Chpter 6 Applictios Itegrtio Sectio 6. Regio Betwee Curves Recll: Theorem 5.3 (Cotiued) The Fudmetl Theorem of Clculus, Prt :,,, the If f is cotiuous t ever poit of [ ] d F is tiderivtive of f o [ ] ( ) = ( ) = ( ) ( ). f d F F F Defiitio: Let f ( ) e fuctio defied o closed itervl [, ] f over [, ] is ( ) lim ( ) f d= f k =. The the defiite itegrl of if the limit eists. Commets: f ( ) d is red the itegrl from to of f of dee or sometimes s the itegrl from to of f of with respect to. Recll from sectio 4.8, is the itegrl sig, f ( ) is the itegrd of the itegrl d is the vrile of itegrtio. is the lower limit of itegrtio d is the upper limit of itegrtio., coverge k Whe the defiitio is stisfied, we s the Riem sums of f o [ ] to the defiite itegrl I = f ( ) d d tht f is itegrle over [, ] The vlue of the defiite itegrl depeds o the fuctio ot the letter we choose to represet the vrile of itegrtio. So if f ( ) d = I, the ( ) f t dt = I or f ( u ) du = I. The vrile of itegrtio is clled dumm vrile.
Are Fid the re etwee the curve = 4, the -is ( 0) = d the -is ( = 0). Ares Betwee Curves Emple: Fid the re eclosed = d =. Emple: Fid the re of the regio ouded = d = 6
Emple: Fid the re of the regio eclosed 3 = d =. Emple: Fid the re eclosed = d = +. Sectio 6.3 Volumes Slicig Three Tpes of Volume Prolem: A. Are of the cross sectio is costt. This occurs for right cliders. A right clider is otied whe ple regio is moved log lie perpediculr to ple regio through distce h. I this cse, the volume is ( re ) B. Are of cross sectio is ot costt. V = R h = Ah whe A is costt. 3
To fid the volume of the solid S tht eteds cross the -is from = to = with cross-sectiol res perpediculr to -is, where ech cross-sectio hs re tht vries from poit to poit.. Divide S ito slices perpediculr to -is, hvig width =.. Volume, V = V+ V + + V = Vi, V i is pproimtel the re of the i= rectgulr fce of the slice times the width of the slice. I other words, V A i i 3. So V Ai where A i is fuctio of. i= i i = 4. Thus = lim ( ) = ( ) V A A d Emple : The se of certi solid is the regio eclosed =, = 0, d = 4. Ever cross sectio perpediculr to the -is is semicircle with its dimeter cross the se of the eclosed regio. Fid the volume of the solid. 4
C: Volumes of revolutio: Volume otied revolvig ple regio out is. Emple : Fid the volume of the solid otied rottig the regio ouded = e, = 0, = 0, d = out the -is. Emple 3: Fid the volume of the solid otied rottig the regio ouded = 6, = 0, d = out the -is. Emple 4: Fid the volume of the solid otied rottig the regio ouded =, = 0, =, d = 3 out the lie =. 5
Emple 5: Fid the volume of the solid otied rottig the regio ouded = d = out the =. Sectio 6.4 Volumes Shells Suppose we cosider revolvig the followig regio out the -is. Notice tht the disk/wsher method does ot provide ice solutio here. But cosider the techique we used. I sectio 6.3, volume is = = ( ) V dv A d, where dv is the volume of slice of the solid. Usig this ide, let s cosider differet w to cut the solid ito pieces. Isted of slicig the solid ito disks, let s cut the solid ito rigs/shells. Just s i sectio 6.3, the volume V = V so let s fid the volume of shell. i= i 6
So the volume of clidricl shell is V = π r h r. Note, π r is the circumferece of the shell, h is the height, d r is the thickess of the shell. For our prolem the dv = π r h r d r is the distce from the shell to the is of rottio, i.e. r h= f, d r = = d. So the volume of the solid V f d. is = π ( ) =, d height is ( ) Notice we c lso just use our origil formul from sectio 6.3. Suppose we slice shell d uroll it so tht it is flt. f ( ) π Now V = A( ) d where A( ) = π f( ) d ( ) urolled shell. A is the re of the fce of the Emple : Fid the volume of the solid geerted whe the regio ouded = is rotted out the -is. =, = 0, d 7
