Application: Volume. 6.1 Overture. Cylinders

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1 Applictio: Volume 6 Overture I this chpter we preset other pplictio of the defiite itegrl, this time to fid volumes of certi solids As importt s this prticulr pplictio is, more importt is to recogize ptter or theme tht will llow us to pply the otio of defiite itegrl to other cotexts For cotiuous fuctios we kow tht lim i= f (x i ) x = f (x) dx Notice tht the Riem sum is ctully sum of products My qutities c be expressed s sum of such products products where the etire qutity hs bee divided ito smller pproximtig pieces (Thik of the rectgles tht pproximte the thi strips of re uder curve which hs bee subdivided by regulr prtitio The re of such pproximtig rectgles is product: b h) Wheever we c pproximte by usig sum of products i this wy, we c compute the etire qutity ot s sum, but s defiite itegrl We explore this powerful ide which I cll subdivide d coquer below Note: The developmet of this mteril is slightly differet th i your text, though the sme results re chieved Cyliders I high school geometry oe is itroduced to shpes kow s cyliders Typiclly we thik of cylider s the shpe of soup c (See Figure 6) A c hs circulr bse which is moved log xis perpediculr to the bse to crete the cylider Where the bse stops movig, the top of the c is formed The volume of cylider is determied by the bse d the legth of the xis perpediculr to the bse More precisely Volume of Cylider = re of the bse height Notice tht this is product Mthemticis tret the otio of cylider more geerlly by llowig the bse to be y fiite ple regio Tke y ple regio B d move it fixed distce h log xis perpediculr to the bse B The resultig solid tht is swept out by this motio is cylider See Figure 62 The volume of y cylider is still the product Figure 6: A circulr cylider is determied by its circulr bse d perpediculr xis (Digrm from wikipedi org/wiki/cylider_(geometry)) Volume of Cylider = re of the bse height (6)

2 mth 3 pplictio: volumes by disks: volume prt i 2 bse height Figure 62: Left: A ple regio Right: Movig the ple regio log xis perpediculr to the regio produces cylider The dshed figure represets cross-sectiol slice perpediculr to the xis Notice tht crdbord crto lso stisfies this more geerl otio of cylider Its bse is the bottom of the crto which we c thik of movig verticlly distce equl to the height of the box to form rectgulr cylider The volume of the crto is (re of the bse) height, where the bse is rectgle The rectgle re is l w, so the volume of the box is is the fmilir formul re of the bse height = l w h Obviously computig the volumes of cyliders (icludig boxes) is esy usig the formul i (6) How do we use this formul i more geerl settigs to obti itegrl? Figure 63: A rectgulr box c be thought of s cylider determied by its rectgulr bse d perpediculr xis A Lof of Bred Cosider ice crusty lof of rtis bred How might we determie its volume? Let s plce the lof o xis suppose the lof lies betwee d b s show i Figure 64 Figure 64: Left: A lof of bred cut ito slices Right: The ith slice is lmost cylider Slice the lof ito equl slices, ech of width x Let V i deote the volume of the ith slice The the volume of the lof is the sum of the volume of ll the slices ( subdivide d coquer ) Volume of Lof = Volume of Slice i = V i i= i= How do we determie the volume of the slice? Whe we extrct the ith slice from the lof (see Figure 64), we see tht it is lmost the shpe of cylider its two fces or cross-sectios re erly ideticl Sice we erly hve cylider, V i (re of the bse) height But the re of the bse is just the cross-sectiol re of the slice d the height is relly the width x of the slice, so V i (re of cross-sectio of ith slice) x Volumetex Versio: Mitchell-25/9/29:46:34

