8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS

Transcription

1 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS 8 Arc Length nd Surfce Are Preliminr Questions Which integrl represents the length of the curve cos between nd π? π π + cos d, + sin d Let cos Then sin, nd + ) + sin Thus, the length of the curve cos between nd π is π + sin d Use the formul for rc length to show tht for n constnt C, the grphs f)nd f)+ C hve the sme length over ever intervl [,b] Eplin geometricll The grph of f)+ C is verticl trnsltion of the grph of f); hence, the two grphs should hve the sme rc length We cn eplicitl estblish this s follows: b [ ] d b length of f)+ C + f ) + C) d +[f d )] d length of f) Use the formul for rc length to show tht the length of grph over [, ] cnnot be less thn Note tht f ), so tht +[f )] Then the rc length of the grph of f) on [, ] is +[f )] d d Eercises Epress the rc length of the curve between nd 6 s n integrl but do not evlute) Let Then nd 6 s + ) d d Epress the rc length of the curve tn for π s n integrl but do not evlute) Let tn Then sec, nd s π/ + sec ) d π/ + sec d Find the rc length of + for Hint: Show tht + ) + ) Let + Then, nd ) ) ) + 6 Mrch,

2 SECTION 8 Arc Length nd Surfce Are 7 Thus, s + ) d ) + d ) since ) + d + > + d ) Find the rc length of + over [, ] Hint: Show tht + ) is perfect squre ) Let + Then ) ) nd ) + ) ) + Hence, s + d ) + d 6 + ) since 55 ) + d + > on[, ] + d In Eercises 5, clculte the rc length over the given intervl 5 +, [, ] Let + Then, nd s + 9 d 6 9, [, ] Let 9 Then, nd s + 9 d 7 /, [, ] Let / Then /, nd s + 9 d ) / 7 8 ) / ) ) / ) / /, [, 8] Let / / Then / /, nd + ) + / ) / + + / + ) / Mrch,

3 8 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS Hence, 8 s 8 + ) d 8 / + / / + / 9 ln, [, e] ) 8 ) d 7 / + ) / d since / + / > 8 / + / d Let ln Then, nd + ) + ) ) Hence, e s e + ) d + ) d ) + e ln e since + ) d e + ln + + > on[, e] e + d lncos ), [, π ] Let lncos ) Then tn nd + ) + tn sec Hence, π/ π/ s + ) d π/ sec d since sec >on π/ ln sec + tn ln + ) sec d [, π ] π/ sec d In Eercises, pproimte the rc length of the curve over the intervl using the Trpezoidl Rule T N, the Midpoint Rule M N, or Simpson s Rule S N s indicted, [, ], T 5 Let Then + ) + ) + 6 Therefore, the rc length over [, ] is + 6 d Now, let f) + 6 With n 5, 5 5 nd { i } 5 i {, 6 5, 7 5, 8 5, 9 5, } Mrch,

4 SECTION 8 Arc Length nd Surfce Are 9 Using the Trpezoidl Rule, + 6 d The rc length is pproimtel units [ sin,, π ], M8 Let sin Then f ) + f i ) + f 5 ) i + + cos Therefore, the rc length over [,π/] is π/ + cos d Now, let f) + cos With n 8, we hve: π/ 8 π 6 Using the Midpoint Rule, π/ nd The rc length is pproimtel 999 units, [, ], S 8 Let Then nd { { } 8 π i i, π, 5π, 7π, 9π, π, π, 5π + cos d 8 fi ) 999 i } Therefore, the rc length over [, ] is Now, let f) + With n 8, ) + + d nd { i } 8 i {, 9 8, 5, 8,, 8, 7, 5 8, } Using Simpson s Rule, + d f ) + f i ) + f i ) + f 8 ) i i The rc length is pproimtel units e, [, ], S 8 Let e Then Therefore, the rc length over [, ] is Now, let f) + e With n 8, 8 + ) + e + e d nd { i } 8 i {,,,,, 5,, 7, } Mrch,

5 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS Using Simpson s Rule, + e d f ) + f i ) + f i ) + f 8 ) 878 i i The rc length is pproimtel 878 units 5 Clculte the length of the stroid / + / Figure ) FIGURE Grph of / + / We will clculte the rc length of the portion of the steroid in the first qudrnt nd then multipl b B implicit differentition so Thus nd The totl rc length is therefore 6 / + /, / / / / + ) + / / / + / / /, s / d 6 Show tht the rc length of the steroid / + / / for >) is proportionl to We will clculte the rc length of the portion of the steroid in the first qudrnt nd then multipl b B implicit differentition so Thus / + /, / / / / + ) + / / / + / / / /, nd / ) s d / / / The totl rc length is therefore 6, which is proportionl to 7 Let,r > Show tht the rc length of the curve r + r r for is proportionl to Using implicit differentition, we find /) r nd + ) + /) r r + r r r + r r ) /r r r ) /r Mrch,

6 SECTION 8 Arc Length nd Surfce Are The rc length is then s r + r r ) /r r r ) /r d Using the substitution u, we obtin u s r + u r ) /r u r ) /r du, where the integrl is independent of 8 Find the rc length of the curve shown in Figure 5 FIGURE Grph of 9 ) Using implicit differentition, 8 ) ) + ) ) ) Hence, ) ) ) 6 ) ) 9 ) ) ) ) ) nd Finll, + ) ) + + ) + ) + s d d + d since + > on[, ] / + ) ) / d / + / 9 Find the vlue of such tht the rc length of the ctenr cosh for equls Let cosh Then sinh nd + ) + sinh cosh Thus, s cosh d sinh) sinh ) sinh Setting this epression equl to nd solving for ields sinh 5) ln5 + 6) Clculte the rc length of the grph of f) m + r over [,b] in two ws: using the Pthgoren theorem Figure ) nd using the rc length integrl mb ) r b b FIGURE Mrch,

7 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS Let h denote the length of the hpotenuse Then, b Pthgors Theorem, h b ) + m b ) b ) + m ), or h b ) + m since b> Moreover, f )) m,so b s Show tht the circumference of the unit circle is equl to + m d b ) + m h d n improper integrl) Evlute, thus verifing tht the circumference is π Note the circumference of the unit circle is twice the rc length of the upper hlf of the curve defined b + Thus, let Then nd + ) + Finll, the circumference of the unit circle is d sin π π) π Generlize the result of Eercise to show tht the circumference of the circle of rdius r is πr Let r denote the upper hlf of circle of rdius r centered t the origin Then + ) + r r r, r nd the circumference of the circle is given b r d C r /r Using the substitution u /r, du d/r,wefind du C r r u sin u π r π )) πr Clculte the rc length of over [,] Hint: Use trigonometric substitution Evlute for Let Then nd s + d Using the substitution tn θ, d sec θdθ,wefind s sec θdθ Net, using reduction formul for the integrl of sec θ, we see tht s sec θ tn θ + ) ln sec θ + tn θ + + ) ln ln + + Mrch,

