Green's theorem and the Planimeter

Oliver Knill and Dale Winter, Harvard University

Part 1. Iterated Integrals and Line Integrals

The graph of the function (x,y) -> Sin[x y]  can be plotted as follows:

Plot3D[
        Sin[x y], {x, -Pi, Pi}, {y, -Pi, Pi}, PlotPoints -> 50, 
        Boxed -> False, Axes -> False];

Problem 1

Determine the  iterated integral of Sin[x y] over that square

Integrate[ Sin[x y],  {x, -Pi, Pi}, {y, -Pi, Pi}]

Mathematica was able to compute this integral exactly. For a numerical integration, we would replaced Integrate[] with NIntegrate:

NIntegrate[ 
      Sin[x y],  {x, -Pi, Pi}, {y, -Pi, Pi}]

Did Mathematica have any difficulty finding the numerical value? What do you attribute this so, if yes?

Problem 2

Enter here your prediction for the value of the integral of f(x,y)=r where r is the length of the vector (x,y):

Now, we plot this function and calculate the double integral with Mathematica:

f[x_, y_] := Sqrt[x^2 + y^2];
Plot3D[f[x, y], {x, -1, 1}, {y, -1, 1}, 
        AspectRatio -> 1];

Problem 3

Compute the double integral.

NIntegrate[
      Sqrt[x^2 + y^2], {x, -1, 1}, {y, -Sqrt[1 - x^2], 
        Sqrt[1 - x^2]}]

Problem 4

Set up and evaluate numerically the double integral of  f(x,y) = log(|1+x y|) on the region G, which is enclosed by the ellipse of width 3 and height 4.

Problem 5

As you can verify, the curve

g[t_] := 
        If[0 ≤ t < 1 , {t, 0}, 
          If[1 ≤ t < 2, {1, t - 1}, 
            If[2 ≤ t < 3, {3 - t, 1}, {0, 4 - t}]]];

is a square:

ParametricPlot[g[t], {t, 0, 4}, Frame -> False, 
      AspectRatio -> 1]

Given a vector field (P(x,y),Q(x,y))=(x^2,y^2), we can evaluate the line integral:

g[t_] := 
        If[0 ≤ t < 1 , {t, 0}, 
          If[1 ≤ t < 2, {1, t - 1}, 
            If[2 ≤ t < 3, {3 - t, 1}, {0, 
                4 - t}]]];
F[{x_, y_}] := {x^2, y^2};
u[t_] := D[g[s], s] /. s -> t;
NIntegrate[F[g[t]] . u[t], {t, 0, 4}]

Problem 6

Express the line integral as a sum of one-dimensional definite integrals.

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Problem 7

Set up and evauate the line integral of the vector field F(x,y)=(P(x,y),Q(x,y))=(x y,x) if C is the boundary of the unit circle centered at (1,1)

Problem 8

Set up and evaluate the iterated integrals and line integrals, in the following two cases:
    Case 1:  The curve/boundary of the region is the unit square

g1[t_] := 
        If[0 ≤ t < 1 , {t, 0}, 
          If[1 ≤ t < 2, {1, t - 1}, 
            If[2 ≤ t < 3, {3 - t, 1}, {0, 4 - t}]]];

Case 2: The curve/boundary of the region is the unit circle:

g2[t_] := {1 + Cos[t], 1 + Sin[t]};

and where the vector fields are from the following three cases:

F1[{x_, y_}] := {x^2*y^2, 0};
F2[{x_, y_}] := {0, Sin[x*y]};
F3[{x_, y_}] := {x^2*y^2, Sin[x*y]};

Summarize your results here.

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Now integrate the following functions over each of the two regions (Case 1 and Case 2):

f1[x_, y_] := -2*x*y^2;
f2[x_, y_] := y Cos[x*y];
f3[x_, y_] := y Cos[x*y] - 2*x*y^2;

Summarize your results here

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Problem 9

What relationships did you notice?

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State your conjecture on the relationship between the values of the iterated integrals and the line integrals.