3 mth 3 pplictio: volumes by disks: volume prt i 3 If we let A(x i ) deote the cross-sectiol re of the ith slice ((see Figure 64), the V i A(x i ) x So the volume of the etire lof is pproximted by Volume of Lof = V i A(x i ) x i= i= The pproximtio is improved by lettig the umber of slices get lrge d the tkig the limit I other words, Volume of Lof = lim i= A(x i ) x = A(x) dx, where we hve used the fct tht if the cross-sectiol re is cotiuous fuctio, the the limit of the Riem sums exists d is defiite itegrl More precisely, we hve proved THEOREM 6 (Volume Formul) Let V be the volume of solid tht lies betwee x = d x = b If for ech x i the itervl [, b] the cross-sectiol re perpediculr to the x-xis is give by the cotiuous fuctio A(x), the the volume V is the solid is V = A(x) dx Note: If the slices re tke perpediculr to the y-xis o the itervl [c, d] d the crosssectiol re is A(y), the d V = A(y) dy c Stop! Notice tht we used the subdivide d coquer process to pproximte the qutity we wish to determie Tht is, we subdivided the volume slicig it ito pproximtig cyliders whose volume we kow how to compute We refied this pproximtio by lettig the umber of slices get lrge Tkig the limit of this process swered our questio Idetifyig tht limit with itegrl mkes it possible to esily (!) compute the volume i questio OK, time for some exmples Exmples EXAMPLE 62 A crystl prism is 2 cm log (figure o the left below) Its cross-sectios re right trigles whose heights re formed from the lie y = 2 x d whose bses re twice the height Fid the volume of the prism Solutio The cross-sectios re right trigles whose heights re 2 x d the bse is twice the height So the cross-sectiol re is A(x) = 2 bh = ( ) 2 2 x x = 4 x2 2 Figure 65: A prism with represettive right trigle cross-sectio Usig Theorem 6 the volume of the prism is V = A(x) dx = 2 4 x2 dx = 2 2 x3 = 2 3 cc Volumetex Versio: Mitchell-25/9/29:46:34

4 mth 3 pplictio: volumes by disks: volume prt i 4 EXAMPLE 63 Fid the volume of the Gret Pyrmid of Cheops which hs squre bse 75 ft o ech edge d height of 5 ft Solutio I order to simplify the mthemtics, it will be useful to drw the pyrmid upside-dow (see Figure 66) A(y) (, ) 75 edge y (, ) 5 Figure 66: Left: The Gret Pyrmid of Cheops upside dow Cross-sectios perpediculr to the y-xis re squres Right: The reltio betwee the height of the cross-sectio d its edge legth The cross-sectios re squres with re A(y) We eed to determie the re of the squre t height y, so we eed to fid the legth of the edge of the squre t height y To do this we c use similr trigles, see the right hlf of Figure 66 We hve edge y Sice the cross-sectio is squre, Therefore, by Theorem 6 V = d c A(y) dy = = 75 5 edge = 3y 2 A(y) = (edge) 2 = 5 9y 2 4 dy = 3 4 y3 ( ) 3y 2 = 9y = 93, 75, cu ft YOU TRY IT 6 Whe I ws i Tsmi i 998, I bought beutiful wedge-shped woode doorstop It is 5 cm log d 5 cm high t its tll ed d 4 cm wide See Figure 67 below Fid the volume of this wedge usig clculus Hit: Fid the equtio of the lie tht forms oe of the top edges (Why should the swer be 5 cu cm?) 5 2 Figure 67: Left: The wedge doorstop for you try it 6 Right: The prism for you try it 62 YOU TRY IT 62 A crystl prism is 2 cm log Its cross-sectios re squres with heights formed by the curve y = x 2 See Figure 67 bove Fid the volume of the prism YOU TRY IT 63 Use Theorem 6 to prove tht the volume formul for coe of height h d rdius r is V = 3 πr2 h Hit: Drw the coe verticlly with its vertex t the origi Determie the equtio of the lie tht forms the right-hd edge of the coe Use tht lier equtio to determie the rdius of the circulr cross-sectios of the coe YOU TRY IT 64 A crystl prism is 2 cm log Its cross-sectios re isosceles right trigles The heights re formed by the curve y = x 2 See Figure 69 below Fid the volume of the prism h Figure 68: Determie the volume of coe of rdius r d height h r Volumetex Versio: Mitchell-25/9/29:46:34