8 SECTION 8 Arc Length nd Surfce Are Thus, when, s 5 + ln 5 + ) 789 Epress the rc length of g) over [, ] s definite integrl Then use the substitution u to show tht this rc length is equl to the rc length of over [, ] but do not evlute the integrls) Eplin this result grphicll Let g) Then + g ) + + nd s + d d With the substitution u, du d, this becomes s + u du Now, let f) Then + f ) +, nd s + d Thus, the two rc lengths re equl This is eplined grphicll b the fct tht for, nd re inverses of ech other This mens tht the two grphs re smmetric with respect to the line Moreover, the grphs of nd intersect t nd t Thus, it is cler tht the rc length of the two grphs on [, ] re equl 5 Find the rc length of e over [,] Hint: Tr the substitution u + e followed b prtil frctions Let e Then + ) + e, nd the rc length over [,] is + e d Now, let u + e Then du e d u d + e u nd the rc length is + e u d u u du u u du + ) u du + u ) du u + u + lnu ) ) lnu + ) [ + e + )] + e ln + e + + e + + e ln + e + + ln + + e + + e ln + e + + ln + ) 6 Show tht the rc length of lnf )) for b is b f) + f ) d f) Let lnf )) Then f ) f) nd + ) f) + f ) f) Mrch,

9 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS Therefore, b f) s + f ) d f) since f)> in order for lnf )) to be defined on [,b] 7 Use Eq ) to compute the rc length of lnsin ) for π π With f) sin, Eq ) ields π/ sin + cos s d π/ sin ln ln ) ln 8 Use Eq ) to compute the rc length of ln With f) e + e, nd Thus, b Eq ), f) + f ) π/ π/ csc d ln csc cot ) π/ π/ ln + ) e ) + e over [, ] f ) e )e e + )e e ) e e ) e ) + e + e e ) e ) + e e ) e + ) e ) Observe tht Therefore, e + s e ) e e + d e + e d e + e e + e e e e + e )/ e e )/ cosh sinh cosh s sinh d lnsinh ) lnsinh ) lnsinh ) 9 Show tht if f ) for ll, then the rc length of f)over [,b] is t most b ) Show tht for f), the rc length equls b ) If f ) for ll, then b b s + f ) d + d b ) If f), then f ) nd b s + d b ) Use the Comprison Theorem Section 5) to prove tht the rc length of / over [, ] is not less thn 5 Note tht f ) / ; for [, ], we hve / so tht f ) Then ) 5 + f ) nd then the rc length is + f ) 5 d d 5 Mrch,

10 SECTION 8 Arc Length nd Surfce Are 5 Approimte the rc length of one-qurter of the unit circle which we know is π ) b computing the length of the polgonl pproimtion with N segments Figure ) FIGURE One-qurter of the unit circle With, the five points long the curve re P, ), P /, 5/), P /, /), P /, 7/), P, ) Then P P 6 + ) 5 59 P P 6 + ) 5 79 P P 6 + ) 7 P P nd the totl pproimte distnce is 555 wheres π/ A merchnt intends to produce specilt crpets in the shpe of the region in Figure 5, bounded b the es nd grph of n units in rds) Assume tht mteril costs $5/d nd tht it costs 5L dollrs to cut the crpet, where L is the length of the curved side of the crpet The crpet cn be sold for 5A dollrs, where A is the crpet s re Using numericl integrtion with computer lgebr sstem, find the whole number n for which the merchnt s profits re miml n 5 A 5 FIGURE 5 The re of the crpet is ) A n )d n+ n + n + n n +, while the length of the curved side of the crpet is L + n n ) d + n n d Using these formuls, we find tht the merchnt s profit is given b 5A 5A + 5L) A 5L n n n n d Using CAS, we find tht the merchnt s profit is mimized pproimtel $ per crpet) when n The tble below lists the profit for n 5 Mrch,

11 6 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS n Profit n Profit In Eercises, compute the surfce re of revolution bout the -is over the intervl, [, ] + ) so tht +, [, ] Let + Then + ) 7 nd SA π d π 6π SA π + ) 7 d π ) 7 + π 7 5, [, ] + ) + 9, so tht SA π + 9 d π d π ) / π 8 6, [, ] Let Then nd SA π + d Using the substitution tn θ, d sec θdθ, we find tht + d sec θ tn θdθ sec 5 θ sec θ) dθ 8 8 Finll, 8 sec θ tn θ + 8 sec θ tn θ + 8 ln sec θ + tn θ sec θ tn θ 6 + ) / + 6 ln + + +C SA π π 6 + ) / 65 65/ 8 + ) 6 ln ln8 + 65) ) 9 65 π π ln8 + 65) ) 5 / ) ln sec θ + tn θ + C 7 / ) /, [, 8] Let / ) / Then nd / / ) /, + ) + / / / Mrch,

12 SECTION 8 Arc Length nd Surfce Are 7 Therefore, 8 ) SA π / ) / / d Using the substitution u /, du / d,wefind SA π u / )du 6π u / du 5 πu5/ 8π 5 8 e, [, ] Let e Then e nd SA π e + e d Using the substitution e tn θ, e d sec θdθ, we find tht e + e d sec θdθ sec θ tn θ ln sec θ + tn θ +C e + e ln + e + e +C Finll, SA πe ) + e π ln + e + e πe + e π ln π πe + e + π ln + e + e ) + π + π ln + ) ) + + e + e 9 ln, [,e] We hve, nd + ) + ) ) Thus, e SA π π π π ) ln + ) e d π ) + 6 ln 8 e + ln ) e 6 e 8 + e 6 8 e ) e ln ) ) 6 ln d π 6 e 9) sin, [,π] Let sin Then cos, nd π SA π sin + cos d Using the substitution cos tn θ, sin d sec θdθ, we find tht sin + cos d sec θdθ sec θ tn θ ln sec θ + tn θ +C Mrch,

13 8 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS cos + cos ln + cos + cos +C Finll, SA π cos + cos ) ln + cos π + cos π ln ) + ) + ln + ) π + ln + ) ) In Eercises, use computer lgebr sstem to find the pproimte surfce re of the solid generted b rotting the curve bout the -is, [, ] SA π + ) d π + d using Mple, [, ] SA π + ) d π d using Mple e /, [, ] SA π e / + e / ) d π e / + e d using Mple [ tn,, π ] Let tn Then sec,+ ) + sec, nd π/ SA π tn + sec d Using computer lgebr sstem to pproimte the vlue of the definite integrl, we find SA Find the re of the surfce obtined b rotting cosh over [ ln, ln ] round the -is Let cosh Then sinh, nd + ) + sinh cosh cosh Therefore, ln SA π ln π ln cosh d π ln We cn simplif this nswer b recognizing tht + cosh ) d π + ) ln sinh ln ln + sinhln) + ln ) sinh ln) π ln + π sinhln) Thus, sinhln) eln e ln SA π ln + 5π Mrch,

