Roerto s Notes on Integrl Clculus Chpter 5: Bsic pplictions of integrtion Section 10 Geometric quntities for polr curves Wht you need to know lredy: How to use integrls to compute res nd lengths of regions ounded y regulr nd prmetric curves Wht you cn lern here: How to use integrls to compute the sme quntities for region ounded y one or more polr curves In the pplictions we hve seen so fr, we hve used slices tht were verticl or horizontl Notice tht this mkes full use of the Crtesin coordintes we normlly use But wht if we re given informtion in polr coordintes? Polr coordintes use rdiclly different pproch to identify Therefore it should not surprise you tht the method needed to compute res of regions ounded y polr curves is sustntil vrition of those we hve used so fr To develop such method, we need to use, once gin, the four step process to construct n integrl, ut in novel wy Since in polr coordintes the independent vrile trces the polr curve rdilly, y rotting round the pole counter clockwise, let us ssume tht the region of interest is ounded y polr curve r r nd two rdii t nd, nd s shown here Let us now slice this region like piece of pie centered t the pole, through severl more rdii In this wy we cn pproximte the re of ech slice y using circulr sector whose ngle is smll nd whose rdius is the distnce from the pole to the curve t some point within the slice r r r r r r Notice tht ech pproximting slice looks like tringle, ut it isn t! Since we work in polr coordintes, we hve to rotte round the pole, so tht wht you re seeing is circulr sector Integrl Clculus Chpter 5: Bsic pplictions of integrtion Section 10: Geometric quntities for polr curves Pge 1
We know (or should know!) tht the re of circulr sector of rdius r nd 1 ngle is given y r, so we use this formul to pproximte the corresponding re ounded y the curve: 1 Ai r i By dding the res of ll pproximting sectors we conclude tht: n 1 A r i i1 As we did in the rectngulr cse, we cn now go from this pproximtion to n exct vlue y tking the limit s the slicing is done into thinner nd more numerous slices This leds to the following method to e ddressed on cse-y-cse The exmples nd prctice questions provide some illustrtion of how this cn e done nd how to hndle these difficulties Exmple: r 31 cos The grph of this polr curve shows n outer loop nd n inner loop: If Technicl fct r r is polr curve defined y function nd ounds, then the re of this ounded region is tht is continuous for rdilly-defined region etween the hlf-lines nd given y: 1 A r d The more generl cse of region ounded y two polr curves, or y two different sections of the sme polr curve, cn e tckled s we hve done in other situtions, y descriing the region s the difference etween regions of the ove, sic type But tht is where is the key difficulty lies: identifying the curves, the limits of integrtion nd the potentil overlps in n pproprite wy These difficulties need To find the re ounded y the inner loop, we strt y noticing tht the curve crosses the pole when: 1 4 31 cos 0 cos, 3 3 This mens tht the loop strts t the first vlue nd ends t the second Since this is exctly the simple sitution, the re is given y: 4 / 3 4 / 3 1 9 A 91 cos d 1 4cos 4cos d / 3 / 3 We cn compute this integrl in the usul wy for trig integrls, thus otining: 9 4 A sin 4sin /3 /3 9 3 3 3 3 4 4 4 4 4 4 4 9 3 6 489 4 Integrl Clculus Chpter 5: Bsic pplictions of integrtion Section 10: Geometric quntities for polr curves Pge
Exmple: r 31 cos Wht out the re of the region etween the outer nd inner loops of the sme polr curve? All we need to do is find the re ounded y the outer loop nd sutrct the re of the inner loop from it Let us compute the re of the region inside oth curves For the outer loop, we cn use symmetry, thus looking t twice the re of the top prt, or simply follow the rottion We see tht such loop strts t nd ends t 3 loop is: Aouter 3 1 Therefore the re ounded y the outer /3 /3 9 1cos d This cn e computed s efore nd then we need to sutrct the inner loop, leding to: Aouter Aouter Ainner /3 4 /3 1 1 91 cos d 91 cos d /3 /3 This region is symmetric, so we cn consider only the portion ove the x- xis Exmple: 1 cos, r cos r By grphing these curves we cn see severl regions Integrl Clculus Chpter 5: Bsic pplictions of integrtion Section 10: Geometric quntities for polr curves Pge 3
