6.4 Green’s theorem (Page 7/12)

Calculus volume 3 Page 7 / 12

The boundary of the upper half of the annulus, therefore, is $P_{1} \cup P_{2} \cup P_{3} \cup P_{4}$ and the boundary of the lower half of the annulus is $− P_{4} \cup P_{5} \cup - P_{2} \cup P_{6} .$ Then, Green’s theorem implies

\begin{matrix} \oint_{\partial D} F \cdot d r & = \int_{P_{1}} F \cdot d r + \int_{P_{2}} F \cdot d r + \int_{P_{3}} F \cdot d r + \int_{P_{4}} F \cdot d r + \int_{− P_{4}} F \cdot d r + \int_{P_{5}} F \cdot d r + \int_{− P_{2}} F \cdot d r + \int_{P_{6}} F \cdot d r \\ = \int_{P_{1}} F \cdot d r + \int_{P_{2}} F \cdot d r + \int_{P_{3}} F \cdot d r + \int_{P_{4}} F \cdot d r + \int_{P_{4}} F \cdot d r + \int_{P_{5}} F \cdot d r + \int_{− P_{2}} F \cdot d r + \int_{P_{6}} F \cdot d r \\ = \int_{P_{1}} F \cdot d r + \int_{P_{3}} F \cdot d r + \int_{P_{5}} F \cdot d r + \int_{P_{6}} F \cdot d r \\ = \oint_{\partial D_{1}} F \cdot d r + \oint_{\partial D_{2}} F \cdot d r \\ = \iint_{D_{1}} (Q_{x} - P_{y}) d A + \iint_{D_{2}} (Q_{x} - P_{y}) d A \\ = \iint_{D} (Q_{x} - P_{y}) d A . \end{matrix}

Therefore, we arrive at the equation found in Green’s theorem—namely,

\oint_{\partial D} F \cdot d r = \iint_{D} (Q_{x} - P_{y}) d A .

The same logic implies that the flux form of Green’s theorem can also be extended to a region with finitely many holes:

\oint_{C} F \cdot N d s = \iint_{D} (P_{x} + Q_{y}) d A .

Using green’s theorem on a region with holes

Calculate integral

\oint_{\partial D} (sin x - \frac{y^{3}}{3}) d x + (\frac{y^{3}}{3} + sin y) d y,

where D is the annulus given by the polar inequalities $1 \leq r \leq 2,$ $0 \leq θ \leq 2 π .$

Although D is not simply connected, we can use the extended form of Green’s theorem to calculate the integral. Since the integration occurs over an annulus, we convert to polar coordinates:

\begin{matrix} \oint_{\partial D} (sin x - \frac{y^{3}}{3}) d x + (\frac{x^{3}}{3} + sin y) d y & = \iint_{D} (Q_{x} - P_{y}) d A \\ = \iint_{D} (x^{2} + y^{2}) d A \\ = \int_{0}^{2 π} \int_{1}^{2} r^{3} d r d θ = \int_{0}^{2 π} \frac{15}{4} d θ \\ = \frac{15 π}{2} . \end{matrix}

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Using the extended form of green’s theorem

Let $F = ⟨ P, Q ⟩ = ⟨ \frac{y}{x^{2} + y^{2}}, - \frac{x}{x^{2} + y^{2}} ⟩$ and let C be any simple closed curve in a plane oriented counterclockwise. What are the possible values of $\oint_{C} F \cdot d r ?$

We use the extended form of Green’s theorem to show that $\oint_{C} F \cdot d r$ is either 0 or $−2 π$ —that is, no matter how crazy curve C is, the line integral of F along C can have only one of two possible values. We consider two cases: the case when C encompasses the origin and the case when C does not encompass the origin.

Case 1: C Does not encompass the origin

In this case, the region enclosed by C is simply connected because the only hole in the domain of F is at the origin. We showed in our discussion of cross-partials that F satisfies the cross-partial condition. If we restrict the domain of F just to C and the region it encloses, then F with this restricted domain is now defined on a simply connected domain. Since F satisfies the cross-partial property on its restricted domain, the field F is conservative on this simply connected region and hence the circulation $\oint_{C} F \cdot d r$ is zero.

