微積分上課ap cal課件合集c

1、Chapter 10ApplicationofIntegral10.1MeanTheMeanValueTheoremforIntegralsValueTheoremforIntegrals(MVTI)If f (x) is continuous on a,b, there is a number c betweenaand b, such thatbòf (x)dx =f (c)(b - a)ay = f (x)f (c)Aca10.1MeanTheMeanValueTheoremforIntegralsValueTheoremforIntegrals(MVTI)Theb.numbe

2、rf(c)givesustheaveragevalueoff(x)ona,y =f (x)f (c) 1(b - a)bò=f (c)f (x)dxaAcaExample1Findtheaveragevalueoff (x) = x2from x = 2 to x = 4.Solution: 14 - 24òf (c) =x2dx23= 1 ( x) |4 = 282233Example210.2AreaVerticalSlicesbòyA =y =f (x)dxf (x)abòA = f (x) - 0dxoroxabaxx + Dx10.2AreaV

3、erticalSlicesyxbax + DxbòA = -f (x)dxoabòA =0 -orf (x)dxy = f (x)a10.2AreaVerticalSlicesx)2bòA = f2 ( x) -f1( x)dxa2 x)RectangularCoordinateSystembA = òbf (x)dxòA = -f (x)dxaax)2bòA = f2 (x) -f1(x)dxaFind the area of the region Rbounded byExample3y = 4 - x2and y = 3x.So

4、lution:Findthepointsofintersection:(-4, -12),(1, 3)dA = (4 - x2 - 3x)dx1òA =(4 - x2 - 3x)dx-4ù112532=4ë2ú.63û-4Example4Example510.2AreaHorizontalSlicesdòA =g( y)dycdòA =g2 ( y) - g1( y)dyoy + DycyFind the area of the region R bounded byy = 4 - x2 and y = 3x.Example

5、6of intersecti(-on4, -12),(1, 3)Find the pointsSolution:y ,x= ±4 - y ,=x123( y +334òòA =4 - y )dy +4 - ydy2-123Choose the right slices!Find the area of the region R bounded byExample7= 2x and y = x - 4.y2Solution:Findthe pointsofintersection,y = xì y 2= 2 xí yÞ (2, -2),

6、 (8, 4).=x - 4îChooseastheintegralyy2= 2xvariable,ææ y + 4 -ö dy = 18.2y4öòy2A =ç÷dA = ç y + 4 -÷ dy2-èø22èø1.WhichofthefollowingintegralscorrectlyExercisecorrespondstotheareaoftheshadedregionin= 1+ x 2thefigureabove?y2ò( x-

7、4)dx2( A)12y=5Bò(4 - x)dx2(B)15ò(x- 4)dx2(C )15ò( x+ 4)dx2(D)x=115ò(4 - x)dx2(E )12.FindtheareaoftheregionbetweenExercisethe twocurvesineachproblem.y = x21) the2) thecurvecurveandthecurvey = 4x - x 2.y = x2 - 4x - 5y = 2x - 5.andthecurve10.2AreaCalculus of Polar CurvesFindIfareai

8、npolarcoordinateswewanttofindtheandareaofaregionr = f (q ), a £ q £ bbetweentheregionthecurvedqoArea = ò 1r 2dqba2Example8Findtheareaoftheregionintheplaner = 4 + 4sinqenclosedbythecardioidy8Solution:1 (4 + 4 sinq )2dq2pòA =202p= ò(8 +16 sin q + 8sin2 q )dq0x-442p0ò(12 +

9、16 sin q - 4 cos 2q )dq=0= 24p10.3VolumeofSolidsofRevolutionCircleCylinderCircle ConeSphereSolidofrevolutionisasolidgeneratedbyaplaneregionthe axis.revolvesaboutafixedline.ThislineiscalledWashersandDisksDiskWashersWashersandDisks(1)Ifbythesolidofrevolutionobtainedy =fyRrevolvingtheplaneregionaboutth

10、ex-axisboundedbyx + dxxxy=f(x)and x=a, x=band xaxis.Whatsthevolumeofit?dV = p f (x)2 dxTheelementofVis:Thenthevolumeofthesolidis:bV = ò p f ( x)2 dxaFindwhenthethevolumeregionofthesolidthatresultsExample9betweenthecurvey =xandthex-axis,fromx=0tox=4,is revolvedaboutthex-axis.Findwhenthethevolume

11、regionofthesolidthatresultsExample10betweenthecurveandthex-axis,fromx=0tox=4,y =xis revolvedaboutthex-axis.Solution:4òV = px )2 dx(02= p x|402Washers (2)andDisksy =fy = g (xy = f (x)y = g (x)TheplaneregionRboundedabovebyy=f(x)andbelowsolidbyofy=g(x)ontheintervalobtainedx=a,x=b.Thentherevolution

12、isbyrevolvingtheregionaboutx-axis.WashersandDisksTheplaneregionRboundedabovebyy=f(x)andbelowbyy=g(x)ontheintervalx=a,x=b.Thenthesolidofrevolutionaxis.isobtainedbyrevolvingtheyregionaboutx-thevolumeofthesolidis:y = fy = g (xxobV = ò p f (x)2 - g(x)2 dxaFindwhen y=xthethe andvolumeregionofthesoli