Moreover, y rotting counter clockwise, we notice tht this region consists of two smller regions: Technicl fct r1 cos r cos If r r, is finite polr curve, then its length is given y: dr L r d d In the first qudrnt, for 0, we use only the second curve, while in the second qudrnt, for, we use only the first curve Therefore the desired formul is: / 1 1 A cos d cos d 0 / / cos d cos d 0 / This is sic integrl whose computtion I leve to you Tht looks confusing! I m frid to sk wht hppens when we tckle lengths, surfce res nd volumes Well, cheer up: lthough surfce res nd volumes cn e tckled when deling with polr curves, they led to very messy formule tht re eyond the scope of our curse I ment course! So, we re left with rc length, which turns out to e surprisingly simple Proof We strt y writing the polr curve in prmetric form: x r cos x ' r 'cos r sin y r sin y ' r 'sin r cos Now we use the formul for the prmetric rc length: ' ' L x y d 'cos sin 'sin cos r r r r d By expnding the two squres, the two doule products cncel ech other nd the remining terms cn e grouped s follows: sin cos ' cos sin L r r d But wht we hve in rckets now is one side of the sic Pythgoren identity, whose other side is 1 Therefore, we cn write the formul s: Integrl Clculus Chpter 5: Bsic pplictions of integrtion Section 10: Geometric quntities for polr curves Pge 4
s climed ' L r r d L / 0 sin 4cos d You my wnt to try nd compute this integrl Once gin, we get tough one, ut despir not: method for hndling these difficult definite integrls is coming Exmple: r sin Just s with this exmple, other exmples tend to e eqully chllenging, so the issue will e only to set up the integrl, which oils down to identifying the limits of integrtion The length of one petl of this rose is given y: Summry The formul for the re of region ounded y polr curve is otined y using the four step process to construct integrls nd is sed on the re of circulr sector, rther thn rectngle The min difficulty in setting up the required integrls is in identifying the polr region, or sometimes regions, involved The formul for rc length of polr curve is otined lgericlly from the prmetric version Generl formule for surfce res nd volumes tend to e complicted enough to e deferred to lter course Common errors to void Don t rush over the identifiction of the polr region: it is the key element nd it cn e tricky! Rememer tht you re deling with polr curves, so don t trce the curve left to right, ut counter clockwise Integrl Clculus Chpter 5: Bsic pplictions of integrtion Section 10: Geometric quntities for polr curves Pge 5
Lerning questions for Section I 5-10 Review questions: 1 Explin how to set up the integrl representing the re of region ounded y polr curves Explin how to set up the integrl representing the length of polr curve Memory questions: 1 Wht geometricl shpes re used to construct the re formul in polr coordintes? Wht is the formul for the re ounded y polr curve of the form r f ( ),? 3 Wht is the integrl formul for the rc length of the grph of polr function r r? 4 When we nlyze polr region to identify the integrtion limits, how do we scn the region? Computtion questions: In questions 1-16, set up the integrl tht provides the re of the region descried there If possile, compute the integrl 1 The region ounded y the first loop of the polr curve polr xis r / e nd the The region elow the polr xis nd contined etween the two loops of the curve r 6sin 5 The region enclosed y one loop of the curve r 1 sin3 6 The region common to the circles r sin nd r cos 7 The figure-8 region ounded y oth r 1 cos nd r 1 cos 3 The region ounded y y x nd r 3 sin 8 The finite region ounded y the loop in the conchoid r 4 sec 9 The region of intersection etween the two circles r sin nd r 1 4 The region inside the circle r 6cos nd outside the crdioid r cos 10 The finite region tht is inside oth the rose r sin 4 nd the circle r 1 Integrl Clculus Chpter 5: Bsic pplictions of integrtion Section 10: Geometric quntities for polr curves Pge 6
11 The region inside oth r sin nd r 3 sin 14 One petl of the rose r 4cos3 1 The i determined y the polr curve r 1 sin 15 The region inside r 6cos nd outside r cos 13 The region ounded y the outermost lyer of the cochleoid shown here: r sin 16 The region outside r 6cos nd inside r cos nd the region common to them In questions 17-1,set up the integrl tht provides the length of the polr curve descried If possile, compute the integrl 17 The first loop of the spirl r / e 0 The limçon r 1 sin 18 One loop of the rose r cos3 1 r 3 sin 19 The rit ers r sin( cos ) Which function of n represents the length of the first n loops of the spirl r? I expect you to evlute ny integrls tht my e needed for the construction of such function Theory questions: 1 In the formul for the re ounded y polr curve, is it importnt tht the limits e used in incresing order? Is it possile for two polr regions to intersect t point tht does NOT corresponds to the sme vlue of for oth curves? Integrl Clculus Chpter 5: Bsic pplictions of integrtion Section 10: Geometric quntities for polr curves Pge 7
3 Given the polr rc length formul, wht would e the formul for the surfce re otined y rotting polr curve round the x xis? 5 Which formul is used to simplify the expression in the formul for the rc length of polr curve? 4 In the formul for the re ounded y polr curve, which vlue goes in the ottom of the integrl nd which on top? Wht questions do you hve for your instructor? Integrl Clculus Chpter 5: Bsic pplictions of integrtion Section 10: Geometric quntities for polr curves Pge 8