14 SECTION 8 Arc Length nd Surfce Are 9 6 Show tht the surfce re of sphericl cp of height h nd rdius R Figure 6) hs surfce re πrh h R FIGURE 6 To determine the surfce re of the cp, we will rotte portion of circle of rdius R, centered t the origin, bout the -is Since the eqution of the right hlf of the circle is R, + ) + R R R, nd R ) SA π R R R h R d πrr R h)) πrh 7 Find the surfce re of the torus obtined b rotting the circle + b) bout the -is Figure 7), b + ), b) FIGURE 7 Torus obtined b rotting circle bout the -is b + gives the top hlf of the circle nd b gives the bottom hlf Note tht in ech cse, + ) + Rotting the two hlves of the circle round the -is then ields SA π π b + ) d + π b d πb b d πb sin ) π πb π )) π b ) d 8 Show tht the surfce re of right circulr cone of rdius r nd height h is πr r + h Hint: Rotte line m bout the -is for h, where m is determined suitbl b the rdius r r m h Mrch,

15 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS From the figure, we see tht m r h,so r h Thus h SA π r h Further Insights nd Chllenges + r πr d h h + r h h d πr h + r ) ) 9 Find the surfce re of the ellipsoid obtined b rotting the ellipse + bout the -is b Tking dvntge of smmetr, we cn find the surfce re of the ellipsoid b doubling the surfce re obtined b rotting the portion of the ellipse in the first qudrnt bout the -is The eqution for the portion of the ellipse in the first qudrnt is Thus, b + ) b + ) + b ), ) nd b SA π + b ) d πb b + ) d We now consider two cses If b >, then we mke the substitution b tn θ, d b sec θdθ, nd find tht SA πb sec θdθ πb b b sec θ tn θ + ln sec θ + tn θ ) b πb + ) + πb b ln b + ) b + πb ) + πb b ln b b + On the other hnd, if >b, then we mke the substitution b sin θ, d cos θdθ, b nd find tht SA πb cos θdθ πb b b θ + sin θ cos θ) πb b ) ) + πb b b sin πb ) + πb b b sin Observe tht in both cses, s pproches b, the vlue of the surfce re of the ellipsoid pproches πb, the surfce re of sphere of rdius b Mrch,

16 SECTION 8 Arc Length nd Surfce Are 5 Show tht if the rc length of f)over [,] is proportionl to, then f)must be liner function s + f ) d For s to be proportionl to, + f ) must be constnt, which implies f ) is constnt This, in turn, requires f)be liner 5 Let L be the rc length of the upper hlf of the ellipse with eqution b Figure 8) nd let η b / ) Use substitution to show tht π/ L η sin θdθ π/ Use computer lgebr sstem to pproimte L for, b FIGURE 8 Grph of the ellipse Let b Then + ) b + ) ) nd b s + ) d ) With the substitution sin t, d cos tdt, cos t nd π/ s cos t b sin t + cos t π/ cos dt t) π/ π/ b sin t + cos tdt Becuse we then hve η b,η b ) η sin t b sin t sin t + b sin t cos t + b sin t which is the sme s the epression under the squre root bove Substituting, we get π/ s η sin tdt π/ When nd b, η Using computer lgebr sstem to pproimte the vlue of the definite integrl, we find s 8 Mrch,

17 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS 5 Prove tht the portion of sphere of rdius R seen b n observer locted t distnce d bove the North Pole hs re A πdr /d + R) Hint: According to Eercise 6, the cp hs surfce re is πrh Show tht h dr/d + R) b ppling the Pthgoren Theorem to the three right tringles in Figure 9 Observer d h R FIGURE 9 Sphericl cp observed from distnce d bove the North Pole Lbel distnces s shown in the figure below d k h R h R B repeted ppliction of the Pthgoren Theorem, we find d + R) R + k R + d + h) + R + d + h) + R R h) Solving for h ields d + dr + R R + d + dh + h + R R + Rh h dr dh + Rh nd thus dr d + R)h h dr d + R ) dr SA πr d + R 5 Suppose tht the observer in Eercise 5 moves off to infinit tht is, d Wht do ou epect the limiting vlue of the observed re to be? Check our guess b clculting the limit using the formul for the re in the previous eercise We would ssume the observed surfce re would pproch πr which is the surfce re of hemisphere of rdius R To verif this, observe: πr d lim SA lim d d R + d lim πr πr d 5 Let M be the totl mss of metl rod in the shpe of the curve f)over [,b] whose mss densit ρ) vries s function of Use Riemnn sums to justif the formul b M ρ) + f ) d Divide the intervl [,b] into n subintervls, which we shll denote b [ j, j ] for j,,,,n On ech subintervl, we will ssume tht the mss densit of the rod is constnt; hence, the mss of the corresponding segment of the rod will be pproimtel equl to the product of the mss densit of the segment nd the length of the segment Specificll, let c j be n point in the jth subintervl nd pproimte the mss of the segment b ρc j ) + f c j ), where + f c j ) is the pproimte length of the segment Thus, M n j ρc j ) + f c j ) Mrch,

18 SECTION 8 Fluid Pressure nd Force As n, this Riemnn sum pproches definite integrl, nd we hve b M ρ) + f ) d 55 Let f)be n incresing function on [,b] nd let g) be its inverse Argue on the bsis of rc length tht the following equlit holds: b + f ) d Then use the substitution u f)to prove Eq 5) fb) f) + g ) d 5 Since the grphs of f)nd g) re smmetric with respect to the line, the rc length of the curves will be equl on the respective domins Since the domin of g is the rnge of f,onf) to fb), g) will hve the sme rc length s f)on to b Ifg) f ) nd u f), then gu) nd du f ) d But Now substituting u f), b s g u) + f ) d f gu)) f ) f ) g u) fb) + f) ) g g u) du u) fb) f) g u) + du 8 Fluid Pressure nd Force Preliminr Questions How is pressure defined? Pressure is defined s force per unit re Fluid pressure is proportionl to depth Wht is the fctor of proportionlit? fluid The fctor of proportionlit is the weight densit of the fluid, w ρg, where ρ is the mss densit of the When fluid force cts on the side of submerged object, in which direction does it ct? Fluid force cts in the direction perpendiculr to the side of the submerged object Wh is fluid pressure on surfce clculted using thin horizontl strips rther thn thin verticl strips? Pressure depends onl on depth nd does not chnge horizontll t given depth 5 If thin plte is submerged horizontll, then the fluid force on one side of the plte is equl to pressure times re Is this true if the plte is submerged verticll? When plte is submerged verticll, the pressure is not constnt long the plte, so the fluid force is not equl to the pressure times the re Eercises A bo of height 6 m nd squre bse of sidemissubmerged in pool of wter The top of the bo is m below the surfce of the wter ) Clculte the fluid force on the top nd bottom of the bo b) Write Riemnn sum tht pproimtes the fluid force on side of the bo b dividing the side into N horizontl strips of thickness 6/N c) To which integrl does the Riemnn sum converge? d) Compute the fluid force on side of the bo Mrch,