Case 2: C Does encompass the origin

In this case, the region enclosed by C is not simply connected because this region contains a hole at the origin. Let $C_{1}$ be a circle of radius a centered at the origin so that $C_{1}$ is entirely inside the region enclosed by C ( [link] ). Give $C_{1}$ a clockwise orientation.

A diagram in two dimensions. A circle C1 oriented clockwise is centered at the origin completely inside a generic curve C that is in all four quadrants. Curve C is oriented counterclockwise. — Choose circle $C_{1}$ centered at the origin that is contained entirely inside C .

Let D be the region between $C_{1}$ and C , and C is orientated counterclockwise. By the extended version of Green’s theorem,

\begin{matrix} \int_{C} F \cdot d r + \int_{C_{1}} F \cdot d r & = \iint_{D} Q_{x} - P_{y} d A \\ = \iint_{D} - \frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}} + \frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}} d A \\ = 0, \end{matrix}

and therefore

\int_{C} F \cdot d r = - \int_{C_{1}} F \cdot d r .

Since $C_{1}$ is a specific curve, we can evaluate $\int_{C_{1}} F \cdot d r .$ Let

x = a cos t, y = a sin t, 0 \leq t \leq 2 π

be a parameterization of $C_{1} .$ Then,

\begin{matrix} \int_{C_{1}} F \cdot d r & = \int_{0}^{2 π} F (r (t)) \cdot r ′ (t) d t \\ = \int_{0}^{2 π} ⟨ - \frac{sin (t)}{a}, - \frac{cos (t)}{a} ⟩ \cdot ⟨ − a sin (t), − a cos (t) ⟩ d t \\ = \int_{0}^{2 π} {sin}^{2} (t) + {cos}^{2} (t) d t = \int_{0}^{2 π} d t = 2 π . \end{matrix}

Therefore, $\int_{C} F \cdot d s = - 2 π .$

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Calculate integral $\oint_{\partial D} F \cdot d r,$ where D is the annulus given by the polar inequalities $2 \leq r \leq 5, 0 \leq θ \leq 2 π,$ and $F (x, y) = ⟨ x^{3}, 5 x + e^{y} sin y ⟩ .$

$105 π$

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Measuring area from a boundary: the planimeter

An MRI image of a patient’s brain with a tumor highlighted in red. — This magnetic resonance image of a patient’s brain shows a tumor, which is highlighted in red. (credit: modification of work by Christaras A, Wikimedia Commons)

Imagine you are a doctor who has just received a magnetic resonance image of your patient’s brain. The brain has a tumor ( [link] ). How large is the tumor? To be precise, what is the area of the red region? The red cross-section of the tumor has an irregular shape, and therefore it is unlikely that you would be able to find a set of equations or inequalities for the region and then be able to calculate its area by conventional means. You could approximate the area by chopping the region into tiny squares (a Riemann sum approach), but this method always gives an answer with some error.

Instead of trying to measure the area of the region directly, we can use a device called a rolling planimeter to calculate the area of the region exactly, simply by measuring its boundary. In this project you investigate how a planimeter works, and you use Green’s theorem to show the device calculates area correctly.

A rolling planimeter is a device that measures the area of a planar region by tracing out the boundary of that region ( [link] ). To measure the area of a region, we simply run the tracer of the planimeter around the boundary of the region. The planimeter measures the number of turns through which the wheel rotates as we trace the boundary; the area of the shape is proportional to this number of wheel turns. We can derive the precise proportionality equation using Green’s theorem. As the tracer moves around the boundary of the region, the tracer arm rotates and the roller moves back and forth (but does not rotate).

Two images. The first shows a rolling planimeter. A horizontal bar has a roller attached to it perpendicularly with a pivot. It does not rotate itself; it only moves back and forth. To the right of the roller is the tracer arm with a wheel and a tracer at the very end. The second shows an interior view of a rolling planimeter. The wheel cannot turn if the planimeter is moving back and forth with the tracer arm perpendicular to the roller. — (a) A rolling planimeter. The pivot allows the tracer arm to rotate. The roller itself does not rotate; it only moves back and forth. (b) An interior view of a rolling planimeter. Notice that the wheel cannot turn if the planimeter is moving back and forth with the tracer arm perpendicular to the roller.

