ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In the figure bove, we know tht = r cos(θ), = r sin(θ) r = +, θ = tn (/). θ is bit trick, tn (/) is in quotes to indicte tht it is not perfectl well-defined. We will emine this in more detil below. We cll (r, θ) the polr coordinte representtion of (, ). The following emple shows is more thn one w to represent n point in polr coordintes. Emple 3.. In the tble below, the (, ) coordintes re t the top of ech column nd vrious possible r,θ representtions re below. (, ) (, ) (, ) (, ) (, ) (, ) (, ) (, ) (r, θ) (, ) (, π/) (, ) (, π/4) (, 3π/4) (, 5π/4) (, π/) (r, θ) (, π) (, 9π/4) (, 3π/4) (, 7.) (r, θ) (, 4π) 3. Double integrls in polr coordintes Recll tht double integrl f(, ) da is sum. Here da is the re of tin region R round the point (, ). In rectngulr coordintes we put tin rectngle t (, ). To compute double integrls using polr coordintes we will put polr rectngle t (, ).
3 POLAR COORDINATES AND DOUBLE INTEGRALS (r + r, θ + θ) r θ θ A r θ (r + r, θ) r r (r, θ) Strting t the point with polr coordintes (r, θ) we form curv rectngle b chnging r b r nd θ b θ. Tht is, with vertices (r, θ), (r +, θ), (r, θ + θ), (r + r, θ + θ). The curv sides re smll rcs of circles nd the stright sides re rdil segments. If r nd θ re ver smll, the curv rectngle is pproimtel rectngle with sides r nd r θ. So, it s re is A r θ r. In the limit we get n ect formul Are in polr coordintes = da = r dr dθ. Emple 3.. The circulr sector R shown below hs ngle nd rdius. Assume its densit is δ(, ) =. Use integrtion with polr coordintes to find its mss. r = nswer: As double integrl the mss is R δ(, ) da. In polr coordintes: δ(, ) = = r cos(θ) sin(θ). To find the limits of integrtion we cover R with rdil lines.: Inner limits (fi θ): < r <, i.e. the rdil lines run from the origin to the circle r =. Outer limits: < θ <, i.e. we hve rdil line for ech θ in this rnge. Thus the mss is M = Inner integrl: Outer integrl: M = R δ(, ) da = θ= r= r 3 cos(θ) sin(θ) dr = r4 cos(θ) sin(θ) 4 Emple 3.3. Compute I = r cos(θ) sin(θ) r dr dθ. 4 cos(θ) sin(θ) dθ = sin θ = 3. nswer: Step is to drw the region. = 4 cos(θ) sin(θ) ( + d d b switching to polr coordintes. ) 3/ The integrl s given hs limits: inner limits:, outer limits:.
3 POLAR COORDINATES AND DOUBLE INTEGRALS 3 In rectngulr coordintes this mens we ve covered the region with verticl lines. This is shown in the figure on the left. = θ = π/4 = = r = sec(θ) r = sec(θ) Region using verticl lines (left) nd rdil lines (right) In polr coordintes, the verticl line = cn be written r cos(θ) =. Tht is, r = sec(θ). So covering the region with rdil lines gives limits (see bove figure on right) : Inner limits: r from sec(θ) to sec(θ); Integrl = π/4 sec(θ) Inner integrl: 4. sec(θ) sec(θ) sec(θ) r dr dθ. r3 r 3 r dr = cos(θ). outer limits: θ from to π/4. π/4 Outer integrl: cos(θ) dθ = sin(θ) Emple 3.4. Find the volume of the region bove the -plne nd below the grph of z =. nswer: As lws, we strt with picture; z π/4 = R As double integrl the volume is z da, where R is the unit disk in the plne below R the grph nd bove R. Since R is the disk nd, in polr coordintes, z = r we will compute this integrl in polr coordintes. For R we hve the following limits. Inner limits: r ; outer limits θ π. volume = π ( r ) rdr dθ.
3 POLAR COORDINATES AND DOUBLE INTEGRALS 4 Inner integrl: Outer integrl: V = ( r ) rdr = 4 = 4. π 4 dθ = π. 3.3 Grphing with polr coordintes We ll eplin wht it mens to grph function r = f(θ) with n emple. Emple 3.5. (Crdiod.) Plot the grph of r = + cos(θ). nswer: This mens to plot ever point (r, θ) in polr coordintes, where r = + cos(θ). To get sense we mke tble of r for vrious θ between π nd π. Since cos(θ) is periodic there is no need to look outside this rnge. Also, since cos( θ) = cos(θ) we get the sme vlue of r for ±θ. θ ±π/6 ±π/4 ± ±π/ ± ±3π/4 ±π r + 3/ + / 3/ / / In the figure, the lbels t the end of rdil segments indicte the polr ngle θ of the segment. Of course, the length of the segment is r. π/ 5π/ π/4 π/6 π/ π/ π/ 5π/ π/4 π/6 This figure is clled crdioid In words, t θ =, r hs its mimum vlue of. As θ increses to π, r decreses to. You cn see this in the top hlf of the crdioid, i.e. s the ngle increses the rdil segment gets shorter. The smmetr mens the bottom hlf (for negtive θ) is the mirror imge of the top. 3.4 Gller of polr functions Lines nd circles cn be written described in polr coordintes. We give formuls in polr coordintes for four of these. In the formuls we use specific constnts, the generl cse should be cler. We lso give brief derivtion of ech formul. Circle of rdius centered t the origin: r =. Derivtion: this is the definition of such circle.