19 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS ) At depth of m, the pressure on the top of the bo is ρgh 98 9,6 P The top hs re 9 m, nd the pressure is constnt, so the force on the top of the bo is 9,6 9 76,N At depth of 8 m, the pressure on the bottom of the bo is ρgh , P, so the force on the bottom of the bo is 78, 9 75,6N b) Let j denote the depth of the j th strip, for j,,,,n; the pressure t this depth is 98 j 98 j P The strip hs thickness m nd length m, so hs re m Thus the force on the strip is 9, j N Sum over ll the strips to conclude tht the force on one side of the bo is pproimtel F N 9, j c) As N, the Riemnn sum in prt b) converges to the definite integrl 9, 8 d d) Using the result from prt c), the fluid force on one side of the bo is j 8 9, d,7 8 88, N A plte in the shpe of n isosceles tringle with bse m nd height missubmerged verticll in tnk of wter so tht its verte touches the surfce of the wter Figure 7) ) Show tht the width of the tringle t depth is f) b) Consider thin strip of thickness t depth Eplin wh the fluid force on side of this strip is pproimtel equl to ρg c) Write n pproimtion for the totl fluid force F on side of the plte s Riemnn sum nd indicte the integrl to which it converges d) Clculte F Δ f) FIGURE 7 ) B similr tringles, f) so f) b) The pressure t depth of feet is ρg P, nd the re of the strip is pproimtel f) m Therefore, the fluid force on this strip is pproimtel ) ρg ρg N c) F ρg j AsN, the Riemnn sum converges to the definite integrl j ρg d d) Using the result of prt c), F ρg d ρg ) N Repet Eercise, but ssume tht the top of the tringle is locted m below the surfce of the wter ) Emine the figure below B similr tringles, f) so f) Mrch,

20 SECTION 8 Fluid Pressure nd Force 5 f) b) The pressure t depth of feet is ρg lb/ P, nd the re of the strip is pproimtel f) ) m Therefore, the fluid force on this strip is pproimtel ) ρg ) ρg ) N c) F N j ρg j j d) Using the result of prt c), F ρg 5 AsN, the Riemnn sum converges to the definite integrl )d ρg ρg 5 )d ) 5 98 [ 5 75 ) 9 7 )] 7, N The plte R in Figure 8, bounded b the prbol nd, is submerged verticll in wter distnce in meters) ) Show tht the width of R t height is f) nd the fluid force on side of horizontl strip of thickness t height is pproimtel ρg) / ) b) Write Riemnn sum tht pproimtes the fluid force F on side of R nd use it to eplin wh c) Clculte F F ρg / )d Wter surfce f), ) R FIGURE 8 ) At height, the thin plte R etends from the point,) on the left to the point,) on the right; thus, the width of the plte is f) ) Moreover, the re of horizontl strip of thickness t height is f) Becuse the wter surfce is t height, the horizontl strip t height is t depth of Consequentl, the fluid force on the strip is pproimtel ρg ) ρg / ) b) If the plte is divided into N strips with j being the representtive height of the jth strip for j,,,,n), then the totl fluid force eerted on the plte is F ρg N j ) j j Mrch,

21 6 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS As N, the Riemnn sum converges to the definite integrl ρg ) d c) Using the result from prt b), F ρg ) d ρg / ) 5 5/ 8 5 ρg Now, ρg 98 N/m so tht F 568 N 5 Let F be the fluid force on side of semicirculr plte of rdius r meters, submerged verticll in wter so tht its dimeter is level with the wter s surfce Figure 9) ) Show tht the width of the plte t depth is r b) Clculte F s function of r using Eq ) r r r FIGURE 9 ) Plce the origin t the center of the semicircle nd point the positive -is downwrd The eqution for the edge of the semicirculr plte is then + r At depth of, the plte etends from the point r,)on the left to the point r,)on the right The width of the plte t depth is then b) With w 98 N/m, r F w r r d 9,6 ) r r r ) / r 9,6r 6 Clculte the force on one side of circulr plte with rdius m, submerged verticll in tnk of wter so tht the top of the circle is tngent to the wter surfce Plce the origin t the point where the top of the circle is tngent to the wter surfce nd orient the positive -is pointing downwrd The eqution of the circle is then + ), nd the width t n depth is ) Thus, F ρg ) d, Using the substitution sin θ, d cos θdθ, the limits of integrtion become π θ π,sowefind F ρg ) d π/ ρg π/ 6ρg 6ρg π/ + sin θ)cos θ)cos θdθ) 6ρg cos θ + sin θ cos θdθ π/ θ + sin θ cos θ ) π/ cos θ π/ π + π ) + ) 8ρgπ 78,π N 7 A semicirculr plte of rdius r meters, oriented s in Figure 9, is submerged in wter so tht its dimeter is locted t depth of m meters Clculte the fluid force on one side of the plte in terms of m nd r Plce the origin t the center of the semicirculr plte with the positive -is pointing downwrd The wter surfce is then t m Moreover, t loction, the width of the plte is r nd the depth is + m Thus, r F ρg + m) r d N Mrch,

22 SECTION 8 Fluid Pressure nd Force 7 Now, Geometricll, r r d r ) / r r r r d represents the re of one qurter of circle of rdius r, nd thus hs the vlue πr Bringing these results together, we find tht F ρg r + π ) r 9,6 r + 9mr N 8 A plte etending from depth mto 5 m is submerged in fluid of densit ρ 85 kg/m The horizontl width of the plte t depth is f) + ) Clculte the fluid force on one side of the plte The fluid force on one side of the plte is given b 5 5 F ρg f ) d ρg + ) d ρg ln + 5 ) ρgln 6 ln 5) 8 ln 6 7 N 5 9 Figure shows the wll of dm on wter reservoir Use the Trpezoidl Rule nd the width nd depth mesurements in the figure to estimte the fluid force on the wll Depth ft) 6 8,8 ft),65,, 9 6 FIGURE Let f)denote the width of the dm wll t depth feet Then the force on the dm wll is F w f ) d Using the Trpezoidl Rule nd the width nd depth mesurements in the figure, F w [ f) + f) + f) + 6 f6) + 8 f8) + f)] w + 66, +, +, +, + 6,),5, lb Clculte the fluid force on side of the plte in Figure A), submerged in wter m m m 7 m A) FIGURE m m m B) The width of the plte vries linerl from meters t depth of meters to 7 meters t depth of 5 meters Thus, t depth, the width of the plte is Finll, the force on side of the plte is 5 F w ) d w + ) ) 5 5w, N Mrch,

