Solutions to Problems Integration in IR 2 and IR 3

Transcription

1 Solutions to Problems Integrtion in I nd I. For ec of te following, evlute te given double integrl witout using itertion. Insted, interpret te integrl s, for emple, n re or n verge vlue. ) dd were is te rectngle,. b) D +) dd, were D is te lf disc. c) +) dd were is te rectngle, b. d) b dd were is te rectngle, b. Solution. ) dd is te re of rectngle wit sides of lengts nd 5. So dd 5. b) D dd becuse is odd under reflection bout te is, wile te domin of integrtion is smmetric bout te is. D dd is te tree times te re of lf disc of rdius. So, D +)dd π 6π. c) dd/ dd is te verge vlue of in te rectngle, nmel. Similrl, dd/ dd is te verge vlue of in te rectngle, nmel b. dd is re of te rectngle, nmel b. So, S +)dd b+b). d) b dd V ddd, were V is te region wit b,, b. + b is clinder of rdius b centered on te is. + b,, is one qurter of tis clinder. It s cross sectionl re πb. V is te prt of tis qurter clinder wit. It s lengt nd cross sectionl re πb. So, b dd π b.. For ec of te following, evlute te given double integrl using itertion. ) + )dd were is te rectngle, b. b) T )dd were T is te tringle wit vertices,),,),,b). c) dd were is te finite region in te first qudrnt bounded b te curves nd. d) Dcosdd were D is te finite region in te first qudrnt bounded b te coordinte es nd te curve. e) e dd were is te region,. f) T + dd were T is te tringle wit vertices,),,),,). Solution. Te following figures sow te domins of integrtion for te integrls in tis problem. ),b),b) +b b,) b) c) d),),) e),) f),)

2 ) b) c) d) e) f) T + )dd )dd D dd cosdd T e dd + dd d d b d + ) b ) d ) b b + b ) ] d d d cos) d d cos d e e +e ] d d b+ b) b+b ) d b ) b ) ] b b b b 6 b d / 6) 7 8) 56 e ) d + rctnt ln+t ) ] d sin ) d e d ) + π ln) ] cos ) d )e dt t +t were t. For ec of te following integrls i) sketc te region of integrtion, ii) write n equivlent double integrl wit te order of integrtion reversed nd iii) evlute bot double integrls. e + ) d d b) d d c) d d + Solution. Te following figures sow te domins of integrtion for te integrls in tis problem. e,5) e ln + + ) b) ) e d d e ln d d d d b) e d e ] e ] e d ln ] ln + ] e d c) e ) /] 8 +, ) c) 5 d d d d d d 5 d ] ] 8 d + ] +] 9 d + + ] 6 + +) ] 5 9

3 . Find te volume ling inside te spere + + nd bove te prboloid +. Solution. Te top surfce + + meets te bottom surfce + wen obes +. Tt is, wen. Te oter root of +, nmel, is inconsistent wit +. Te top nd bottom surfces meet t te circle,, +. In polr coordintes, te top surfce is r nd te bottom surfce is r, so Vol In crtesin coordintes dr π r ) / r Vol r r r ] π dr r r r ] ] π + /].6 d d ] Te integrl cn be done using te substitution cost, but it is esier to use polr coordintes. 5. Find te volume of te solid inside te clinder + 8, bove te plne nd below te plne 8. Solution. Looking down from te top, we see te clinder + 8. Tt gives te bse region. Te top of te solid, bove n fied,) in te bse region, is t 8 tis is lws positive becuse never gets bigger tn 8). Te bottom of te solid, below n fied,) in te bse region, is t tis is lws negtive becuse is lws between nd ). So te eigt of te solid t n,) is top bottom 8 ) ). Te volume is 8 d d ) 8 ecll from first er tt if f) is n odd function mening tt f ) f) for ll ), ten f) d becuse te two integrls f) d nd f) d ve te sme mgnitude but opposite signs). Appling tis twice gives 8 8 d nd 8 d d d 8 8 since nd 8 re bot odd. Tus 8 8 d d ) volume d d 8 8 so tt te volume is just times te re of te ellipse + 8, wic is π 8 ) 8 π

4 6. Use polr coordintes to evlute ec of te following integrls. ) S +)dd were S is te region in te first qudrnt ling inside te disc + nd under te line. b) S dd, were S is te disc segment +,. c) T + )dd were T is te tringle wit vertices,),,) nd,). d) + ln + )dd Solution. ) In polr coordintes te domin of integrtion, +,, becomes Te integrl is r, rsinθ rcosθ or r, θ rctn π S +)dd dr rrcosθ +rsinθ) dr r sinθ cosθ ]π + ] b) In polr coordintes te domin of integrtion, +,, becomes ] 6 + r, rcosθ or cosθ r,) r For cosθ r to be nonempt, we need cosθ or θ π. smmetr under, te integrl is S dd dr rrcosθ) cos θ cosθ r / ] cosθ ] ] π / cosθ cos θ / sinθ tnθ c) In polr coordintes, te tringle wit vertices,),,) nd,) s sides θ, θ π nd r cosθ wic is te polr coordintes version of ). Te integrl is T + )dd cosθ r t+ t dr rr ) cos θ sec θ +tn θ ) ] cos θ sec θ dt +t ) were t tnθ