Emple : Fid the volume of the solid geerted whe the regio ouded =, = 0, d = is rotted out the -is. Emple 3: Fid the volume of the solid geerted whe the regio ouded =, = 0, d = 5 is revolved out the lie = 3. Emple 4: Which method should ou use? Fid the volume of the solid geerted whe the regio ouded rotted out the -is = d = is 8
Sectio 6.5 Legths of Curves ( ) ( ) To fid the legth of curve from, f ( ) to, ( ) Crete smll rcs Legth is the sum of the smll rcs Wht is the pproimte legth of segmet? f : Δs d Δ Δ So s ( ) + ( ) = ( ) + = + ( ( )) = + f ' Me Vlue Theorem Arc Legth is = lim i = lim + ( '( )) = + ( '( )) i= i= L ds f f d Emple: Fid the legth of the curve l = for 4 4 Emple: Fid the legth of the curve 3 3/ / = for 9 9
Sectio 6.6 Phsicl Applictios From phsics, force is the ifluece tht teds to cuse motio i od. Work Doe Costt Force If od moves distce, d, i the directio of pplied costt force, F, the the work doe is W = F d For emple, the work to lift 90 l g of cocrete 3 feet is W = Fd = 90l 3 ft = 70 ft l ( )( ) Commo Uits of Work d Force Mss Distce Force Work kg M ewto Joules g cm de Erg slug ft poud Ft-l Work Doe with Vrile Force Suppose F is cotiuous fuctio d F( ) is vrile force tht cts o oject movig log the -is from To determie work: = to =. Prtitio the itervl [, ] ito suitervls of legth. If the the force is essetill costt o the suitervl. So ( ) suitervl d i i = ( ) ( ) W = lim F = F d is sufficietl smll, W F for ech Emple: A oject locted ft from fied strtig positio is moved log stright rod force F ( ) = ( 3 + 5) l. Wht work is doe the force to move the oject from = to = 4 feet? i 0
Hooke s Lw Whe sprig is pulled uits pst it equilirium (rest) positio, there is restorig F = k tht pulls the sprig ck towrd equilirium. force ( ) Emple: The turl legth of certi sprig is 0 cm. If it requires joules of ergs of work to stretch the sprig to 8 cm, how much work is will e eeded to stretch the sprig to 0 cm? Emple: A ucket tht weighs 4 l d rope of egligile weight re used to drw wter from well 80 ft deep. The ucket is filled with 40 l of wter d is pulled up t rte of ft/sec, ut the wter leks from hole i the ucket t rte of 0. ls/sec. Fid the work to pull the ucket to the top of the well. Emple: A clidricl tk of rdius 3 feet d height of 0 feet is filled to depth of 3 feet with liquid of desit ρ = 40 l/ft. Fid the work doe i pumpig ll the liquid to height of feet ove the top of the tk.
Emple: You re i chrge of the evcutio d repir of hemisphericl storge tk (with se dow, see picture i tet). The tk hs rdius 0 ft d is full of ezee 3 weighig 56 l/ft. A firm ou cotcted ss it c empt the tk for ½ cet per footpoud of work. Fid the work required to empt the tke pumpig the ezee to outlet feet ove the top of the tk. If ou hve $5000 udgeted for the jo, c ou fford to hire the firm? Force d Pressure ( re of strip)( pressure) ( ) ρ ( ) F = = w g So force is ( ) ( ) F = ρg w d Emple: The lower edge of dm is defied the prol = /6, with the top t = 5. Determie the force o the dm. Legths re mesured i meters.
Sectio 6.7 Logrithmic d Epoetil Fuctios Revisited. Defiitio: The Nturl Logrithm The turl logrithm of umer > 0 deoted l ( ) is defied s l = dt t Theorem 6.4 Properties of Nturl Logrithms. The domi d rge of = l re ( ) d u' ( ). ( l ( u( ) )) =, u( ) 0 d u ( ) 3. l ( ) = l + l, ( > 0, > 0) = > > l p = pl > 0, p re rel umer d = l + C 4. l l l, ( 0, 0) 5. ( ) 6. 0, d (, ), respectivel. Defiitio: Bse of the Nturl Logrithm The turl logrithm is the logrithm with se of ( ) / follows tht000 l e =. e= lim + h.7888. It Theorem 6.5 Properties of e The epoetil fuctio f ( ) = e stisfies the followig properties, ll of which follow from the itegrl defiitio of l. Let d e rel umers +. e = ee e. e = e 3. ( ) e = e 4. l ( e ) 5. e l = =, > 0 h 0 h 3
Theorem 6.6 Derivtive d Itegrl of the Epoetil Fuctio For rel umer, d ( ) ( ) ( e u u ) e u '( = ) d e d = e + C d Geerl Logrithmic d Epoetil Fuctio Chge of Bse Rules (pge 3) Let e positive rel umer with. The = e, for ll d log l l =, for > 0 l 4