Mathematics 162                              Laboratory 14                        Week of May 3, 1993
	  
Name: _____________________________   Lab Partner: ___________________________
	  
Consulted with:  ____________________________________________________________

© Lafayette College, 1994

Area in Polar Coordinates

In today's laboratory, you will evaluate some areas in polar coordinates.  Before we begin the lab, enter the following command to load some special commands we will use:

<<Graphics`Graphics`; <<Graphics`Colors`

Part 1: Areas in Polar Coordinates

In each of the following, polar equations describe a plane region R.  Set up (and write down) a definite integral to compute the area of R and then let Mathematica evaluate the integral for you.  Write your answers at the end.

Example: R is the region inside the cardioid r = 1 – sinq.

In order to compute the area in R, first we plot the region using PolarPlot:

PolarPlot[1 - Sin[t], {t,0,2Pi}]

Looking at the plot, we see that the plot is symmetric about the y-axis, so we will attempt to find the area to the right of the axis.  A good guess for the limits of integration are: –π/2 ≤ q ≤ π/2. To check these we can use PolarPlot again, using these limits:

PolarPlot[1 - Sin[t], {t,-Pi/2,Pi/2}]

This yields the desired region, so we can ask Mathematica to perform the integration.

2 Integrate[(1 - Sin[t])^2/2, {t,-Pi/2, Pi/2}]

Here, the multiple of 2 in front takes into account the symmetry, and we have used the formula for area in polar coordinates:

To find a numerical estimate of this area we can use the command

N[%]

as long as we enter this immediately after the command above; it numerically evaluates the last output Mathematica has given.

Now you find the areas for the following regions:

(a) R is the region inside the graph of r = 1 – sinq and outside the graph of r = 1.

Hint #1:  The command below will plot both curves simultaneously.

PolarPlot[{1 - Sin[t], 1}, {t, 0, 2Pi},
    PlotStyle->{Magenta, Coral}]

Hint #2: You can use the command below to find one of the points where the curves intersect, if you have any doubt:

Solve[1 - Sin[t] == 1,t]

(b) R is the region inside the cardioid r = 1 – sinq and outside r = – sinq.

(c) R is the region which lies inside the graph of r = – sinq and outside the graph of r = cos2q.

Hint: The Solve command won't work this time, how can you find the points of intersection?

(d) R is the shaded region in the figure below.  In the figure, R is bounded by graphs with the equations given.

plot1 = PolarPlot[Sqrt[4Cos[2t]],{t,-Pi/4,Pi/4}];

plot2 = PolarPlot[Sqrt[4Cos[2t]],{t,3Pi/4+0.000001,5Pi/4}];

plot3 = PolarPlot[Sqrt[2],{t,0,2Pi}];

Show[Graphics[{
Text["r = Sqrt2",{1.25,1.25}],
Text["r = Sqrt(4cos2t)",{2.75,0.5}]}],
Graphics[{GrayLevel[0.5],Polygon[
Join[Table[{Sqrt[4Cos[2t]]Cos[t],Sqrt[4Cos[2t]]Sin[t]},
{t,-Pi/6,Pi/6,Pi/96}],Table[{Sqrt[2]Cos[t],Sqrt[2]Sin[t]},
{t,Pi/6,-Pi/6,-Pi/60}]]]}],
plot1,plot2,plot3,Axes->True,AspectRatio->Automatic,
PlotRange->{{-2.1,3.6},All}];