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# Calculus MAT 021B

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This 61 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 021B at University of California - Davis taught by Qinglan Xia in Fall. Since its upload, it has received 25 views. For similar materials see /class/187344/mat-021b-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15

65 Areas of Surfaces of Revolution Surface Area Suppose a curve is revolved along an axis and get a surface What is the area ofthat surface I Rotate a horizontal line segment pamllel to the axis of revolu tion We get a cylinder The smface area is S 27ryAx FIGURE 642 3 A cylindrical surface genemwd by rotating me horizontal line segment18 a length it about the Hxis has area ZWyAX 1le cm and rolled oul cylmdncal surface as a reclanglcl o Rotate a slanted line segment The surface area is S ZhylAs where y w and As length ofthe segment 0 Rotate a curve along an axis 3 6 sic 2k hymn Let the limit we have S 27r7ds b S 27rradiushandwidth a 2w bl nouns 643 31110 rmhim nfa cane generamd by mmihg the slanted line segment AB oflcngth A aboul the xaxis has area zhy39Ai b The area oflhercctangle for y39 y 2 ihe average height out above the x axis V b s z 4 A W dx Axis of nzvolnlioll FIGURE 651 The area oflhc surface swept out by revolving an AB mm the axis shown here is f 21m k The exam expression depends on ihc formulas for p and tin Here 0 the band width d5 is the arc length differential That is d5 xdac2 dy2 o p is the radius from the aXis of revolution to an element of arc length d5 Example Revolution about the x axis General forrn Here p lyl and the surface of revolution is S 7 27ryds dx dy 2 2 2dz wlyndt ltdtgt b S 27rpds 2dt In particular ifthe cuwe is given by y 06 Z 0 Then ac iH f a7das d d5 1d a and the surface ofrevolution is b S 2 1 77d 27ryt MM 0 b ZMxNT Home FIGURE am The surface generated by revolving the gmph urn nonuagativa function y x a s x s b abounhe xaxis The surface is a union of bands like Ihc one swept cut by Ihc am PQ Example Find the area ofthe surface generated by revolving the cuwey 2 1 g a g 4 about the xiaxis Example Revolution abom me liaxis General form b S 27r7ds Here p 2 and the sulface ofrevolution is s 2alalds 2w z l7 grim In particular ifthe curve is given by 2 m 2 0 Id ds d71dy ag y71dy 5 dz 2 s 7127r2dy1dy b27 y y2 My Example Findthe area ofthe surface generatedby revolving the curve 2 slny0 g y 3 7r about the yiaxis Then and the surface ofrevolution is Example The line segment a 1 7 y 0 g y g 1 is revolved about the yiaxis to generate aeone Find its lateral surface area which excludes the base area FIGURE 6149 Revolving line segment AB about the yraxis generates a cone whose lmeml surface area we can now calculate in No differs ways Example 2 male 39 H 39 J 39 quot 7y7 39 linexl7withl7gtagt0 Le surface area ofatorus The lheoi39ems of Pappus 3rd century THEOREM 1 Pappus39s Theorem for Volumes If n plane region is revolved once about a line in the plant thal does not out through the region s interior then the volume ufthe solid it generates is equal In the region s area times the distance traveled by the region s centroid during the revolution Ifn is the distance from the axis ofrevolution to the centroid then V 2mm 9 Idea of proof By the shell method V ZwyL7Jdy But for the centroid f ydm f mm A g fdm Therefore v 2mm Example The disk 22 y a2 is revolved about the vertical line 2 bWith b gt a gt O to generate atorus Find the volume of the solid We have calculated it before But indeed the centroid is clearly O 0 here so THEOREM 2 Pappus39s Theorem for Surface Areas 3911 1 u or n F L c luv an thrrmmlminn 39 39 39 39 39 mm S 2m 11 Idea of proof The surface area is y S 27rde But for the centroid 92 ofthe arc 7 7 f dm 7 fyds y f d5 L Therefore I7 6 s 2ng f Example Find the area of the surface generated by revolving the circle 22 y a2 about the vertical line 2 b With