APPLICATIONS OF DEFINITE INTEGRALS


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1 Chpter 6 APPICATIONS OF DEFINITE INTEGRAS OVERVIEW In Chpter 5 we discovered the connection etween Riemnn sums ssocited with prtition P of the finite closed intervl [, ] nd the process of integrtion. We found tht for continuous function ƒ on [, ], the limit of S P s the norm of the prtition 7P7 pproches zero is the numer S P = n k = ƒsc k d k ƒsd d = Fsd  Fsd where F is n ntiderivtive of ƒ. We pplied this to the prolems of computing the re etween the is nd the grph of = ƒsd for, nd to finding the re etween two curves. In this chpter we etend the pplictions to finding volumes, lengths of plne curves, centers of mss, res of surfces of revolution, work, nd fluid forces ginst plnr wlls. We define ll these s limits of Riemnn sums of continuous functions on closed intervls tht is, s definite integrls which cn e evluted using the Fundmentl Theorem of Clculus. 6. Volumes Slicing nd Rottion Aout n Ais In this section we define volumes of solids whose crosssections re plne regions. A crosssection of solid S is the plne region formed intersecting S with plne (Figure 6.). Suppose we wnt to find the volume of solid S like the one in Figure 6.. We egin etending the definition of clinder from clssicl geometr to clindricl solids with ritrr ses (Figure 6.). If the clindricl solid hs known se re A nd height h, then the volume of the clindricl solid is Volume = re * height = A # h. This eqution forms the sis for defining the volumes of mn solids tht re not clindricl the method of slicing. If the crosssection of the solid S t ech point in the intervl [, ] is region R() of re A(), nd A is continuous function of, we cn define nd clculte the volume of the solid S s definite integrl in the following w. 96
2 6. Volumes Slicing nd Rottion Aout n Ais 97 P Crosssection R() with re A() S FIGURE 6. A crosssection of the solid S formed intersecting S with plne P perpendiculr to the is through the point in the intervl [, ]. A se re h height Plne region whose re we know Clindricl solid sed on region Volume se re height Ah FIGURE 6. The volume of clindricl solid is lws defined to e its se re times its height. S We prtition [, ] into suintervls of width (length) k nd slice the solid, s we would lof of red, plnes perpendiculr to the is t the prtition points = 6 6 Á 6 n =. The plnes P k, perpendiculr to the is t the prtition points, slice S into thin sls (like thin slices of lof of red). A tpicl sl is shown in Figure 6.. We pproimte the sl etween the plne t k  nd the plne t k clindricl solid with se re As k d nd height k = k  k  (Figure 6.). The volume V k of this clindricl solid is As k d # k, which is pproimtel the sme volume s tht of the sl: Volume of the k th sl V k = As k d k. FIGURE 6. solid S. k k A tpicl thin sl in the The volume V of the entire solid S is therefore pproimted the sum of these clindricl volumes, V n k = V k = n This is Riemnn sum for the function A() on [, ]. We epect the pproimtions from these sums to improve s the norm of the prtition of [, ] goes to zero, so we define their limiting definite integrl to e the volume of the solid S. k = As k d k.
3 98 Chpter 6: Applictions of Definite Integrls Plne t k Approimting clinder sed on R( k ) hs height k k k DEFINITION Volume The volume of solid of known integrle crosssectionl re A() from = to = is the integrl of A from to, V = Asd d. k The clinder s se is the region R( k ) with re A( k ) NOT TO SCAE k Plne t k FIGURE 6. The solid thin sl in Figure 6. is pproimted the clindricl solid with se Rs k d hving re As k d nd height k = k  k . This definition pplies whenever A() is continuous, or more generll, when it is integrle. To ppl the formul in the definition to clculte the volume of solid, tke the following steps: Clculting the Volume of Solid. Sketch the solid nd tpicl crosssection.. Find formul for A(), the re of tpicl crosssection.. Find the limits of integrtion.. Integrte A() using the Fundmentl Theorem. Tpicl crosssection (m) FIGURE 6.5 The crosssections of the prmid in Emple re squres. HISTORICA BIOGRAPHY Bonventur Cvlieri (598 67) EXAMPE Volume of Prmid A prmid m high hs squre se tht is m on side. The crosssection of the prmid perpendiculr to the ltitude m down from the verte is squre m on side. Find the volume of the prmid. Solution. A sketch. We drw the prmid with its ltitude long the is nd its verte t the origin nd include tpicl crosssection (Figure 6.5).. A formul for A(). The crosssection t is squre meters on side, so its re is. The limits of integrtion. The squres lie on the plnes from = to =.. Integrte to find the volume. EXAMPE V = Asd d = d = d = 9 m Cvlieri s Principle Asd =. Cvlieri s principle ss tht solids with equl ltitudes nd identicl crosssectionl res t ech height hve the sme volume (Figure 6.6). This follows immeditel from the definition of volume, ecuse the crosssectionl re function A() nd the intervl [, ] re the sme for oth solids.
