Application: Volume. 6.1 Overture. Cylinders


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1 Applictio: Volume 61 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 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 61) 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 61: 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 (61)
2 mth 131 pplictio: volume, prt ii 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 crosssectiol 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 (61) 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 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 crosssectios re erly ideticl Sice we erly hve cylider, V i (re of the bse) height But the re of the bse is just the crosssectiol re of the slice d the height is relly the width x of the slice, so V i (re of crosssectio of ith slice) x
3 mth 131 pplictio: volume, prt ii 3 If we let A(x i ) deote the crosssectiol 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 The pproximtio is improved by lettig the umber of slices get lrge d the tkig the limit I other words, Volume of Lof = lim A(x i ) x = A(x) dx, where we hve used the fct tht if the crosssectiol re is cotiuous fuctio, the the limit of the Riem sums exists d is defiite itegrl More precisely, we hve proved THEOREM 61 (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 crosssectiol re perpediculr to the xxis 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 yxis 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 61 A crystl prism is 20 cm log (figure o the left below) Its crosssectios re right trigles whose heights re formed from the lie y = 2 1 x d whose bses re twice the height Fid the volume of the prism SOLUTION The crosssectios re right trigles whose heights re 1 2 x d the bse is twice the height So the crosssectiol re is A(x) = 1 2 bh = 1 ( ) x x = 1 4 x2 20 Figure 65: A prism with represettive right trigle crosssectio Usig Theorem 61 the volume of the prism is 20 1 V = A(x) dx = 0 4 x2 dx = x3 0 = EXAMPLE 62 Fid the volume of the Gret Pyrmid of Cheops which hs squre bse 750 ft o ech edge d height of 500 ft cc
4 mth 131 pplictio: volume, prt ii 4 A(y) (0, 0) 750 edge y (0, 0) 500 Figure 66: Left: The Gret Pyrmid of Cheops upside dow Crosssectios perpediculr to the yxis re squres Right: The reltio betwee the height of the crosssectio d its edge legth SOLUTION I order to simplify the mthemtics, it will be useful to drw the pyrmid upsidedow (see Figure 66) The crosssectios 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 crosssectio is squre, Therefore, by Theorem 61 = edge = 3y 2 A(y) = (edge) 2 = ( ) 3y 2 = 9y2 2 4 d 500 9y V = A(y) dy = 2 c 0 4 dy = y3 = 93, 750, 000 cu ft 0 YOU TRY IT 61 Whe I ws i Tsmi i 1998, I bought beutiful wedgeshped woode doorstop It is 15 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 150 cu cm?) 15 2 Figure 67: Left: The wedge doorstop for you try it 61 Right: The prism for you try it 62 YOU TRY IT 62 A crystl prism is 2 cm log Its crosssectios 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 61 to prove tht the volume formul for coe of height h d rdius r is V = 1 3 πr2 h Hit: Drw the coe verticlly with its vertex t the origi Determie the equtio of the lie tht forms the righthd edge of the coe Use tht lier equtio to determie the rdius of the circulr crosssectios of the coe YOU TRY IT 64 A crystl prism is 2 cm log Its crosssectios 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 Figure 69: Left: The prism for you try it 64 Right: The prism for you try it 65
5 mth 131 pplictio: volume, prt ii 5 YOU TRY IT 65 A crystl prism is 4 cm log Its crosssectios 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 300 meters Its height is 150 meters Fid its volume Hit: Review the Pyrmid of Cheops problem The equtios re simpler if you tur the prymid upside dow d use crosssectios perpediculr to the yxis (Aswer: 4,500,000 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 40 cm itervls over the rge from 0 to 200 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 = 1 + [ f (x)] 2 x, k=1 where f (x) is cotiuous fuctio Express lim S s itegrl We will use this i clss i couple of dys
s = 1 2 at2 + v 0 t + s 0
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