b gt a gt O ie surface area of a torus We have calculated it before But indeed the centroid is clearly O 0 here so 65 Areas of Surfaces of Revolution Surface Area Suppose a curve is revolved along an axis and get a sur face What is the area of that surface o Rotate a horizontal line segment parallel to the axis of revolution We get a cylinder The surface area rs S 27ryAx Try a 3 area ZwyAx a The GUI and rolled out cylmdrical surface as a rectangle o Rotate a horizontal line segment parallel to the aXis of revolution We get a cylinder The surface area is 27y S 27ryAx area ZvyAx a The cut and rolled cu cylindrical surface as a rectangle As 0 Rotate a slanted line segment The surface area is 2 quot a S 27ry A5 where y g and A5 length on Of the segment FIGURE 613 aThc Cruslumofa com generalcd by mutinglhc slanted inc segment1E of length Ax about hexaxis has area Jquot 12th avemge 27039AL b The area oflh rcclangle for y height 0MB above the x axis o Rotate a curve along an axis 5 25k ZanZAsk k 16 Let the limit we have FIGURE 5 51 Th area ofthc surface swept out by b S 27139 pd s a rcvulving an AB about the axis shawl here is ffzirp dx b S 27f ra dius ban d Width The cxaclexprcssion depends on me formulas fur mm d a Axh or mwlmmn Here 0 the band width d5 is the arc length differential That is d5 idz2 6132 2dt o p is the radius from the axis of revolution to an element of arc length d5 Example Revolution about the w axis General form b S 27Tpds a Here y p lyl and the surface of revolution is H S 27Tyds dx dy H W ltEgt2ltEgt2dt L In particular ifthe curve is given by y f 96 2 0 Then and the surface of revolution is b d y 2 FIGURE 5 M The surface i generated by S a 27W 1 dx d9 revolving um gmph ofaxmnnegzlive b fu 39 ncllon y m u s r s In about the 7 2 XaxisTl surfaec isaumcnofbandshkc 27TH 1 f d9 heoncswcptumbylhearcPQ Example Find the area of the surface generated by revolving the curve y Z g x g 4 about the xiaxis We evaluate the formula 4 dy S 2 1 2d A WW Hm 56 4 1 27T2 1 dx 1 IE 4 47nx1dx 1 47r ltx1jl 87T 32 32 35 2 Example Revolution about the yiaxis General form I x S 27Tpds 1 Here p and the surface of revolution is d d S 27T I d827T IM 22dt In paiticular if the curve is given by I 99 2 0 d2 7 2 7 2 d571dy ldyiqgy ldy Then and the surface of revolution is 5 dx 8 27m 2ldy a tltdygt 9 b27rgltygt lltygt2 1dy Example Find the area of the surface generated by revolving the curve x siny 0 g y 3 7T about the y aXis and the surface of revolution is 5 dx S27Tx 2ldy a Ugly 9 gramme Example Find the area of the surface generated by revolving the curve x siny 0 g y 3 7T about the y axis We evaluate the formula 7T 5 2 d2 0 dy 7T 27T sinltygt xoos2y ldy 0 Example The line segment x l y 0 g y g l is revolved about the y axis to generate a cone Find its lateral surface area Which excludes the base area 7m 2ldy Example Find the area of the surface generated by revolving the circle x2 32 a2 about the vertical line x b with b gt a gt O ie surface area of a torus Example Find the area of the surface generated by revolving the circle x2 32 a2 about the vertical line x b with b gt a gt O ie surface area of a torus The circle may be expressed parametrically as x aoosty asint 0 g t g 27T The radius p b x Thus the surface area is S 27Tpds AZ 0 7T27rb x 22dt 0 27Tb a oost a2 sin2t a2 oos2tdt 27Tb a ooslttgtgtadt The theorems of Pappus 3rd century THEOREM 1 Pappus s Theorem for Volumes If a plane region is revolved once about a line in the plane that does not cut through the region s interior then the volume of Ihc solid il generates is equal to the region39s area times the dismnee traveled by the region39s centroid during the revolution pr is the distance from the axis ofrcvolution to the centroid than V 2 2mm 9 Idea of proof By the shell method V 27TyLydy a eeeeee WH But for the centroid 7117 P Lm g f Wm W p f dm A F Therefore 7 W V 27TQJA Example The disk x2 y2 a2 is revolved about the vertical line x b With b gt