23 8 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS Clculte the fluid force on side of the plte in Figure B), submerged in fluid of mss densit ρ 8 kg/m Becuse the fluid hs mss densit of ρ 8 kg/m, w 8)98) 78 N/m For depths up to meters, the width of the plte t depth is ; for depths from meters to 6 meters, the width of the plte is constnt meters Thus, 6 F w )d + w d w + w 6 8w + w w 85,6 Find the fluid force on the side of the plte in Figure, submerged in fluid of densit ρ kg/m The top of the plce is level with the fluid surfce The edges of the plte re the curves / nd / N Fluid level 8 / / 8 FIGURE At height, the plte etends from the point,)on the left to the point,)on the right; thus, the width of the plte is f) ) Becuse the wter surfce is t height, the horizontl strip t height is t depth of Consequentl, F ρg ) )d ρg ) 5 5 6ρg ,6 N 5 5 Let R be the plte in the shpe of the region under sin for π in Figure A) Find the fluid force on side of R if it is rotted counterclockwise b 9 nd submerged in fluid of densit kg/m with its top edge level with the surfce of the fluid s in B) Fluid level sin R p Fluid level R A) FIGURE B) Plce the origin t the bottom corner of the plte with the positive -is pointing upwrd The fluid surfce is then t height π, nd the horizontl strip of the plte t height is t depth of π nd hs width of sin Now, using integrtion b prts we find π/ π ) [ π ) ] F ρg sin d ρg π/ π ) cos sin ρg π ) N In the nottion of Eercise, clculte the fluid force on side of the plte R if it is oriented s in Figure A) You m need to use Integrtion b Prts nd trigonometric substitution Plce the origin t the lower left corner of the plte Becuse the fluid surfce is t height, the horizontl strip t height is t depth of Moreover, this strip hs width of π sin cos Thus, F ρg )cos d Mrch,

24 SECTION 8 Fluid Pressure nd Force 9 Strting with integrtion b prts, we find )cos d ) cos + d cos + d d d Now, d For the remining integrl, we use the trigonometric substitution sin θ, d cos θdθnd find Finll, d sin θdθ θ sin θ cos θ) ) sin π 8 F ρg π ) 98 π ) 6567 N Clculte the fluid force on one side of plte in the shpe of region A shown Figure The wter surfce is t, nd the fluid hs densit ρ 9 kg/m B A e ln FIGURE Becuse the fluid surfce is t height, the horizontl strip t height is t depth of Moreover, this strip hs width of e e Thus, Now, F ρg )e e )d eρg )d ρg )e d nd using integrtion b prts )d ), )e d )e + e ) e Combining these results, we find tht ) F ρg e e ) ρg ) e 9 98 ) e 5657 N 6 Clculte the fluid force on one side of the infinite plte B in Figure, ssuming the fluid hs densit ρ 9 kg/m Becuse the fluid surfce is t height, the horizontl strip t height is t depth of Moreover, this strip hs width of e Thus, F ρg )e d Mrch,

25 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS Using integrtion b prts, we find )e d [ )e + e ] Thus, F ρg ,6 N 7 Figure 5A) shows rmp inclined t leding into swimming pool Clculte the fluid force on the rmp A horizontl strip t depth hs length 6 nd width sin Thus, F ρg 6d 96ρg If distnces re in feet, then ρg w 65 lb/ft nd F 6 lb; if distnces re in meters, then ρg 98 N/m nd F 9,8 N 8 Clculte the fluid force on one side of the plte n isosceles tringle) shown in Figure 5B) Wter surfce Wter surfce 6 f) 6 Verticl chnge Δ A) FIGURE 5 A horizontl strip t depth hs length f) nd width sin 6 Thus, F 5 w B) d w If distnces re in feet, then w 65 lb/ft nd F 7688 lb; if distnces re in meters, then w 98 N/m nd F,,665 N 9 The mssive Three Gorges Dm on Chin s Yngtze River hs height 85 m Figure 6) Clculte the force on the dm, ssuming tht the dm is trpezoid of bse m nd upper edge m, inclined t n ngle of 55 to the horizontl Figure 7) m m m FIGURE 6 Three Gorges Dm on the Yngtze River FIGURE 7 Let be t the bottom of the dm, so tht the top of the dm is t 85 Then the width of the dm t height is + 85 The dm is inclined t n ngle of 55 to the horizontl, so the height of horizontl strip is sin 55 Mrch,

26 SECTION 8 Fluid Pressure nd Force so tht the re of such strip is Then + ) F ρg + ) 85 d ρg + 66 d ρg + 85 ) 85 55,78,ρg 55,78, N A squre plte of sidemissubmerged in wter t n incline of with the horizontl Clculte the fluid force on one side of the plte if the top edge of the plte lies t depth of 6 m Becuse the plte is meters on side, is submerged t horizontl ngle of, nd hs its top edge locted t depth of 6 meters, the bottom edge of the plte is locted t depth of 6 + sin 5 meters Let denote the depth t n point of the plte The width of ech horizontl strip of the plte is then sin, nd 5/ F ρg )d ρg) 595,5 N 6 The trough in Figure 8 is filled with corn srup, whose weight densit is 9 lb/ft Clculte the force on the front side of the trough h b d FIGURE 8 Plce the origin long the top edge of the trough with the positive -is pointing downwrd The width of the front side of the trough vries linerl from b when to when h; thus, the width of the front side of the trough t depth feet is given b b + b h Now, h F w b + b ) h d w b + b ) h b h w 6 + ) h 5b + )h lb Clculte the fluid pressure on one of the slnted sides of the trough in Figure 8 when it is filled with corn srup s in Eercise b h θ b The digrm bove displs side view of the trough From this digrm, we see tht Thus, F w sin θ h sin θ ) b + h ) h 9 b + h dh ) b d d h 5dh + h Mrch,

27 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS Further Insights nd Chllenges The end of the trough in Figure 9 is n equilterl tringle of side Assume tht the trough is filled with wter to height H Clculte the fluid force on ech side of the trough s function of H nd the length l of the trough l H FIGURE 9 Plce the origin t the lower verte of the trough nd orient the positive -is pointing upwrd First, consider the fces t the front nd bck ends of the trough A horizontl strip t height hs length of nd is t depth of H Thus, H F w H ) H d w ) H 9 wh For the slnted sides, we note tht ech side mkes n ngle of 6 with the horizontl If we let l denote the length of the trough, then F wl H H )d lwh A rectngulr plte of side l is submerged verticll in fluid of densit w, with its top edge t depth h Show tht if the depth is incresed b n mount h, then the force on side of the plte increses b wa h, where A is the re of the plte Let F be the force on side of the plte when its top edge is t depth h nd F be the force on side of the plte when its top edge is t depth h + h Further, let b denote the width of the rectngulr plte Then h+l ) h+l l ) + lh F w b d bw bw h h h+l+ h ) h+l+ h F w b d bw bw l + lh + l h h+ h h+ h nd F F bwl h wa h 5 Prove tht the force on the side of rectngulr plte of re A submerged verticll in fluid is equl to p A, where p is the fluid pressure t the center point of the rectngle Let l denote the length of the verticl side of the rectngle, denote the length of the horizontl side of the rectngle, nd suppose the top edge of the rectngle is t depth m The pressure t the center of the rectngle is then p w m + l ), nd the force on the side of the rectngulr plte is l+m F w d w [ m l + m) m ] wl ) l l + m) Aw + m Ap 6 If the densit of fluid vries with depth, then the pressure t depth is function p) which need not equl w s in the cse of constnt densit) Use Riemnn sums to rgue tht the totl force F on the flt side of submerged object submerged verticll is F b f)p)d, where f)is the width of the side t depth Suppose the object etends from depth of to depth of b Divide the object into N horizontl strips, ech of width Let p) denote the pressure within the fluid t depth nd f)denote the width of the flt side of the submerged object t depth The pproimte force on the jth strip j,,,,n)is p j )f j ), Mrch,