13、dthatresultsExample11bounded by thecurvey=x2,fromx=0tox=1,isrevolvedaboutthex-axis.y = xSolution:yy = x21òV = p- x4 )dx2(x0x2px3x5= p (0-) | =103515WashersandDisks(3)Ifthe solidofrevolutionobtainedbyrevolvingtheplaneregion Rby x=g(y)y=c,y=dyboundedandandaxis.Choose yasintervalvariable,y dy + Dy

14、ycoy,y+dy,andThechoosesubintervalx = g( y)element ofVis:dV = p g( y)2 dydV = ò p g ( y)2 dycFindwhenthethevolumeregionofthesolidthatresultsExample12boundedbythecurvex=2/y,fromy=1toy=4,isrevolvedaboutthey-axis.FindwhenthethevolumeregionofthesolidthatresultsExample13boundedbythecurvex=2/y,fromy=1

15、toy=4,isrevolvedaboutthey-axis.2y4Solution:òV = p2() dy144(1= p ò)dyy21p (-p= 4) | = 341yWashersandDisksIfthesolidofrevolutionobtainedbyrevolvingtheplane(4)Rx=f(y)regionboundedabovebyandbelowbyx=g(y)interval y=c,y=d.ontheyy = g (x)y =thevolumeofthesolidis:fodV = ò p f ( y)2 - g( y)2dy

16、cFindwhen x=y2thethe andvolumeregionofthesolidthatresultsExample14bounded by thecurvex=y3,fromy=0toy=1,isrevolvedaboutthey-axis.x = y3ySolution:x = y21òV = p- y6 )dy4( y0x2py5y7= p (0-) | =105735Findwhen y=x2thethe andvolumeregionofthesolidthatresultsExample15bounded bythex=1,curvey=x,fromx=0to

17、isrevolvedaboutthelineyy=-2.y = xSolution:y = x2xy=-21òV = p+- (x2+ 2)2 )dx2(x2)0= 4p05Findwhen y=x2thethe andvolumeregionofthesolidthatresultsExample16bounded bythey=1,curvey=x,fromy=0toisrevolvedaboutthelinex=-2.y = xSolution:yy = x2x1òV = py + 2)- ( y + 2) dy22(0= 5p06x=-2Exercise 1.Aso

18、lidisgeneratedwhentheregioninthefirstquadrantenclosedbythegraphofy = (1+ x2 )3thelinex=1,thex-axis,andthey-axisIsrevolvedaboutthethex-axis.Itsvolumeisfoundbyevaluating+ 1)3dxwhichoffollowingintegrals? ( A)p ò8 ( x 2(B)p ò(C )Bò0(D)1òp+ 1)6dx2(E )2( x0Exercise 2.Thevolumegenerated

19、byrevolvingaboutthexaxisthe= 9 - x2and y = 9 - 3xregionenclosedbythegraphs y,for0 £ x £ 3isB( A) - 8p(B) 4p(C) )8p(D)24p(E )48pCylindricalShellsy = g(x)yregion RTheplaneboundedy=f(x)abovebybyandbelowy =f (x)x=a,they=g(x)ontheintervalxx=b.Supposewerevolveab0regionaboutthey-axis,thenwegetthe

20、volumeofthesolid:CylindricalShellsy = g(x)yg(x) - f (x)yxab0bV = òa 2p × xg(x) - f (x)dx= f (x)CylindricalShellsregion Rby x=f(y)Theplaneboundedaboveandbelowbyinterval y=c,y=d.x=g(y)ontheSupposewerevolvetheregionsolid:aboutthex-axis,thenwegetthevolumeofthedV = òc2p × yg( y) - f (

21、 y)dyFindwhenthey =volumeoftheregionthatExample17boundedbythecurvexthex-axis,andthethelinex=9is.Set up butrevolvedabouty-axisdo not evaluate theIntegral.Solution:y9xV = ò (2p xx=9x )dx003òV = p ´ 9´ 3 -(p y )dy240Findthevolumeoftheregionthatthewhenx-axis,Example18y =xboundedbythe

22、curveandx=-2thelinex=9isrevolvedaboutthe.Set up but do not evaluate the Integral.Solution:yxx=90x=-29òp (xV =+ 2)2x dx0Exercise1.Usethemethodofcylindricalshellstofindthevolumeofthesolidthatresultsy =whentheisregionboundedbyxy = x -1and x = 0axis.revolvedaroundthey2.Usethemethodofcylindricalshel

23、lstofindthevolumeofthesolidthatresultswhentheisregionboundedbyy = 4 andaxis.x = 0revolvedaroundthexy = x2VolumesofSolids withKnownCross-sectionsConsiderasolidwiththepropertythatcrosssectionsxperpendiculartoaxishaveknownarea.A(x) is the area of thecross section at xx + dxabx,x+dx ,andchoosesubinterva

24、ldV = A(x)dx,bòV =A( x)dx.aFindthevolumeofthesolidwhosebaseExample19istheregionbetweenthesemi-circleandthex-y =16 - x2axis,andwhosecross-sectionsperpendiculartotheSolution:x-axis aresquares withasideonthebase.= 2564òV =(16 - x2 )dx-43ExerciseFindthevolumeofthesolidwhosebaseistheregionbetweenperpendicularandy=4,andwhosecross-sectionstoX-axisaretheisoscelesbase.righttriangleswiththehypotenuseony1