a gt O to generate a torus Find the volume of the solid Example The disk x2 y2 a2 is revolved about the vertical line x b With b gt a gt O to generate a torus Find the volume of the solid We have calculated it before But indeed the centroid is clearly O 0 here so V 27Tb7ra2 27TZCL2Z THEOREM 2 Papnus39s Theorem for Surface Areas not cut through the arc s interior Illen the area of the surface genemled by the arc equals lhe lenglh of L 39 L 39 L quot quot 39 the revolution If r 39 39 ua Acu by p 39 then S 217pL 11 Idea of proof The surface area is S 27Tyds But for the centroid 7117 of the arc 7 f dm fyds y 7 7 fds L Therefore S 2m L Example Find the area of the surface generated by revolving the circle x2 32 a2 about the vertical line x b with b gt a gt O ie surface area of a torus Example Find the area of the surface generated by revolving the circle x2 32 a2 about the vertical line x b with b gt a gt O ie surface area of a torus We have calculated it before But indeed the centroid is clearly 0 0 here so V 27TpL 27Tb27ra 4W2ab 64 Mements and Centers of Mass Center of Mass Suppose two masses m1 m2 are located on a rigid line supported by a fulcrum X l 0 To get balanced Where shall we put the fulcrum Center of Mass Suppose two masses m1 m2 are located on a rigid line supported by a fulcrum X l 0 To get balanced Where shall we put the fulcrum o If m1 m2 then Center of Mass Suppose two masses m1 m2 are located on a rigid line supported by a fulcrum X l 0 1 2 To get balanced where shall we put the fulcrum o If m1 m2 then 1 l2 0 If m1 y m2 eg a baby and an adult on a seesaw then Center of Mass Suppose two masses m1 m2 are located on a rigid line supported by a fulcrum X l 0 1 2 To get balanced where shall we put the fulcrum o If m1 m2 then 1 l2 0 If m1 y m2 eg a baby and an adult on a seesaw then migli 7712912 where g is the acceleration of gravity That is m1l1 mglg m N on the line What happens when there are a lot of masses m1 mg m We usually want to know where to put the fulcrum to make the system balance that is at what point 97 to place it to make the torques add to zero For each mass mk the torque of mk about i is 90k mm The system torque about 9 equals to Zkxk 90mm Thus The system is balanced at f ltgt System torque about 97 0 ltgt 97ka 0 42gt 52mk 0 ltgt Zkak Zimk0 ltgt Zkmeka 42gt i m 2 2 Wk M Here the number M0 Z mkf k is called the moment of the system about the origin MIka M0 the moment of the system about the origin is the system s total mass and j M the system s total mass is called the center of mass Example Find the center of mass for the following masses mas III ll 6 System 5 total mass Example Find the center of mass for the following masses mas III ll 6 System 5 total mass System In om ent Example Find the center of mass for the following masses mas III ll 6 System s total mass 1 l l M7 7 71i 236 System moment 1 l l l l M772 771 727177 77li 0 2lt gt3lt gt6 33 Therefore the center of mass is Example Find the center of mass for the following masses mas III ll 6 System s total mass 1 l l M7 7 71i 236 System moment 1 l l l l M772 771 727177 77li 0 2lt gt3lt gt6 33 Therefore the center of mass is Wires and Thin Rods How to nd the center of mass of a rod or a thin strip of metal Wires and Thin Rods How to nd the center of mass of a rod or a thin strip of metal Idea Approximation Cut it into small pieces of mass Amk by a partition of the interval a b Wires and Thin Rods How to nd the center of mass of a rod or a thin strip of metal Idea Approximation Cut it into small pieces of mass Amk by a partition of the interval a b Then the total mass of the system is approximated by Z Amk Z 5ltkgtALEk k k Where 6 is the density function of the rod at x Also the system moment is approximated by k k Now we let the norm of partition P approaches 0 then M Ab x ix 1 M0 6 a and the center of mass is j M0 fabw x ix M f 6xdx Example Show that the center of mass of a straight thin strip or rod of constant