28 SECTION 8 Center of Mss where j is depth ssocited with the jth strip Summing over ll of the strips, N F p j )f j ) j As N, this Riemnn sum converges to definite integrl, nd b F p)f)d 8 Center of Mss Preliminr Questions Wht re the - nd -moments of lmin whose center of mss is locted t the origin? Becuse the center of mss is locted t the origin, it follows tht M M A thin plte hs mss Wht is the -moment of the plte if its center of mss hs coordintes, 7)? The -moment of the plte is the product of the mss of the plte nd the -coordinte of the center of mss Thus, M 7) The center of mss of lmin of totl mss 5 hs coordintes, ) Wht re the lmin s - nd -moments? The -moment of the plte is the product of the mss of the plte nd the -coordinte of the center of mss, wheres the -moment is the product of the mss of the plte nd the -coordinte of the center of mss Thus, M 5) 5, nd M 5) Eplin how the Smmetr Principle is used to conclude tht the centroid of rectngle is the center of the rectngle Becuse rectngle is smmetric with respect to both the verticl line nd the horizontl line through the center of the rectngle, the Smmetr Principle gurntees tht the centroid of the rectngle must lie long both of these lines The onl point in common to both lines of smmetr is the center of the rectngle, so the centroid of the rectngle must be the center of the rectngle Eercises Four prticles re locted t points, ),, ),, ),, ) ) Find the moments M nd M nd the center of mss of the sstem, ssuming tht the prticles hve equl mss m b) Find the center of mss of the sstem, ssuming the prticles hve msses,, 5, nd 7, respectivel ) Becuse ech prticle hs mss m, M m) + m) + m) + m) m; M m) + m) + m) + m) 9m; nd the totl mss of the sstem is m Thus, the coordintes of the center of mss re M M, M ) 9m M m, m ) ) 9 m, b) With the indicted msses of the prticles, M ) + ) + 5) + 7) ; M ) + ) + 5) + 7) 6; nd the totl mss of the sstem is 7 Thus, the coordintes of the center of mss re M M, M ) 6 M 7, ) 7 Mrch,

29 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS Find the center of mss for the sstem of prticles of msses,, 5, locted t, ),, ),, ),, ) With the indicted msses nd loctions of the prticles M ) + ) + 5 ) + ) 7; M ) + ) + 5) + ) ; nd the totl mss of the sstem is Thus, the coordintes of the center of mss re M M, M ), 7 ) M Point msses of equl size re plced t the vertices of the tringle with coordintes, ), b, ), nd,c) Show tht the center of mss of the sstem of msses hs coordintes + b), c) Let ech prticle hve mss m The totl mss of the sstem is then m nd the moments re Thus, the coordintes of the center of mss re M M, M ) M M m) + m) + cm) cm; nd M m) + bm) + m) + b)m + b)m m, cm ) + b, c ) m Point msses of mss m, m, nd m re plced t the points, ),, ), nd, ) ) Suppose tht m 6 Find m such tht the center of mss lies on the -is b) Suppose tht m 6 nd m Find the vlue of m such tht CM With the given msses nd loctions, we find M m ) + m ) + m ) m ; M m ) + m ) + m ) m m ; nd the totl mss of the sstem is m + m + m Thus, the coordintes of the center of mss re ) m m m, m + m + m m + m + m ) For the center of mss to lie on the -is, we must hve m m, or m m Given m 6, it follows tht m b) To hve CM requires m or m m + m m + m + m Given m 6 nd m, it follows tht m 5 Sketch the lmin S of constnt densit ρ g/cm occuping the region beneth the grph of for ) Use Eqs ) nd ) to compute M nd M b) Find the re nd the center of mss of S A sketch of the lmin is shown below ) Using Eq ), Using Eq ), 9 M )d 9 M )d ) / 79 Mrch,

30 SECTION 8 Center of Mss 5 b) The re of the lmin is A d 9cm With constnt densit of ρ g/cm, the mss of the lmin is M 7 grms, nd the coordintes of the center of mss re M M, M ) / M 7, 79/ ) 9 7, 7 ) 6 Use Eqs ) nd ) to find the moments nd center of mss of the lmin S of constnt densit ρ g/cm occuping the region between nd 9 over [, ] Sketch S, indicting the loction of the center of mss With ρ g/cm, M ) 9) ) ) d 5, nd M 9 )d The mss of the lmin is M 9 )d 6 g, so the coordintes of the center of mss re M M, M ) M 6, ) 5 A sketch of the lmin, with the loction of the center of mss indicted, is shown below Find the moments nd center of mss of the lmin of uniform densit ρ occuping the region underneth for With uniform densit ρ, M ρ ) d 6ρ nd M ρ )d ρ 7 5 The mss of the lmin is M ρ d ρ, so the coordintes of the center of mss re M M, M ) 8 M 5, 6 ) 7 8 Clculte M ssuming ρ ) for the region underneth the grph of for in two ws, first using Eq ) nd then using Eq ) B Eq ), M d Mrch,

31 6 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS Using the substitution u, du d, wefind B Eq ), M u) udu u/ ) 5 u5/ 5 M ) d + 5 ) Let T be the tringulr lmin in Figure 7 ) Show tht the horizontl cut t height hs length nd use Eq ) to compute M with ρ ) b) Use the Smmetr Principle to show tht M nd find the center of mss 6 FIGURE 7 Isosceles tringle ) The eqution of the line from, ) to, 6) is + 6, so The length of the horizontl cut t height is then ), nd 6 M ) d b) Becuse the tringulr lmin is smmetric with respect to the -is, cm, which implies tht M The totl mss of the lmin is M + 6)d, so cm / Finll, the coordintes of the center of mss re, ) In Eercises 7, find the centroid of the region ling underneth the grph of the function over the given intervl f) 6, [, ] The moments of the region re M 6 ) d 8 nd M 6 )d 9 The re of the region is A 6 )d 9, so the coordintes of the centroid re M A, M ), ) A f), [, ] The moments of the region re M d 5 nd M d 6 5 Mrch,

