Integration in polar coordinatesWorksheet by Mike May, S.J.- maymk@slu.edurestart:
<Text-field layout="Heading 2256" style="Heading 2256">A review of plotting in polar coordinates:</Text-field>The first problem in trying to do double integrals in polar coordinates is to be able to sketch graphs in of functions described in polar coordinates. On your calculators you switch to polar mode. With Maple you use the coord=polar option on the plot command. You can either plot the curve either with r as a function of theta, or with both r and theta described as functions of a parameter t. When we plot in polar coordinates it is probably wise to use the scaling=CONSTRAINED option so the axes have the same scale.plot(cos(2*theta),theta=0..2*Pi, coords=polar, scaling=CONSTRAINED); plot([2+sin(2*t),Pi*sin(t),t=0..Pi], coords=polar, scaling=CONSTRAINED);To plot several curves together at the same time we use set notation, just like we did in Cartesian coordinates.plot({1,2*sin(theta)}, theta=0..2*Pi, coords=polar, scaling=CONSTRAINED);Plotting several curves together lets up plot regions made of a number of curves. The region plotted below is the portion of an annulus between two specified angles.plot({[1+t,Pi/6,t=0..1],[1+t,Pi/3,t=0..1],[1,t,t=Pi/6..Pi/3], [2,t,t=Pi/6..Pi/3],[0,0,t=0..0]},coords=polar, scaling=CONSTRAINED);Exercise:1) Plot the curves NiMvJSJyRyomIiIkIiIiLSUkY29zRzYjJSZ0aGV0YUdGJw== and NiMvJSJyRywmIiIiRiYtJSRjb3NHNiMlJnRoZXRhR0Ym. Find the points of intersection. (You may want to use either your calculator or a piece of paper to find the points of intersection.)
Once we can sketch curves, the problems involved in setting up an integral in polar coordinates are similar to the problems involved in setting up a double integral in Cartesian coordinates. The biggest problem is finding the correct limits of integration. We will also be concerned with switching the order of integration.
<Text-field layout="Heading 2" style="Heading 2"><Font bold="true" foreground="[0,0,0]" italic="false" size="14" style="_cstyle256" underline="false">Finding the limits of integration: </Font> </Text-field>
<Text-field layout="Heading 3" style="Heading 3">Setting up drd<Equation input-equation="theta" style="2D Comment">NiMlJnRoZXRhRw==</Equation> integrals</Text-field>Consider first the case of integrals using drdNiMlJnRoZXRhRw==. Since dr is on the inside we have a region bounded by curves NiMvJSJyRy0lImdHNiMlJnRoZXRhRw== and NiMvJSJyRy0lImhHNiMlJnRoZXRhRw== with the value of NiMlJnRoZXRhRw== bound by two constants. The integration with respect to r for a particular NiMlJnRoZXRhRw== integrates along a radial line. The following block of code is designed to help you visualize what limits of integration mean.r:='r': theta := 'theta': lowtheta := Pi/6; hightheta := 3*Pi/2; lowr := theta -> 1.5-sin(theta); highr := theta -> 3 + cos(theta); print(`Region of integration for `, Int(Int(f(r, theta)*r,r=lowr(theta)..highr(theta)), theta=lowtheta..hightheta)); inside := plot([lowr(theta), theta,theta=lowtheta..hightheta], color=red, coords=polar, thickness=2): outside := plot([highr(theta), theta,theta=lowtheta..hightheta], color=green, coords=polar, thickness=2): line := {}: for i from 0 to 10 do tval := evalf(lowtheta + i/10*(hightheta-lowtheta)): line := line union {[r,tval, r=evalf(lowr(tval))..evalf(highr(tval))]}; od: plotlines := plot(line,coords=polar, color=blue) : plots[display]({inside, outside, plotlines},scaling=CONSTRAINED);Note that since we integrate with respect to r first, the r-limits are functions of theta while the theta limits are constants. WE integrate by first integrating, from the inside (red) curve to the outside (green) curve, along the radial lines.Exercises:2) Find the limits of integration to integrate over the region inside the curve NiMvJSJyRywmIiIiRiYtJSRjb3NHNiMlJnRoZXRhR0Ym and outside the curve r=1. Modify the code above to show that you have the correct region.3) Find the limits of integration to integrate over the region inside the curve NiMvJSJyRyomIiIkIiIiLSUkY29zRzYjJSZ0aGV0YUdGJw== and outside the curve NiMvJSJyRywmIiIiRiYtJSRjb3NHNiMlJnRoZXRhR0Ym. Modify the code above to show that you have the correct region.