4 6. Volumes Slicing nd Rottion Aout n Ais 99 Sme volume EXAMPE Volume of Wedge Sme crosssection re t ever level FIGURE 6.6 Cvlieri s Principle: These solids hve the sme volume, which cn e illustrted with stcks of coins (Emple ). A curved wedge is cut from clinder of rdius two plnes. One plne is perpendiculr to the is of the clinder. The second plne crosses the first plne t 5 ngle t the center of the clinder. Find the volume of the wedge. 5 9, 9 FIGURE 6.7 The wedge of Emple, sliced perpendiculr to the is. The crosssections re rectngles. Solution We drw the wedge nd sketch tpicl crosssection perpendiculr to the is (Figure 6.7). The crosssection t is rectngle of re Asd = sheightdswidthd = sda9  B The rectngles run from = to =, so we hve V = Asd d = 9  d = s9  d > d = + s9d> = 8. = 9 . et u = 9 , du =  d, integrte, nd sustitute ck. Solids of Revolution: The Disk Method The solid generted rotting plne region out n is in its plne is clled solid of revolution. To find the volume of solid like the one shown in Figure 6.8, we need onl oserve tht the crosssectionl re A() is the re of disk of rdius R(), the distnce of the plnr region s oundr from the is of revolution. The re is then So the definition of volume gives Asd = psrdiusd = p[rsd]. V = Asd d = p[rsd] d.
5 Chpter 6: Applictions of Definite Integrls R() () Disk () R() FIGURE 6.8 The region () nd solid of revolution () in Emple. This method for clculting the volume of solid of revolution is often clled the disk method ecuse crosssection is circulr disk of rdius R(). EXAMPE A Solid of Revolution (Rottion Aout the Ais) The region etween the curve =,, nd the is is revolved out the is to generte solid. Find its volume. Solution We drw figures showing the region, tpicl rdius, nd the generted solid (Figure 6.8). The volume is EXAMPE 5 The circle Volume of Sphere is rotted out the is to generte sphere. Find its volume. = 8p. Solution We imgine the sphere cut into thin slices plnes perpendiculr to the is (Figure 6.9). The crosssectionl re t tpicl point etween  nd is Therefore, the volume is V = Asd d = ps  d d = p c  d =  p.  V = p[rsd] d = pc D d = p Asd = p = ps  d.  d = p d = p sd + = (, ) Rsd = A() ( ) FIGURE 6.9 The sphere generted rotting the circle + = out the is. The rdius is Rsd = =  (Emple 5).
6 6. Volumes Slicing nd Rottion Aout n Ais The is of revolution in the net emple is not the is, ut the rule for clculting the volume is the sme: Integrte psrdiusd etween pproprite limits. EXAMPE 6 A Solid of Revolution (Rottion Aout the ine = ) Find the volume of the solid generted revolving the region ounded = nd the lines =, = out the line =. Solution We drw figures showing the region, tpicl rdius, nd the generted solid (Figure 6.). The volume is V = p[rsd] d = pc  D d = p C  + D d = p c  # > + d = 7p 6. R() R() (, ) () () (, ) FIGURE 6. The region () nd solid of revolution () in Emple 6. To find the volume of solid generted revolving region etween the is nd curve = Rs d, c d, out the is, we use the sme method with replced. In this cse, the circulr crosssection is As d = p[rdius] = p[rs d]. EXAMPE 7 Rottion Aout the Ais Find the volume of the solid generted revolving the region etween the is nd the curve = >,, out the is.