32 SECTION 8 Center of Mss 7 The re of the region is so the coordintes of the centroid re A d, M A, M ) 9 A 5, 5 ) 56 f), [, ] The moments of the region re The re of the region is M 6 d nd M d 5 so the coordintes of the centroid re A d, M A, M ) A 5, ) 7 f) 9, [, ] The moments of the region re The re of the region is M 9 ) d 5 nd M 9 )d 8 so the coordintes of the centroid re A 9 )d 8, M A, M ) 9 A 8, 8 ) 5 f) + ) /, [, ] The moments of the region re M + d tn tn nd M + d The re of the region is A d ln ln + ), so the coordintes of the centroid re M A, M ) A ln + ), tn ) ln + ) 5 f) e, [, ] The moments of the region re M e d e 8) nd M e d e + ) 5e Mrch,

33 8 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS The re of the region is so the coordintes of the centroid re A e d e, M A, M ) 5e A e, e 8 ) e ) 6 f) ln, [, ] The moments of the region re M ln ) d ln ) ln + ) ln ) ln+ ; nd M ln d ln ) ln The re of the region is A ln d ln ) ln, so the coordintes of the centroid re M A, M ) ) ln A ln, ln ) ln+ ln 7 f) sin, [,π] The moments of the region re The re of the region is so the coordintes of the centroid re M π sin d π sin cos ) π ; nd π π M sin d cos + sin ) π π A sin d, M A, M ) π A, π ) 8 8 Clculte the moments nd center of mss of the lmin occuping the region between the curves nd for The moments of the lmin re M )d nd M )d 5 The re of the lmin is A )d 6, so the coordintes of the centroid re M A, M ) A, ) 5 Mrch,

34 SECTION 8 Center of Mss 9 9 Sketch the region between + nd for Using smmetr, eplin wh the centroid of the region lies on the line Verif this b computing the moments nd the centroid A sketch of the region is shown below The region is clerl smmetric bout the line, so we epect the centroid of the region to lie long this line We find M + ) ) ) d ; M + ) )) d 8 ; nd A + ) )) d 8 Thus, the coordintes of the centroid re 7 6, ) In Eercises 5, find the centroid of the region ling between the grphs of the functions over the given intervl,, [, ] The moments of the region re M )d nd M )d 5 The re of the region is A )d 6, so the coordintes of the centroid re 6 5, ),, [, ] The moments of the region re M )d nd M )d The re of the region is A )d, so the coordintes of the centroid re 9, 9 ) Note: This mkes sense, since the functions re inverses of ech other This mkes the region smmetric with respect to the line Thus, b the smmetr principle, the center of mss must lie on tht line Mrch,

35 5 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS,, [, ] The moments of the region re M [ ) ] ) d ) nd M ) d The re of the region is so the coordintes of the centroid re ) A ) d ln, ) 6ln, ln 6 e,, [, ] The moments of the region re M e )d e nd M e )d e e ) The re of the region is A e )d e, so the coordintes of the centroid re e ), e ) e ) ln,, [, ] The moments of the region re M [ ) ln ) ] ) d ln ) + ln ln ln ) ; nd M ) ln )d ln ) 9 ln The re of the region is ) A ln )d ln ln, so the coordintes of the centroid re ) 7 ln 8 ln, 8 ln 9ln ) 8 ln 5 sin, cos, [,π/] The moments of the region re M π/ cos sin )d π/ cos d ; nd π/ π/ M cos sin )d [ ) sin + + ) cos ] The re of the region is π/ A cos sin )d, so the coordintes of the centroid re π ) ), ) π Mrch,

36 SECTION 8 Center of Mss 5 6 Sketch the region enclosed b +, nd ), nd find its centroid A sketch of the region is shown below 5 The moments of the region re M + ) ) d + ) ) )5 5 M + ) ) )d d ) The re of the region is A + ) ) d so tht the coordintes of the centroid re 7 9, 6 5 ) 9, 8 ) ) d + ) 9 s 7 Sketch the region enclosed b, + ), nd ), nd find its centroid A sketch of the region is shown below The moments of the region re M ) + ) 6 d + ) 6 d 7 ; nd M b the Smmetr Principle The re of the region is A + ) d + ) d, so the coordintes of the centroid re, ) 7 In Eercises 8, find the centroid of the region ) ) 8 Top hlf of the ellipse + The eqution of the top hlf of the ellipse is 6 Thus, M 6 ) 6 d B the Smmetr Principle, M The re of the region is one-hlf the re of n ellipse with mjor is nd minor is ; ie, π)) π Finll, the coordintes of the centroid re, 6 ) π Mrch,

37 5 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS ) ) 9 Top hlf of the ellipse + for rbitrr,b > b Thus, The eqution of the top hlf of the ellipse is b b M b b d b B the Smmetr Principle, M The re of the region is one-hlf the re of n ellipse with es of length nd b; ie, πb Finll, the coordintes of the centroid re, b ) π Semicircle of rdius r with center t the origin The eqution of the top hlf of the circle is r Thus, M r r ) r d r B the Smmetr Principle, M The re of the region is one-hlf the re of circle of rdius r; ie, πr Finll, the coordintes of the centroid re, r ) π Qurter of the unit circle ling in the first qudrnt B the Smmetr Principle, the center of mss must lie on the line in the first qudrnt Therefore, we need onl clculte one of the moments of the region With,wefind M d The re of the region is one-qurter of the re of unit circle; ie, π Thus, the coordintes of the centroid re π, ) π Tringulr plte with vertices c, ),,c),, b), where,b,c >, nd b<c B smmetr, the center of mss must lie on the line connecting c, ) nd the midpoint /,b+ c)/) of the opposite side: l : b + c + c) + c Also b smmetr, the center of mss must lie on the line connecting,c) nd the midpoint c)/,b/) of the opposite side: l : b c c + c These lines intersect t one point cm, cm ) Equting the formuls for the two lines nd solving for ields c Substituting this vlue for into the eqution for l gives Hence, the coordintes of the centroid re b c c c c + c b + c, b + c ) Mrch,

38 SECTION 8 Center of Mss 5 Find the centroid for the shded region of the semicircle of rdius r in Figure 8 Wht is the centroid when r nd h? Hint: Use geometr rther thn integrtion to show tht the re of the region is r sin h /r ) h r h ) r h FIGURE 8 From the smmetr of the region, it is obvious tht the centroid lies long the -is To determine the -coordinte of the centroid, we must clculte the moment bout the -is nd the re of the region Now, the length of the horizontl cut of the semicircle t height is ) r r r Therefore, tking ρ, we find r M r d h r h ) / Observe tht the region is comprised of sector of the circle with the tringle between the two rdii removed The ngle of the sector is θ, where θ sin h /r, so the re of the sector is r θ) r sin h /r The tringle hs bse r h nd height h, so the re is h r h Therefore, Y CM M A r h ) / r sin h /r h r h When r nd h /, we find Y CM /)/ sin π Sketch the region between n nd m for, where m>n nd find the COM of the region Find pir n, m) such tht the COM lies outside the region A sketch of the region for nd is below Since m>n, the grph of n lies bove tht of m for between nd Thus the moments re M n m d ) n + n+ m + m+ ) n + m n m + n + )m + ) M n m )d n+ m+ d n + n+ ) m + m+ n + m + m n n + )m + ) Mrch,