<Text-field layout="Heading 3" style="Heading 3">Setting up d<Equation input-equation="theta" style="2D Comment">NiMlJnRoZXRhRw==</Equation>dr integrals</Text-field>Similarly we can set up integrals using drdNiMlJnRoZXRhRw==. Now the region of integration is bounded by curves NiMvJSZ0aGV0YUctJSJnRzYjJSJyRw== and NiMvJSZ0aGV0YUctJSJoRzYjJSJyRw== with the value of r being bound by two constants. Instead of integrating first on radial lines, we start by integrating along circular arcs with a fixed value of r. r:='r': theta := 'theta': lowr := 1; highr := 5; lowtheta := r -> Pi*r/6; hightheta := r -> Pi*(2-r/12); print(`Region of integration for `, Int(Int(f(r, theta)*r, theta=lowtheta(r)..hightheta(r)),r=lowr..highr)); lowthetacurve := plot ([r,lowtheta(r), r=lowr..highr], color=red, coords=polar, thickness=2) : highthetacurve := plot ([r, hightheta(r), r=lowr..highr], color=green, coords=polar, thickness=2) : arcs := {} : for i from 0 to 10 do tempr := evalf(lowr + i/10*(highr-lowr)): arcs := arcs union {[tempr, theta, theta=lowtheta(tempr)..hightheta(tempr)]}: od: grapharcs := plot(arcs,coords=polar, color=blue) : plots[display] ( {lowthetacurve, highthetacurve, grapharcs } ,scaling=CONSTRAINED) ;Exercises:4) Find the limits of integration to integrate over the region inside both curves NiMvJSJyRywmIiIiRiYtJSRjb3NHNiMlJnRoZXRhR0Ym and r=1. Modify the code above to show that you have the correct region. Explain why you want to use dthetadr rather than drdtheta for this problem.5) Find the limits of integration to integrate over the region inside the curve NiMvJSJyRyomIiIjIiIiLSUkc2luRzYjJSZ0aGV0YUdGJw== and outsidethe curve NiMvJSJyRyomIiIiRiYiIiMhIiI=. Modify the code above to show that you have the correct region.
<Text-field layout="Heading 3" style="Heading 3">Changing order of integration</Text-field>Some regions can be described to use either drdNiMlJnRoZXRhRw== or dNiMlJnRoZXRhRw==dr. When we switch the order of integration we need to switch the limits as well.Exercise:6) The region inside the curve r=1 and outside the curve NiMvJSJyRywmIiIiRiYtJSRjb3NHNiMlJnRoZXRhR0Ym can be set up in either order. Find the limits of integration both ways. Show that you have the correct region.
<Text-field layout="Heading 2" style="Heading 2"><Font bold="true" foreground="[0,0,0]" italic="false" size="14" style="_cstyle257" underline="false">Integrating over a region in polar coordinates:</Font> </Text-field>Recall that dA is rdrdNiMlJnRoZXRhRw== or r dNiMlJnRoZXRhRw==dr. Thus to find the area of the region bounded by r=1, r=2, NiMvJSZ0aGV0YUcqJiUjUGlHIiIiIiInISIi, and NiMvJSZ0aGV0YUcqJiUjUGlHIiIiIiIkISIi, we evaluate the integral NiMtJSRJbnRHNiQtRiQ2JCUickcvRig7IiIiIiIjLyUmdGhldGFHOyomJSNQaUdGKyIiJyEiIiomRjFGKyIiJEYz.Int(Int(r,r=1..2),theta=Pi/6..Pi/3)=int(int(r,r=1..2),theta=Pi/6..Pi/3);Exercises:7) Find the area inside both curves r=1 and NiMvJSJyRywmIiIiRiYtJSRjb3NHNiMlJnRoZXRhR0Ym.8) Integrate the function NiMtJSRzaW5HNiMqJCUickciIiM= over the disk of radius 2 centered at the origin.
<Text-field layout="Heading 2257" style="Heading 2257">Changing coordinates systems to integrate </Text-field>One of the reasons we want to be able to integrate in polar coordinates is that some integrals work out nicely in one coordinate system and are ugly in another. To change an integral in Cartesian into polar, we need to do several things. First sketch the region with its boundary curves. Then change the formulas of the boundary curves, the function to be integrated and dA into polar form. We are then ready to set up the integral and integrate.Consider the integral NiMtJSRJbnRHNiQtRiQ2JComIiIiRiksKCokJSJhRyIiI0YpKiQlInhHRi1GKSokJSJ5R0YtRikhIiIvRjE7LCQtJSVzcXJ0RzYjLCYqJCUiYkdGLUYpRi5GMkYyRjYvRi87LCRGO0YyRjs=. int(int(1/(a^2+x^2+y^2),y=-sqrt(b^2-x^2)..sqrt(b^2-x^2)),x=-b..b);Depending on the version of Maple you are using, it either chokes on this integral, or gives an anwer involving functions we don't know how to evaluate. However the integral above converts to NiMtJSRJbnRHNiQtRiQ2JComJSJyRyIiIiwmKiQlImFHIiIjRioqJEYpRi5GKiEiIi8lJnRoZXRhRzsiIiEqJkYuRiolI1BpR0YqL0YpO0Y0JSJiRw== which can easily be done by hand using the substitution NiMvJSJ1RyokJSJyRyIiIw==. Maple also has no problem with it.int(int(r/(a^2+r^2),theta=0..2*Pi),r=0..b);Exercise:9) Convert the integral NiMtJSRJbnRHNiQtRiQ2JComJSJ4RyIiIiUieUdGKi9GKTtGKy0lJXNxcnRHNiMsJiIiJUYqKiRGKyIiIyEiIi9GKzsiIiEtRi82I0Y0 to polar form and evaluate.
<Text-field layout="Heading 3" style="Heading 3">Extra credit -</Text-field> At the beginning of this worksheet we plotted a region described by the parametric curve [2+sin(2*t),Pi*sin(t),t=0..Pi]. I am interested in finding the area of the region bounded by this curve to 3 decimal places. Find the area of the region and cleanly write up your work carefully justifying your method.