $\Omega $ be the closed disc $0\le $$x^2+y^2\leq a^2$ With

$P(x,y)=xy^2$ and $Q(x,y)=x^2y,$

we have
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By Green's theorem
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Problem 2. Use Green's theorem to evaluate
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where $C$ is the square with vertices MATH.

Solution. Let $\Omega $ be the square region enclosed by $C$. With

$P(x,y)=1+10xy+y^2$ and $Q(x,y)=6xy+5x^2$,

we have
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By Green's theorem the line integral equals
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where $y$ is the $y$-coordinate of the centroid of the square region. Obviously $\bar y=\frac 12a$ and the integral equals $2a^3.$

Problem 3. Use Green's theorem to evaluate
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where $C$ is the circle $(x+1)^2+(y-2)^2=9.$

Solution. The curve bounds the closed disc MATH of radius $3$ centered at $(-1,2)$. Here
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Therefore the line integral over $C$ equals
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With Green's theorem we can show that, if $C$ is a piecewise-smooth Jordan curve, then

the area enclosed by $C=$MATH

Proof. Let $\Omega $ be the region enclosed by $C$. In the first integral
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Therefore
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Thus by Green's theorem
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That the second integral also gives the area of $\Omega $ can be verified in a similar manner.

Problem 4. Let $\Omega $ be a Jordan region of area $A$ with a piecewise-smooth boundary $C$. Show that the coordinates of the centroid of $\Omega $ are given by
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Solution.
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Regions Bounded by Two or More Jordan Curves

An annular region $\Omega $ is not a Jordan region: the boundary consists of two Jordan curves $C_1$ and $C_2.$ We cannot apply Green's theorem to $\Omega $ directly, but we can break up $\Omega $ into two Jordan regions $\Omega _1$ and $\Omega _2$ and then apply Green's theorem to each piece.
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When we add the double integrals, we get the double integral over $\Omega $. When we add the line integrals, the integrals over the crosscuts cancel and we are left with the counterclockwise integral over $C_{\QTR{group}{1}}$ and the clockwise integral over $C_2.$ Thus, for the annular region,
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As a corollary to this we see that, if MATH throughout $\Omega $, then the double integral on the left is $0$, the sum of the integrals on the right is $0$, and therefore
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Problem 5. Let $C_{1}$ be a Jordan curve that does not pass through the origin $(0,0)$. Show that
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Solution. In this case
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Thus
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If $C_{\QTR{group}{1}}$ does not enclose the origin, then MATH throughout the region enclosed by $C_{\QTR{group}{1}}$, and, by Green's theorem, the line integral is $0$.

If $C_{\QTR{group}{1}}$ does enclose the origin, we draw within the inner region of $C_{\QTR{group}{1}}$ a small circle centered at the origin
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Since MATH on the annular region bounded by $C_{\QTR{group}{1}}$ and $C_{2}$, we conclude that the line integral over $C_{\QTR{group}{1}}$ equals the line integral over $C_{2}$. All we have to show now is that the line integral over $C_{2}$ is $2\pi .$ This is straightforward. Parametrizing the circle by
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we have
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For a region bounded by three Jordan curves: $C_{2}$ and $C_{3}$ both exterior to one another, both within MATH Green's theorem gives
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To see this, break up $\Omega $ into two regions by making the crosscuts. The general formula for configurations of this type reads
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Exercise-Green's Theorem

Evaluate the line integral using Green's Theorem.

1. MATH, where $C$ is the square with vertices $(0,0),(1,0),(1,1),$ and $(0,1)$ oriented counterclockwise. $[0]$

2. MATH where $C$ is the rectangle bounded by $x=-2,x=4,y=1,$ and MATH

3. MATH where $C$ is the square with vertices MATH and MATH

4. MATH where $C$ is the circle MATH

5. MATH where $C$ is the triangle

with vertices $(0,0),(2,0),$ and MATH

6. MATH where $C$ is the square with vertices $(0,0),(0,1),(1,1),$ and MATH

7.MATH where $C$ is the boundary of the region enclosed by $y=x^{2}$ and MATH

8. Use the formula MATH to find the area of the region swept out by the line from the origin to the ellipse MATH if $t$ varies from $t=0$ to MATH

Evaluate by Green's theorem.

9.MATH

10.$\oint_{C}x^{2}dy$ where $C$ is the rectangle with vertices MATH

11. MATH

$[7\pi ]$

12. MATH

13.MATH

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