3 POLAR COORDINATES AND DOUBLE INTEGRALS 5 R t ngle from the origin: θ =. Derivtion: see the figure below. Verticl line = : r = sec(θ). Derivtion: = r cos(θ) = r = sec(θ). Horizontl line = : r = csc(θ). Derivtion: = r sin(θ) = r = csc(θ). Here re plots of ech of the bove. r θ = Circle r =. line = ; r = sec(θ) line = ; r = csc(θ) 3.4. Circles through the origin We ll stte the conclusions up front nd then derive them. The grph of r = cos(θ) for π/ θ π/ is circle of rdius nd center (, ). Tht is, it is circle through the origin with center on the -is. Likewise, the grph of r = sin(θ) for θ π is circle of rdius nd center (, ). Tht is, it is circle through the origin with center on the -is. Circle r = cos(θ) π/ θ π/. Circle r = sin(θ). θ π. Let s prove this for r = cos(θ). We cn convert this eqution to rectngulr coordintes b first multipling it b r to get: r = r cos(θ). Then, since r = + nd = r cos(θ) we cn write this in rectngulr coordintes s + =. Now little lgebric mnipultion nd completing the squre gives + = + + = ( ) + =.
3 POLAR COORDINATES AND DOUBLE INTEGRALS 6 This lst formul shows we hve circle of rdius nd center on the -is t (, ), just s climed. We still hve to show tht we cover the entire circle ectl once s θ goes from π/ to π/. To do this, we ll first check the endpoints:. t θ = π/ we hve r = cos( π/) =,. t θ = π/ we hve r = cos(π/) =. So, s θ goes through the rnge from π/ to π/, the grph strts nd ends t the origin. Tht is bsicll enough to show wht we wnt, but let s go through some other vlues of θ. π/4 π/6 π/ π/ π/6 π/4 Circle r = cos(θ) π/ θ π/. In the figure the ngle θ is mrked t the end of ech rdil segment. It illustrtes tht s θ goes from π/ to π/ the point (r, θ) with r = cos(θ) strts t nd goes once round the circle in counterclockwise direction. The picture for r = sin(θ) is similr. Since the circle is bove the -is, the ngle θ runs from to π. 3.4. Plotting negtive vlues of r. When plotting, we cn llow r to be negtive. If the formul produces negtive vlue for r, we just drw the rdil line segment bckwrds. For emple, if r = when θ = π/4, then we plot this t the point (, ) = ( /, ). r =, θ = π/4 θ = π/4 Wrning. While this works for plots, ou should never use it in integrtion. Bd things cn hppen if ou let r be negtive in n integrl. It is better to mke use of smmetr nd onl integrte over regions where r is positive.
3 POLAR COORDINATES AND DOUBLE INTEGRALS 7 Emple 3.6. A limçon is curve with eqution r = ( + b cos(θ)), where b >. Plot the limçon r = + cos(θ). nswer: We strt with tble of vlues. Notice tht t θ =, r = nd for < θ π, r is negtive. As with the crdioid, we know tht the limçon is smmetric cross the is. θ ±π/6 ±π/4 ± ±π/ ± ±3π/4 ±π r 3 + 3 +.4 π/4 π/ π/6 3π/4 3π/4 π π/ π/6 π/4 Limçon, r = + cos(θ) 3.5 More polr plots Here re two more nice plots. We show them without n comment. Four leved rose r = sin(θ). Lemniscte: r = cos(θ) 3.6 Ellipse Let E be the ellipse given b the eqution + =. This is n ellipse centered t nd b oriented with the coordinte es.
3 POLAR COORDINATES AND DOUBLE INTEGRALS 8 b (, ) = (r cos(θ), r sin(θ)) Ellipse + b =. Clim. In polr coordintes, E hs the eqution r = b b cos (θ) + sin (θ). Proof. First, rewrite the formul for E s b + = b. Now, use = r cos (θ) nd = r sin (θ). So, b + = b b r cos (θ)+ r sin (θ) = b r = b b cos (θ) + sin (θ). QED.