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Part 2. A Link between Line Integrals and Iterated Integrals

Problem 1

The curves and vector fields to check the theorem: we start with a circle

g[t_] := {Cos[t], Sin[t]};
F[{x_, y_}] := {Sqrt[x^2 + y^2], Sqrt[x^2 + y^2]};

Next, the curve is the polar curve r^2=9 cos[theta]:

g[t_] := {3*Sqrt[Abs[Cos[t]]]* Cos[t], 
          3*Sqrt[Abs[Cos[t]]]*Sin[t]};
ParametricPlot[g[t], {t, 0, 2*Pi}, Frame -> False, 
        Axes -> False];
F[{x_, y_}] := {Cos[y], Sin[y]};

Next, the curve is a triangle:

g[t_] := 
        If[0 ≤ t < 1 , {1 + t, 1 + t}, 
          If[1 ≤ t < 2, {2, 2 + (t - 1)}, {2 - (t - 2), 
              3 - 2 (t - 2)}]];
ParametricPlot[g[t], {t, 0, 3}, Frame -> False, 
        Axes -> False];
F[{x_, y_}] := {Sqrt[x^2 + y^2], Sqrt[x^2 + y^2]};

Finally the curve is a polar curve r=1+cos(t):

g[
          t_] := {(1 + Cos[t])*Cos[t], (1 + Cos[t])*
            Sin[t]};
ParametricPlot[g[t], {t, 0, 2  Pi}, Frame -> False, 
        Axes -> False];
F[{x_, y_}] := {Sqrt[x^2 + y^2], Sqrt[x^2 + y^2]};

Problem 2

Parameterize the triangle C with vertices (0,0),(2,0),(0,2) and form the line integral with the vector field F(x,y)=(cos(x),sin(y)) as well as converting the line integral into 
a double integral and evaluating it.

Problem 3

Calculate the iterated integral of f(x,y)=(x-y) over the annular region (as a difference of two itereated integrals over disks).

Calculate the lineintegral over the boundaryof the annular region (as a sum of two line integrals over circles). Note the orientation of the paths! When walking along the path, the region
has to be at your "left".

Problem 4

List some potential uses for the theorem

Problem 5

Set up procedures so that you can evaluate line integrals and double integrals over a rectangular region [a,b] x [c,d] with a vector field F(x,y)=(P(x,y),0).

Problem 6

Set up procedures so that you can evaluate line integrals and double integrals over a rectangular region [a,b] x [c,d] with a vector field F(x,y)=(0,Q(x,y)).

Problem 7

Do you expect the relationships to work for curves other than rectangles and squares? Explain.

Part 3: Calculating Areas in the Plane

Problem 1

For each of the three vector fields

F1[{x_, y_}] := {-y, 0};

F2[{x_, y_}] := {0, x};

F3[{x_, y_}] := {-y, x}/2;

calculate the line integral over the each of the following three curves.

g1[t_] := {Cos[t], Sin[t]};
ParametricPlot[g1[t], {t, 0, 2  Pi}, Frame -> False, 
        Axes -> False];

g2[t_] := 
        If[0 ≤ t < 1 , {t, 0}, 
          If[1 ≤ t < 2, {1, t - 1}, 
            If[2 ≤ t < 3, {3 - t, 1}, {0, 
                4 - t}]]];
ParametricPlot[g2[t], {t, 0, 4}, Frame -> False, 
        Axes -> False];

g3[t_] := 
        If[0 ≤ t < 1, {-1 + 2*t, 0}, 
          If[1 ≤ t < 
              2, {1 - (t - 1), (t - 1)}, {-(t - 2), 
              1 + (2 - t)}]];
ParametricPlot[g3[t], {t, 0, 3}, Frame -> False, 
        Axes -> False];

You can find an example how to compute a line integral in the boxed Input Cell of Part 2. NOTE: don't forget to put the correct limits of integration.
List your results from evaluating the integrals here.

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Problem 2

What similarities do you see in the values of the integarls that you have calculated? How can you account for these similarities?

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Problem 3

What do these line integrals appear to be actually calculating? Summarize your findings by stating a conjecture. Test your conjecture by defining several curves and evaluating the line integrals on those curves.

