] dx (3) = [15x] 2 0

Transcription

1 Leture 6. Double Integrls nd Volume on etngle Welome to Cl IV!!!! These notes re designed to be redble nd desribe the w I will eplin the mteril in lss. Hopefull the re thorough, but it s good ide to hve other referenes, like stndrd tetbook. I like to strt eh setion with poem. This first one is little lme, but mbe ou will understnd wh I put it in this setion. (It s not just beuse m wife likes ts, but tht m hve tipped the sle.) There one were two ts from Kilkenn. Eh thought tht ws one t too mn, so the strted to fight nd to srth nd to bite. Now, insted of two ts, there ren t n. -Anonmous Students re usull prett good t evluting iterted integrls, whih is wht we will strt with. Itertion is onept tht ppers often in mth, (see leture 4.6 on Newton s Method) nd involves repetition of proess. If ou n integrte one, ou n do it twie. Work from the inside out. Emple. 4 2 d d ( 4 ) 2 d d () [ ] 2 4 d (2) [ 4 2 2] d (3) 5 d (4) [5] 2 (5) 3. (6) A student who hs mstered Clulus I will find the bove emple firl es, one the understnd the itertion, step (). Slightl trikier is when the integrnd is funtion of multiple vribles, so ou tret them s onstnts, eept the vrible of integrtion. Emple 2. ( ) d ] [ () 2 + () 2 ()4 4 (7) + 7() (8) (9)

2 In the previous emple is onstnt, so pretend it s our fvorite onstnt (mine is 8) if ou need help, like this Cl I emple: ( ) d ] [ (This emple ould be in Cl book, but it s slightl weird, onl beuse 8 2 is usull written 64 nd ) Ok, we sort of know how to do double integrls now. But wh do we do them? This wh goes deeper thn ou might suspet. First of ll, wh do we do integrls t ll? You know from Cl I tht integrls give res, but wh does tht work? The tools of mth re onl s powerful s our understnding of them, nd ou should go bk right now nd rered setions 5. through 5.4 (just refresh our memor). emember iemnn sums? emember wh we write b f() d to represent re? (Hints: f( k ), the letter S for Sigm, Sum, wht letter does look like, FTC, et...) Are is 2 dimensionl 2D. We re going to bump it up to 3D. Insted of re we get volume. Insted of retngles we get boes (lso lled retngulr prisms or prllelepipeds, nd I often ll them 3D retngles beuse tht mkes sense to me. The ren t ubes though, eept in the rre se when ll 3 dimensions hve equl length.). Insted of single integrl, we need double integrl. b b b () intervl [, b] (b) urve, grph of f() () retngles pproimte re () region (b) surfe, grph of f(, ) () boes pproimte volume

3 Don t get greed; strt with the bsis nd hve ptiene. You should know how to find the volume of bo (length times width times height). You should know how to pproimte volume under funtion f(, ) over region b dding up volumes of boes. The piture filittes understnding. Emple 3. Approimte the volume under f(, ) on the retngle with verties (, ), (,.5), (, ), nd (,.5) using 6 boes with squre bses of equl re. For heights, use () the vlue of f t the lower left orner of eh squre. (b) the vlue of f t the midpoint of eh squre. Solution: Alws drw piture! I epet ou to drw the 2D piture of nd its prtition, but the 3D ones might be trik to drw it s importnt to visulize them in our hed but I usull won t mrk points off if ou don t drw them. Here ou see the retngulr region in the plne, then in z-spe with the funtion f(, ) bove it. Note tht the grph of f is plne, nd lthough the domin of f is ll points in the -plne it is onl grphed for (, ) in. is split into 6 squres, whih re the bses of the boes..5 () in 2D (b) the grph of f in blue, restrited to