Let C denote the boundary of region D , the area to be calculated. As the tracer traverses curve C , assume the roller moves along the y -axis (since the roller does not rotate, one can assume it moves along a straight line). Use the coordinates $(x, y)$ to represent points on boundary C , and coordinates $(0, Y)$ to represent the position of the pivot. As the planimeter traces C , the pivot moves along the y -axis while the tracer arm rotates on the pivot.

Watch a short animation of a planimeter in action.

Begin the analysis by considering the motion of the tracer as it moves from point $(x, y)$ counterclockwise to point $(x + d x, y + d y)$ that is close to $(x, y)$ ( [link] ). The pivot also moves, from point $(0, Y)$ to nearby point $(0, Y + d Y) .$ How much does the wheel turn as a result of this motion? To answer this question, break the motion into two parts. First, roll the pivot along the y -axis from $(0, Y)$ to $(0, Y + d Y)$ without rotating the tracer arm. The tracer arm then ends up at point $(x, y + d Y)$ while maintaining a constant angle $ϕ$ with the x -axis. Second, rotate the tracer arm by an angle $d θ$ without moving the roller. Now the tracer is at point $(x + d x, y + d y) .$ Let $l$ be the distance from the pivot to the wheel and let L be the distance from the pivot to the tracer (the length of the tracer arm).

A diagram in quadrants 1 and 2 showing the motion of the planimeter. Two points are labeled on the y axis: (0, Y) and (0, Y + dY), where Y is less than Y + dY. The first point is the pivot. Three points are labeled further up and to the right in quadrant 1: (x, y), (x, y + dy), and (x + dx, y + dy). Note that the uppercase Y and the lowercase y are not the same; y is much larger. A line segment is drawn between (0,Y) and (x,y). About midway down this line is a mark labeled for the wheel, and the (x,y) endpoint is labeled for the tracer. Let l be the distance from the pivot to the wheel, and let L be the distance from the pivot to the tracer. Line segments are also drawn from (0, Y + dY) to each of the other points in quadrant 1. The angle between the line segment with (0,Y) as an endpoint and the y axis is labeled phi. The angle between the line segments with (0, Y+dY) as an endpoint is “d theta.” A curve is drawn going through the wheel, the tracer, and the three points in quadrant 1, up and across the y axis, down and back across the y axis at a smaller y value lose to the height of the tracer, and down across the line segments and back to the wheel. — Mathematical analysis of the motion of the planimeter.

Explain why the total distance through which the wheel rolls the small motion just described is $sin ϕ d Y + l d θ = \frac{x}{L} d Y + l d θ .$
Show that $\oint_{C} d θ = 0.$
Use step 2 to show that the total rolling distance of the wheel as the tracer traverses curve C is
Total wheel roll $= \frac{1}{L} \oint_{C} x d Y .$
Now that you have an equation for the total rolling distance of the wheel, connect this equation to Green’s theorem to calculate area D enclosed by C .
Show that $x^{2} + {(y - Y)}^{2} = L^{2} .$
Assume the orientation of the planimeter is as shown in [link] . Explain why $Y \leq y,$ and use this inequality to show there is a unique value of Y for each point $(x, y) :$ $Y = y = \sqrt{L^{2} - x^{2}} .$
Use step 5 to show that $d Y = d y + \frac{x}{\sqrt{L^{2} - x^{2}}} d x .$
Use Green’s theorem to show that $\oint_{C} \frac{x}{\sqrt{L^{2} - x^{2}}} d x = 0.$
Use step 7 to show that the total wheel roll is
Total wheel roll $= \frac{1}{L} \oint_{C} x d y .$
It took a bit of work, but this equation says that the variable of integration Y in step 3 can be replaced with y .
Use Green’s theorem to show that the area of D is $\oint_{C} x d y .$ The logic is similar to the logic used to show that the area of $D = \frac{1}{2} \oint_{C} − y d x + x d y .$
Conclude that the area of D equals the length of the tracer arm multiplied by the total rolling distance of the wheel.
You now know how a planimeter works and you have used Green’s theorem to justify that it works. To calculate the area of a planar region D , use a planimeter to trace the boundary of the region. The area of the region is the length of the tracer arm multiplied by the distance the wheel rolled.

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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