Problem 4

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Problem 5

Use your conjecture to find the area enclosed by the curve x^2/a^2+y^2/b^2=1 using a line integral. Supply a mathematical argument to verify the correctness of your conjecture.

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Part 4: Experiments with the Planimeter

Problem 1

What is the ratio for the area of the circle over the area of the square ?

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Problem 2

List the measurements that you made using the planimter applet for the circle and the square.

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Problem 3

What is the ratio of these measurements.

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Problem 4

What is your equation for converting the output of the planimeter? What does your equation convert the output of the planimeter into?

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Problem 5

List the planimeter results for the triangle and ellipse. Do these validate the conversion equation you have found?

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Part 5: The Mathematics of the Planimeter

Problem 1

Which of the following three vector fields represents the Planimeter vector field?

<< Graphics`PlotField`
a[x_, 
          y_] := (x^3 + x*y^2 - 
              Sqrt[4*x^2*y^2 + y^6])/(2*(x^2 + y^2));
b[x_, 
          y_] := (-x^2*y + y^3 + 
              Sqrt[4*x^4 - x^6 + x^2*y^4])/(2*(x^2 + 
                  y^2));
PlotVectorField[
      If[x < -0.1 || y < -0.1, {0, 0}, {-(y^2 - b[x, y]), 
          x^2 - a[x, y]}], {x, 0.1, 1}, {y, 0.1, 1}]

<< Graphics`PlotField`
a[x_, 
          y_] := (x^3 + x*y^2 + 
              Sqrt[4*x^2*y^2 - x^4*y^2 + 4*y^4 - 2*x^2*y^4 - 
                  y^6])/(2*(x^2 + y^2));
b[x_, 
          y_] := (x^2*y + y^3 - 
              Sqrt[4*x^4 - x^6 + 4*x^2*y^2 - 2*x^4*y^2 - 
                  x^2*y^4])/(2*(x^2 + 
                  y^2));
PlotVectorField[
      If[x < -0.1 || y < -0.1, {0, 0}, {-(y - b[x, y]), 
          x - a[x, y]}], {x, 0.1, 1}, {y, 0.1, 1}]

<< Graphics`PlotField`
a[x_, 
          y_] := (x^3 + x*y^2 + 
              Sqrt[4*x^2*y^2 + y^6])/(2*(x^2 + y^2));
b[x_, 
          y_] := (x^2*y + y^3 + 
              Sqrt[4*x^4 - x^6 + x^2*y^4])/(2*(x^2 + 
                  y^2));
PlotVectorField[
      If[x < -0.1 || y < -0.1, {0, 0}, {-(y^2 - b[x, y]), 
          x^2 - a[x, y]}], {x, 0.1, 1}, {y, 0.1, 1}]

Problem2

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Problem3

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Problem4

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Problem 5

The elbow coordinates (a,b) are obtained from  (x,y) by solving a system of quadratic equations:

Solve[{(x - a)^2 + (y - b)^2 == 1, 
        a^2 + b^2 == 1}, {a, b}]

Problem 6

Consider the quantity that you plan to integrate in the double integral that you found to be equal to the line integral.

a[x_, 
        y_] := (x^3 + x*y^2 + 
            Sqrt[4*x^2*y^2 - x^4*y^2 + 4*y^4 - 2*x^2*y^4 - 
                y^6])/(2*(x^2 + y^2))
b[x_, 
        y_] := (x^2*y + y^3 - 
            Sqrt[4*x^4 - x^6 + 4*x^2*y^2 - 2*x^4*y^2 - 
                x^2*y^4])/(2*(x^2 + y^2))

Compute the curl of the planimeter vector field mens computing the curl of (x,y) -> (-(y-b),(x-a)) which is 2-(a_x+b_y). We compute now a_x+b_y:

D[a[x, y], x] + D[b[x, y], y]

Simplify[%]

Can you help Mathematica to see what the result is?

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Problem 7

What property does the Planimeter calculate?

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Problem 8

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Part 6: Summary

Problem 1

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Problem 2

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Problem 3

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Problem 4

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Problem 5

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6/5/2001, Oliver Knill and Dale Winter, Harvard University, Copyright CCP and the Authors