5 mth 3 pplictio: volumes by disks: volume prt i YOU TRY IT 65 A crystl prism is 4 cm log Its cross-sectios re right trigles The heights re formed by the curve y = 2 x d the bses by the curve y = x 2 See Figure 69 bove Fid the volume of the prism YOU TRY IT 66 (The Gret Pyrmid of Geev) The pyrmid t the Pyrmid Mll i Geev t the crest of Be s Hill hs squre bse with edges tht mesure 3 meters Its height is 5 meters Fid its volume Hit: Review the Pyrmid of Cheops problem The equtios re simpler if you tur the prymid upside dow d use cross-sectios perpediculr to the y-xis (Aswer: 4,5, cu m) YOU TRY IT 67 A field biologist is doig survey of smll wooded forest She is iterested i fidig the volume of tree truks from the forest floor to poit 2 meters bove the groud Sice she cot mesure the volume directly, she uses pir of tree clipers to mesure the rdius of the tree t 4 cm itervls over the rge from to 2 cetimeters She brigs the dt to you (see tble below) d sks you to provide relible estimte o the volume of the tree truk i cubic cetimeters How c you do so usig Riem sums? Wht estimte should you use to get resobly good pproximtio? Expli your resoig Height (h) Rdius (r) YOU TRY IT 68 (Theory) Suppose we form regulr prtitio of the itervl [, b] d crete the Riem sum: S = + [ f (x)] 2 x, k= where f (x) is cotiuous fuctio Express lim S s itegrl We will use this i clss i couple of dys Figure 69: Left: The prism for you try it 64 Right: The prism for you try it 65 Volumetex Versio: Mitchell-25/9/29:46:34

6 mth 3 pplictio: volumes by disks: volume prt i 6 62 Volumes of Revolutio: The Disk Method Oe of the simplest pplictios of itegrtio (Theorem 6) d the ccumultio process is to determie so-clled volumes of revolutio I this sectio we will cocetrte o method kow s the disk method Solids of Revolutio If regio i the ple is revolved bout lie i the sme ple, the resultig object is solid of revolutio, d the lie is clled the xis of revolutio The followig situtio is typicl of the problems we will ecouter Solids of Revolutio from Ares Uder Curves Suppose tht y = f (x) is cotiuous (o-egtive) fuctio o the itervl [, b] Rotte the regio uder the f betwee x = d x = b roud the the x-xis d determie the volume of the resultig solid of revolutio See Figure 6 f (x i ) y = f (x) f (x i ) x i b x i b Oce we kow the cross-sectiol res of the solid, we c use Theorem 6 to determie the volume But s Figure 6 shows, whe the poit (x i, f (x i )) o the curve is rotted bout the x-xis, it forms circulr cross-sectio of rdius R = f (x i ) Therefore, the cross-sectiol re t x i is Figure 6: Left: The regio uder the cotiuous curve y = f (x) o the itervl [, b] Right: The solid geerted by rottig the regio bout the x-xis Note: The poit (x i, f (x i )) o the curve trces out circulr crosssectio of rdius r = f (x i ) whe rotted A(x i ) = πr 2 = π[ f (x i )] 2 Sice f is cotiuous, so is π[ f (x)] 2 d cosequetly Theorem 6 pplies Volume of Solid of Revolutio = A(x) dx = π[ f (x)] 2 dx Of course, we could use this sme process if we rotted the regio bout the y-xis d itegrted log the y-xis We gther these results together d stte them s theorem THEOREM 62 (The Disk Method) If V is the volume of the solid of revolutio determied by rottig the cotiuous fuctio f (x) o the itervl [, b] bout the x-xis, the V = π [ f (x)] 2 dx (62) If V is the volume of the solid of revolutio determied by rottig the cotiuous fuctio f (y) o the itervl [c, d] bout the y-xis, the d V = π [ f (y)] 2 dy (63) c Aother Developmet of the Disk Method Usig Riem Sums Isted of usig Theorem 6, we could obti Theorem 62 directly by usig the subdivide d coquer strtegy oce gi Sice we will use this strtegy i lter situtios, let s quickly go through the rgumet here There re ofte severl wys to prove result i mthemtics I hope oe of these two will resote with you Volumetex Versio: Mitchell-25/9/29:46:34