4 () boes hit funtion t lower lefts (in 2D) (b) boes hit funtion t midpoints of squres Just s we n use left-endpoints, right-endpoints, midpoints, et. in iemnn sums, we n hoose different points in eh squre where we will mesure the height of eh bo. The height of the bo is just the funtion s vlue t the hosen point. For prt () the points re (,), (.5,), (,), (,.5), (.5,.5) nd (,.5). Plot them in the 2D drwing bove if ou don t understnd where the me from! The funtion vlues re f(, ), f(.5, ).5, f(, )., f(,.5).25, f(.5,.5).3, nd f(,.5).35. The re of eh squre is A (.5)(.5).25. Therefore, the sum of volumes of the boes is f(, ) A ( )(.25) Note tht f(, ) is not ver ler nottion; I onl wnt ou to ompre these emples, nd the nottion, to the single vrible emples in hpter 5. For prt (b) the points re (.25,.25), (.75,.25), (.25,.25), (.25,.75), (.75,.75), nd (.25,.75). The pproimte volume should be f(, ) A ( )(.25) The height of the plne given b f is inresing in both the positive nd diretions. Therefore, the points we used in prt () give lower sum for the volume under f, whih is n pproimtion less thn the tul volume. When the boes heights re ll greter thn or equl to the funtion s vlues, we get n upper sum. In prt (b), the midpoints of the squres lnded us prtl bove nd prtl below the surfe. (Solution end) As in Cl, using finer prtition gives more urte pproimtion. Infinitel mn boes with infinitel smll bses! We wnt both nd to shrink to, nd informll we n think

5 of this s A. Like the single vrible se in hpter 5, we get n integrl when we tke limit of sums. The 3D equivlent of definite integrl is f(, ) A f(, ) da Volume under f on. lim A In Cl, we proved the Fundmentl Theorem of Clulus, whih reltes res nd ntiderivtives. A 3D version is lled Fubini s Theorem, whih proves ou n find volumes with integrls. Fubini s Theorem on retngulr region Let be the retngle in the -plne with b nd d. Then for n integrble funtion f(, ), f(, ) da d b f(, ) d d b d f(, ) d d, where f(, ) da is the net volume (whih n be negtive) under the grph of f on. We won t see mn funtions in this lss tht re not integrble, so I ll point it out if we do. Students tend to fous on the order-of-integrtion-swithing prt of Fubini s Theorem, but the importnt prt is tht it proves volume f(, ) da n be lulted with iterted integrls. The theorem is nmed fter Fubini beuse he ontributed to desription of wht it mens to be integrble, but he ws not the first person to figure out tht volumes n be lulted with iterted integrls. To get glimpse of the proof, think of sliing the solid S ( solid hs volume, like region hs re) infinitesimll thinl. If ou slie prllel to the -is, eh slie is solid so thin tht it is lmost 2D, nd its min fes (the sides ou would put butter on if it were bred) hve lmost identil shpe nd re. (See figure (b) below.) The tops of slies of S re urves on the grph of f. The re of slie s fe is A() b f(, ) d. The re A() is funtion of, whih hnges, rnging from to d, depending on where ou slie it. The thikness of the slie is, whih will beome the differentil d in the limiting proess. The volume of slie is re times thikness, i.e., A(). Adding up ll the slies volumes gives the totl volume. In the limiting proess, the sum turns into n integrl. We derive the et totl volume s n integrl: A() ( b Thus, ) f(, ) d f(, ) da d ( b d b ) f(, ) d d f(, ) d d. d b f(, ) d d.

6 Sine there is no reson tht sliing prllel to the -is is better thn sliing prllel to the -is, we n repet the sme proof tking slies of re A() d f(, ) d nd thikness. The sme rgument ields f(, ) da whih ompletes the frmed result bove. b d f(, ) d d, () The solid S. (b) Vol. of S A() () Vol. of S A() Emple 4. Find the volume under f(, ) on the retngle with verties (, ), (,.5), (, ), nd (,.5). Note tht this is the sme volume we pproimted erlier in this setion with 6 boes. Fubini s Theorem will give us the et volume, nd the solid is shown below. The volume is f(, ) da ( ) d d [ ] d [( ) ( + + )] d (.25 +.) d [ ].5 (.25)(.5) + (.5)(.5)

7 The volume of the solid is.9875, whih is mirulousl the sme s we got with our midpoint pproimtion using 6 boes. Wow! (but tht doesn t usull hppen think bout wh it did nd wh it might not.) Another mirle (whih lws hppens for integrble funtions) is tht we n reverse the order of integrtion nd get the sme nswer s well: f(, ) da.5 ( ) d d [ ].5 d (.5 +.5(.5) 2 +.5(.5)()) d ( ) d ] [ (.75) (.75)/ Double wow!