5 d) ln + )dd + dr rlnr π ] π slns s π dr rlnr π ds lns were s r 7. Find te volume ling inside te clinder + ) nd between te upper nd lower lves of te cone +. Solution. For tis region nd run over te interior of te clinder + ). For ec,) inside te clinder, runs from + to +. As + ) if nd onl if +, te clinder s eqution r rsinθ, or equivlentl, r sinθ, in polr coordintes. Tus r,θ) runs over θ π, r sinθ nd for ec r,θ) in tis region runs from r to r. smmetr under, te volume is V sinθ t t sinθ) ] dr r r r) ] 6 9 sinθ dr r sinθ cos θ) 8. Find te volume of te region in te first octnt below te prboloid b Solution. Te prboloid its te plne t + b. r sinθ dt t ) were t cosθ Volume b b d d ) b d dv v ) were bv Tink of tis integrl s being of te form b d g) wit g) dv v ). Volume b b u du dv u v ) u +v u,v dudv u v ) were u Now switc to polr coordintes u rcosθ, v rsinθ. Volume b dr r r ) ] r b r π 8 b 5

6 9. Find te volume common to te clinders + nd. Use polr coordintes. Solution. Te figure below sows te top view of te specified solid.,) runs over te interior of te circle +. For ec fied,) in tis disk, runs from to +. In polr coordintes, te circle is r rcosθ or r cosθ. Te solid is smmetric under nd, so we cn restrict to, nd multipl b. Te volume is Volume cosθ dr r rcosθ cosθ 5 r5/ cosθ cosθ ) t t dt t ) ] 8 5 r cosθ cosθ sin θ ) were t sinθ. A smmetricl coffee percoltor olds cups wen full. Te interior s circulr cross-section wic tpers from rdius of t te centre to t te bse nd top, wic re prt. Te bounding surfce is prbolic. Were sould te mrk indicting te 6 cup level be plced? Solution. Let r) be te rdius of te urn t eigt bove its middle. ecuse te bounding surfce of te urn is prbolic, r) must be qudrtic function of tt vries between t nd t ±6. Te function r) 6) does te job. Slice te urn into orontl slices, wit te slice t eigt disk of rdius r) nd tickness d nd ence of volume πr) d. Te volume to eigt is V) 6 We wis to coose so tt π d πr) 6 d π ] 6 π ] π ] ] 6 6 or ] ] Since ].95.6 nd , tere is solution.9.9 to two deciml plces). Te mrk sould be bout.5 bove te bottom. 6 6

7 . Evlute dv were is te tetredron bounded b te coordinte plnes nd te plne + b + c. Solution. Te domin of integrtion is,,, + b + c. In tis region, tkes ll vlues between nd c. For ec fied c,,) tkes ll vlues stisfing,, + b c. Tis is pictured in te figure on te rigt below.,,c),b c )),,),b,) b c ),) dv c c bc b d d b 6 c ) d b c b c ) d ) ] b c ) c c b d d b 6 c ) c d b ) c ) bc ] c c). Evlute dv were is te portion of te cube,, ling bove te plne + nd below te plne + +. Solution. Te domin of integrtion is,,,,. In te figure on te left below, te more drkl sded region is prt of nd te more ligtl sded region is prt of. Te figure on te rigt sows section wit constnt. + dv d d d d d du u)+u ) du u +u) d d ) were u d + ) 7

8 . For ec of te following, epress te given iterted integrl s n iterted integrl in wic te integrtions re performed in te order: first, ten, ten. ) d d d f,,) b) d d d f,,) Solution. ) Te domin of integrtion is,, or equivlentl,,,, +, +. In tis region, tkes ll vlues between nd. For ec fied,,) tkes ll vlues stisfing,,, +. Tis is pictured in te figure on te rigt below. In te figure on te left, te front fce is + nd te ligtl sded fce rigt is +., ) In te new order, te integrl is,,) d d d f,,)+ d d d f,,) b) Te domin of integrtion is,, or equivlentl,,. In tis region, tkes ll vlues between nd. For ec fied,,) tkes ll vlues stisfing,. In te new order, te integrl is d d d f,,). Use clindricl coordintes to evlute te volumes of ec of te following regions. ) Above te plne, inside te cone + nd inside te clinder +. b) Above te plne, under te prboloid nd in te wedge. c) Above te prboloid + nd below te plne. Solution.) Te figures below sow te prts of te cone, te clinder nd te intersection, respectivel, tt re in te first octnt. r r r r sinθ r sinθ r 8