density 6 C lies halfway between its endpoints FIGURE 630 The center 0f mass of a straight thin rod Cur strip of constant density lies halfway between its ends Example I Example Show that the center of mass of a straight thin strip or rod of constant density 6 C lies halfway between its endpoints FIGURE 630 The center 0f mass of a straight thin rod Cur strip of constant density lies halfway between its ends Example I M EleCMf2ba Al omC 2 j Example Find the center of mass of a rod of density 6 x2 over 0 10 Example Find the center of mass of a rod of density 6 x2 over 0 10 10 4 10 4 4 103 4 3 0 m 10 2 J M x dx D 0 FIGURE 631 We can treata and of SO center OfmaSS IS V rlatblf thickness as a rod of variable M0 104 3 15 densrty Example 2 LE M 4 103 2 Masses distributed over a plane region Suppose that we have a nite collection of masses located in the plane with mass mk at the point wk Each mass mk has a moment about each axis 0 moment about x axis mkyk ml a moment about y aXis mkxk quotllquot V k 3 System moment about 0 x aXis x Mm Z mkyk k FIGURE 632 Each mass mk has a i mc39m nt aboutcaczh axis 9 3X 5 My 2 WM 19 System mass is M Z mk k F The center ofmass 92 y39 is quote 2kakxk M 2k mic 7 2k mkyk y M 2k mic FIGURE 63 A vbdimensional army ofmasses balances on us ccnlcr ofmass Thin at plates How to nd the center of mass of a thin at plate Idea Cut 1t Into strlps parallel to one v I of the axes Then for each strip nd Stnp ol mans lm 1ts mass Amk and center wk So 0 Moment about the y axis b A Amik qimn k a o Moment about the x axis FIGURE 634 A plate cut into thin strips b parallel to theiraxis The moment exerted Mm gdm by a typttal strip abnut each am IS the a moment its mass Am would exert if M concentrated at the slrip s center of mass ass b an MZdm a Center of mass My fjidm x b M fadm gf5dm M ffdm Thus to calculate the center of mass we need i g and dm 0 Sketch a strip parallel to one of the axes 0 nd the strip s mass dm and center of mass i 0 Use the formula to nd Mm My M j g Example Find the center of mass of a thin plate of constant density 6 cov ering the region bounded above by the parabola y 4 2 and below by the x axis Example Find the center of mass of a thin plate of constant density 6 cov ering the region bounded above by the parabola y 4 x2 and below by the x aXis We cut the region into small pieces of vertical strips The typical verti cal strip has 393 4 x2 Center of mass 0 center of mass is I a mass is gtIdxllt dm 6dA 64 x2dr 2 2 32 M dm 64 r2dx 6 2 2 3 So because the integrand is an odd function on a symmetric interval Therefore My 0 x M Mx 68 yM 3735 5 The plate s center of mass is the point f g 0 5 In the above example what happens if the density at the point x y is 6x y 2x2 because the integrand is an odd function on a symmetric interval Therefore My 0 x M Mx 68 yM 3735 5 The plate s center of mass is the point f g 0 5 In the above example What happens if the density at the point x y is 6x y 2x2 Here i g is still same but dm 6dA 2x24 x2dx So 2 2 Mdm 2x24 x2dx 2 15 So O2M8 w8 x y1 67 Example Find the center of mass of a thin plate of constant density 6 cov ering the region bounded above by y Rltxgt and below by y x over the interval a b Where Rltxgt gt Mr for each x E a b Recall that My abidm x b M fadm gf dm M fjdm So we need to nd i g and dm for a typical strip For a typical yertical strip a center of mass i x R 1 g 6 Rltx22rlt2 rltgt wa My 6Rx 2 71 dm 6 dA 6R rltxgtgtdx Thus M dm lb 6Rx rltxgtgtdx 6A Therefore fbRx rltxgtgtdx a foam 7mm f R5L Cm mam rltxgtgtdx Remark When the density is a constant 6 the center of mass f g is independent of 6 Thus the location of a center of mass is a feature of the geometry of the object and not of the material from which it is made In such cases engineers may call the center of mass the centroid of the shape To nd a centroid one may simply set 6 l and then nd E and g as before j g

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