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 cross-sections re plne regions. A cross-section 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 cross-section 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

6. Volumes Slicing nd Rottion Aout n Ais 97 P Cross-section R() with re A() S FIGURE 6.

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.

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.

2 6. Volumes Slicing nd Rottion Aout n Ais 97 P Cross-section R() with re A() S FIGURE 6. A cross-section 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.

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 cross-sectionl re A() from = to = is

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.

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 cross-sectionl 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 cross-section.. Find formul for A(), the re of tpicl cross-section.. Find the limits of integrtion.. Integrte A() using the Fundmentl Theorem. Tpicl cross-section (m) FIGURE 6.5 The cross-sections 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 cross-section 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 cross-section (Figure 6.5).. A formul for A(). The cross-section 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 cross-sectionl res t ech height hve the sme volume (Figure 6.6). This follows immeditel from the definition of volume, ecuse the cross-sectionl re function A() nd the intervl [, ] re the sme for oth solids.

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.

4 6. Volumes Slicing nd Rottion Aout n Ais 99 Sme volume EXAMPE Volume of Wedge Sme cross-section 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 cross-sections re rectngles. Solution We drw the wedge nd sketch tpicl cross-section perpendiculr to the -is (Figure 6.7). The cross-section 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 cross-sectionl 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.

Chpter 6: Applictions of Definite Integrls R() () Disk () R() FIGURE 6.8 The region () nd solid of revolution () in Emple.

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.

The volume is EXAMPE 5 The circle Volume of Sphere is rotted out the -is to generte sphere. Find its volume. = 8p.

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 cross-section 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 cross-sectionl 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. 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.

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 cross-section 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.

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.

Solution We drw figures showing the region, tpicl rdius, nd the generted solid (Figure 6.). Note tht the cross-sections re perpendiculr to the line =.

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 cross-sections 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.

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

8 6. Volumes Slicing nd Rottion Aout n Ais (, R()) (, r()) R() r() Wsher FIGURE 6. The cross-sections 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 cross-sections 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 Cross-Section (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.

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.

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 Cross-Section (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 cross-sectionl 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.

5.2 Volumes: Disks and Washers

5.2 Volumes: Disks and Washers 4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of cross-section or slice. In this section, we restrict