7 mth 3 pplictio: volumes by disks: volume prt i 7 As bove, we strt with cotiuous fuctio o [, b] This time, though, we crete regulr prtitio of [, b] usig itervls d drw the correspodig pproximtig rectgles of equl width x I left hlf of Figure 6 we hve drw sigle represettive pproximtig rectgle o the ith subitervl f (x i ) x b y = f (x) x b f (x i ) y = f (x) Rottig ech represettive rectgle cretes represettive disk (cylider) of rdius R = f (x i ) (See the right hlf of Figure 6) The volume of this cylider is give by (6) Volume of Cylider = (re of the bse) height I this cse whe the disk is situted o its side, we thik of the height s the width x of the disk Moreover, sice the bse is circle, its re is πr 2 = π[ f (x i )] 2 so Volume of represettive disk = V i = π[ f (x i )] 2 x To determie the volume of etire solid of revolutio, we tke ech pproximtig rectgle, form the correspodig disks (see the middle pel of Figure 62) d sum the resultig volumes, it geertes represettive disk whose volume is V = πr 2 x = π[r(x i )] 2 x Figure 6: Left: The regio uder the cotiuous curve y = f (x) o the itervl [, b] d represettive rectgle Right: The disk (cylider) of rdius R = f (x i ) geerted by rottig the represettive rectgle bout the x-xis The volume or this disk is πr 2 w = π[ f (x i )] 2 x Figure 62: A geerl solid of revolutio d its pproximtio by series of disks (Digrm from Lrso & Edwrds) Approximtig the volume of the etire solid by such disks (see the righthd pel of Figure 62) of width x d rdius f (x i ) produces Riem sum Volume of Revolutio i= π[ f (x i )] 2 x = π i= [ f (x i )] 2 x (64) As usul, to improve the pproximtio we let the umber of subdivisios d tke limit Recll from our erlier work with Riem sums, this limit exists Volumetex Versio: Mitchell-25/9/29:46:34

8 mth 3 pplictio: volumes by disks: volume prt i 8 becuse [ f (x)] 2 is cotiuous o [, b] sice f (x) is cotiuous there Volume of Revolutio = lim π i= [ f (x i )] 2 x = π [ f (x)] 2 dx (65) where we hve used the fct tht the limit of Riem sum is defiite itegrl This is the sme result we obtied i Theorem 62 Ee could use this sme process if we rotted the regio bout the y-xis d itegrted log the y-xis Stop! Notice how we used the subdivide d coquer process to pproximte the qutity we wish to determie Tht is we hve subdivided the volume ito pproximtig disks whose volume we kow how to compute We hve the refied this pproximtio by usig fier d fier subdivisios Tkig the limit of this process provides the swer to our questio Idetifyig tht limit with itegrl mkes it possible to esily (!) compute the volume i questio OK, time for some exmples I ll dmit it is hrd to drw figures like Figure 62 However, drwig represettive rectgle for the regio i questio, s i the left hlf of Figure 6 is usully sufficiet to set up the required volume itegrl Exmples Let s strt with couple of esy oes EXAMPLE 622 Let y = f (x) = x 2 o the itervl [, ] Rotte the regio betwee the curve d the x-xis roud the x-xis d fid the volume of the resultig solid Solutio Usig Theorem 62 Wow, tht ws esy! V = π [ f (x)] 2 dx = π [x 2 ] 2 dx = πx5 5 = π 5 EXAMPLE 623 Let y = f (x) = x 2 o the itervl [, ] Rotte the regio betwee the curve d the y-xis roud the y-xis d fid the volume of the resultig solid Solutio The regio is ot the sme oe s i Exmple 622 It lies betwee the y-xis d the curve, ot the x-xis See Figure 64 Sice the rottio is bout the y-xis, we eed to solve for x s fuctio of y Sice y = x 2, the x = y Notice tht the regio lies over the itervl [, ] o the y-xis ow Usig Theorem 62 d V = π [g(y)] 2 dy = π [ y] 2 dy = πy2 c 2 = π 2 EXAMPLE 624 Fid the volume of sphere of rdius r which c be obtied by rottig the semi-circle f (x) = r 2 x 2 bout the x-xis f (x) = x 2 f (x) = x 2 Figure 63: Left: A represettive rectgle for the curve y = x 2 Right: A represettive circulr slice for the curve y = x 2 rotted bout the x-xis f (x) = x 2 x = y Figure 64: Left: A represettive rectgle for the regio betwee the curve y = x 2 (x = y) d the y-xis Right: A represettive circulr slice for the curve x = y rotted bout the x-xis Volumetex Versio: Mitchell-25/9/29:46:34

9 mth 3 pplictio: volumes by disks: volume prt i 9 r r Figure 65: Left: A represettive rectgle for the curve y = r 2 x 2 Right: A represettive circulr slice for the sphere tht results whe rottig the semi-circle bout the x-xis Solutio Usig Theorem 62 r V = π [ f (x)] 2 dx = π [ r 2 x 2 ] 2 dx r r = π r 2 x 2 dx r ) = π (r 2 x x3 r 3 r [( ) ( )] = π r 3 r3 r 3 + r3 3 3 = 4πr3 3 Amzig! We hve derived the volume formul of sphere from the volume by disks formul Volumetex Versio: Mitchell-25/9/29:46:34