39 5 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS The re of the region is A n m d n + m + Thus the center of mss hs coordintes ) n + )m + ) n + )m + ), n + )m + ) n + )m + ) In the cse grphed bove, for n, m, the center of mss is, ) 6, ) 6 nd Thus the point, 8 7 ) 8 7 < 6 m n n + )m + ) ) lies on nd then the curve lies below the center of mss of the region In Eercises 5 7, use the dditivit of moments to find the COM of the region 5 Isosceles tringle of height on top of rectngle of bse nd height Figure 9) FIGURE 9 The region is smmetric with respect to the -is, so M b the Smmetr Principle The moment bout the -is for the rectngle is M rect d 8, wheres the moment bout the -is for the tringle is The totl moment bout the -is is then M tringle 5 )d M M rect + M tringle Becuse the re of the region is + 6, the coordintes of the center of mss re, 9 ) 6 An ice crem cone consisting of semicircle on top of n equilterl tringle of side 6 Figure ) 6 FIGURE Mrch,

40 SECTION 8 Center of Mss 55 The region is smmetric with respect to the -is, so M b the Smmetr Principle The moment bout the -is for the tringle is M tringle d 5 For the semicircle, first note tht the center is, ), so the eqution is + ) 9, nd + M semi 9 ) d Using the substitution w, dw d, wefind M semi w + ) 9 w dw w 9 w dw w dw 8 + 7π, where we hve used the fct tht 9 w dw represents the re of one-qurter of circle of rdius The totl moment bout the -is is then M M tringle + M semi 7 + 7π Becuse the re of the region is 9 + 9π, the coordintes of the center of mss re ), 6 + π π + 7 Three-qurters of the unit circle remove the prt in the fourth qudrnt) B the Smmetr Principle, the center of mss must lie on the line Let region be the semicircle bove the -is nd region be the qurter circle in the third qudrnt Becuse region is smmetric with respect to the -is, M b the Smmetr Principle Furthermore M d Thus, M M + M + ) The re of the region is π/, so the coordintes of the centroid re 9π, ) 9π 8 Let S be the lmin of mss densit ρ obtined b removing circle of rdius r from the circle of rdius r shown in Figure Let M S nd MS denote the moments of S Similrl, let Mbig nd M smll be the -moments of the lrger nd smller circles r r FIGURE ) Use the Smmetr Principle to show tht M S b) Show tht M S Mbig M smll using the dditivit of moments c) Find M big nd M smll using the fct tht the COM of circle is its center Then compute M S d) Determine the COM of S ) Becuse S is smmetric with respect to the -is, M S using b) Mrch,

41 56 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS b) Becuse the smll circle together with the region S comprise the big circle, b the dditivit of moments, M S + Msmll M big Thus M S Mbig M smll c) The center of the big circle is the origin, so cm big ; consequentl, M big On the other hnd, the center of the smll circle is r, ), socm smll r; consequentl M smll smll cm Asmll r πr πr B the result of prt b), it follows tht M S πr ) πr d) The re of the region S is πr πr πr The coordintes of the center of mss of the region S re then πr ) r ) πr,, 9 Find the COM of the lmins in Figure obtined b removing squres of side from squre of side 8 8 FIGURE 8 Strt with the squre on the left Plce the squre so tht the bottom left corner is t, ) B the Smmetr Principle, the center of mss must lie on the lines nd 8 The onl point in common to these two lines is, ), so the center of mss is, ) Now consider the squre on the right Plce the squre s bove B the smmetr principle, cm Now, let s denote the squre in the upper left, s denote the squre in the upper right, nd B denote the entire squre Then M s 8 6 )d 8; M s )d 8; nd 6 M B 8 8 d 56 B the dditivit of moments, M Finll, the re of the region is A 6 56, so the coordintes of the center of mss re, ), 5 ) 56 7 Further Insights nd Chllenges A medin of tringle is segment joining verte to the midpoint of the opposite side Show tht the centroid of tringle lies on ech of its medins, t distnce two-thirds down from the verte Then use this fct to prove tht the three medins intersect t single point Hint: Simplif the clcultion b ssuming tht one verte lies t the origin nd nother on the -is Orient the tringle b plcing one verte t, ) nd the long side of the tringle long the -is Lbel the vertices, ),, ), b, c) Thus, the equtions of the short sides re c b nd b c b c Now, M b c/b) d + ) c c d c b b 6 ; b ) c c c + b) M c/b) d + d ; nd b b 6 M c Mrch,

42 + b so the center of mss is, c the medin from, ) nd the medin from, ) We find ) ) ) SECTION 8 Center of Mss 57 ) To show tht the centroid lies on ech medin, let be the medin from b, c), c b /), so c + b, so c b ), so + b + b + b ) c ; ) c ; ) c This shows tht the center of mss lies on ech medin We now show tht the center of mss is of the w from ech verte For, note tht b gives the verte nd gives the midpoint of the opposite side, so two-thirds of this distnce is b + ) b + b, the -coordinte of the center of mss Likewise, for, two-thirds of the distnce from to +b is +b, nd for, the two-thirds point is + ) b + b A similr method shows tht the -coordinte is lso two-thirds of the w long the medin Thus, since the centroid lies on ll three medins, we cn conclude tht ll three medins meet t single point, nmel the centroid Let P be the COM of sstem of two weights with msses m nd m seprted b distnce d Prove Archimedes Lw of the weightless) Lever: P is the point on line between the two weights such tht m L m L, where L j is the distnce from mss j to P Plce the lever long the -is with mss m t the origin Then M m d nd the -coordinte of the center of mss, P,is m d m + m Thus, nd L m d, L d m d m d, m + m m + m m + m m L m m d m m d L m m + m m + m Find the COM of sstem of two weights of msses m nd m connected b lever of length d whose mss densit ρ is uniform Hint: The moment of the sstem is the sum of the moments of the weights nd the lever Let A be the cross-sectionl re of the rod Plce the rod with m t the origin nd rod ling on the positive -is The -moment of the rod is M ρad, the -moment of the mss m is M m d, nd the totl mss of the sstem is M m + m + ρad Therefore, the -coordinte of the center of mss is m d + ρad m + m + ρad Smmetr Principle Let R be the region under the grph of f)over the intervl [,], where f) Assume tht R is smmetric with respect to the -is ) Eplin wh f)is even tht is, wh f) f ) b) Show tht f ) is n odd function c) Use b) to prove tht M d) Prove tht the COM of R lies on the -is similr rgument pplies to smmetr with respect to the -is) ) B the definition of smmetr with respect to the -is, f) f ),sof is even b) Let g) f ) where f is even Then nd thus g is odd g ) f ) f ) g), Mrch,