7 Chpter 6: Applictions of Definite Integrls () R(), R() Solution We drw figures showing the region, tpicl rdius, nd the generted solid (Figure 6.). The volume is EXAMPE 8 V = p[rs d] d = = p = p. p d d = p c d = p c d Rottion Aout Verticl Ais Find the volume of the solid generted revolving the region etween the prol = + nd the line = out the line =. Solution We drw figures showing the region, tpicl rdius, nd the generted solid (Figure 6.). Note tht the crosssections re perpendiculr to the line =. The volume is V = p[rs d] d  = p[  ] d  Rsd =  s + d =  FIGURE 6. The region () nd prt of the solid of revolution () in Emple 7. = p [  + ] d  = p c d  = 6p. 5 R() ( ) R() (, ) 5 5 (, ) () () FIGURE 6. The region () nd solid of revolution () in Emple 8.
8 6. Volumes Slicing nd Rottion Aout n Ais (, R()) (, r()) R() r() Wsher FIGURE 6. The crosssections of the solid of revolution generted here re wshers, not disks, so the integrl Asd d leds to slightl different formul. R() (, 5) r() Intervl of integrtion () (, ) Solids of Revolution: The Wsher Method If the region we revolve to generte solid does not order on or cross the is of revolution, the solid hs hole in it (Figure 6.). The crosssections perpendiculr to the is of revolution re wshers (the purplish circulr surfce in Figure 6.) insted of disks. The dimensions of tpicl wsher re The wsher s re is Asd = p[rsd]  p[rsd] = ps[rsd]  [rsd] d. Consequentl, the definition of volume gives Outer rdius: Rsd Inner rdius: rsd R() (, 5) r() (, ) V = Asd d = ps[rsd]  [rsd] d d. This method for clculting the volume of solid of revolution is clled the wsher method ecuse sl is circulr wsher of outer rdius R() nd inner rdius r(). Wsher cross section Outer rdius: R() Inner rdius: r() () FIGURE 6. () The region in Emple 9 spnned line segment perpendiculr to the is of revolution. () When the region is revolved out the is, the line segment genertes wsher. EXAMPE 9 A Wsher CrossSection (Rottion Aout the Ais) The region ounded the curve = + nd the line =  + is revolved out the is to generte solid. Find the volume of the solid. Solution. Drw the region nd sketch line segment cross it perpendiculr to the is of revolution (the red segment in Figure 6.).. Find the outer nd inner rdii of the wsher tht would e swept out the line segment if it were revolved out the is long with the region.
9 Chpter 6: Applictions of Definite Integrls Intervl of integrtion R() r() () (, ) or or These rdii re the distnces of the ends of the line segment from the is of revolution (Figure 6.).. Find the limits of integrtion finding the coordintes of the intersection points of the curve nd line in Figure 6... Evlute the volume integrl. Outer rdius: Rsd =  + Inner rdius: r sd = + V = ps[rsd]  [rsd] d d +  = s + ds  d = = pss + d  s + d d d  = ps d d  + =  + = , = = p c d  = 7p 5 Vlues from Steps nd r() R() To find the volume of solid formed revolving region out the is, we use the sme procedure s in Emple 9, ut integrte with respect to insted of. In this sitution the line segment sweeping out tpicl wsher is perpendiculr to the is (the is of revolution), nd the outer nd inner rdii of the wsher re functions of. EXAMPE A Wsher CrossSection (Rottion Aout the Ais) The region ounded the prol = nd the line = in the first qudrnt is revolved out the is to generte solid. Find the volume of the solid. () FIGURE 6.5 () The region eing rotted out the is, the wsher rdii, nd limits of integrtion in Emple. () The wsher swept out the line segment in prt (). Solution First we sketch the region nd drw line segment cross it perpendiculr to the is of revolution (the is). See Figure 6.5. The rdii of the wsher swept out the line segment re Rs d =, rs d = > (Figure 6.5). The line nd prol intersect t = nd =, so the limits of integrtion re c = nd d =. We integrte to find the volume: d V = ps[rs d]  [rs d] d d c = = p  d = p c.  d = 8 p p c d  c d d
10 6. Volumes Slicing nd Rottion Aout n Ais 5 Summr In ll of our volume emples, no mtter how the crosssectionl re A() of tpicl sl is determined, the definition of volume s the definite integrl V = Asd d is the hert of the clcultions we mde.
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