9 smmetr under, te full volume is twice te volume in te first octnt. Vol sinθ sinθ dr r r dr r r) d sin θ 8 sin θ ] 8 π dt t )] were t cosθ 8 π ) ] ) π 9 b) Te boundries of te wedge correspond, in polr coordintes, to θ tn ) π nd θ tn π. In clindricl coordintes, te prboloid becomes r. Tis prboloid intesects te plne on te circle r. Vol π r dr r d π + π ) dr r r ) 7 π ) 7 8 π c) Te region is +. In clindricl coordintes, te bottom is r nd te top is rsinθ. Te top nd bottom intersect on te circle r sinθ or + ). smmetr under, te full volume is twice te volume in te first octnt. Vol sinθ rsinθ dr r d r ) sin θ 6 π π sinθ dr rrsinθ r ) 5. Te centre of mss,ȳ, ) of bod ving densit ρ,,) units of mss per unit volume) t,,) is defined to be were M ρ,,) dv ȳ M M ρ,,) dv M ρ,,) dv ρ,,) dv is te mss of te bod. So, for emple, is te weigted verge of over te bod. Find te centre of mss of te prt of te solid bll + + wit, nd, ssuming tt te densit ρ is constnt. Solution. smmetr, ȳ, so it suffices to compute, for emple,. Te mss of te bod is te densit, ρ, times its volume, wic is one eigt of te volume of spere. So M ρ 8 π 9

10 In clindricl coordintes, te eqution of te spericl surfce of te bod is r +. Te prt of te bod t eigt bove te plne is one qurter of disk of rdius. Te numertor of is ρ dv ρ ρ d d / / dr r ρ ) π ρ ] π 6 ρ Dividing b M π 6 ρ gives ȳ 8. d / r π ρ d ) 6. Evlute te volume of circulr clinder of rdius nd eigt b mens of n integrl in spericl coordintes. Solution. cosϕ ϕ ϕ ϕ sinϕ ϕ tn ϕ < tn ϕ > tn Te top of te clinder s eqution i.e. cosϕ. Te side of te clinder s eqution + i.e. sinϕ. Te bottom of te clinder s eqution i.e. ϕ π. For ec fied ϕ, θ runs from to π nd runs from to eiter cosϕ if ϕ < tn ) or sinϕ if ϕ > tn ). Vol tn π π dϕ tn { dt { π /cosϕ dϕ sinϕ cos ϕ +π t ) + } d sinϕ+ dϕ tn tn dϕ sinϕ sin ϕ /sinϕ d sinϕ } ds were t tnϕ, dt sec ϕdϕ, s cotϕ, ds csc ϕdϕ { } π 6 + π 7. Use spericl coordintes to find ) Te volume inside te cone + nd inside te spere + +. b) dv nd dv over te prt of te spere of rdius tt lies in te first octnt. c) Te mss of spericl plnet of rdius wose densit t distnce from te center is ρ A/ + ). d) Te volume enclosed b cosφ). Here nd φ refer to te usul spericl coordintes. Solution. ) Vol d π cosφ ] π π { dϕ sinϕ d }{ ) }{ } dϕ sinϕ

11 b) smmetr, te two integrls re equl. d dϕ sinϕ {}}{ cosϕ π π 8 π 6 dϕ sinϕcosϕ dt t were t sinϕ, dt cosϕdϕ c) mss πa d dϕ sinϕ πa πa d + ) / πa ) tn densit {}}{ { A + πa }{ dϕ sinϕ ds +s were s, d ds d + } d) Te volume in question is invrint under rottions bout te is. Its intersection wit te rigt lf of te plne is given in te figure below. Volume π π dϕ sinϕ cosϕ) dϕ sinϕ cosϕ) dt t π 8 π d were t cosϕ, dt sinϕdϕ 8. A torus of mss M is generted b rotting circle of rdius bout n is in its plne t distnce b from te centre b > ). Te torus s constnt densit. Find te moment of inerti bout te is of rottion. definition te moment of interti is r dm were dm is te mss of n infinitesml piece of te solid nd r is its distnce from te is. Solution. Te torus is constructed b rotting te circle b) + viewed s ling in te plne) bout te is. On tis circle, runs from b to b+. In clindricl coordintes, te torus s eqution r b) +. On tis torus, r runs from b to b+. For ec fied r, runs from r b) to r b). As te torus is smmetric bout te plne, its volume is twice tt of te volume of te prt wit. Volume π b+ b b+ b r b) dr r d dr r r b) ds s+b) s were s r b

12 As s s is odd under s s, ds s s. Also, ds s is precisel te re of te top lf of circle of rdius. So So te mss densit of te torus is M π b moment of inerti Volume bπ ds s π b M so tt dm π b dv M π brdrd nd b+ b b+ M π b b M π b M π b r b) dr r ds s+b) s d M π b r dr r r b) were s r b ds s +s b+sb+b ) s Agin, b oddness, te s nd sb integrls re ero. For te oters, sub in s sint, ds cost. moment M π b M π M π π π costdt) bsin t+b )cost M π dt cos t+b cos t+ cos t) ] π 8 +bπ +π M +b ) π dt sin t+b )cos t