AP Calculus BC Review Applications of Integration (Chapter 6) noting that one common instance of a force is weight


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1 AP Clculus BC Review Applictions of Integrtion (Chpter Things to Know n Be Able to Do Fin the re between two curves by integrting with respect to x or y Fin volumes by pproximtions with cross sections: isks (cyliners, wshers, n other shpes Fin volume by cylinricl shells: (rius r, height h, n thickness r gives volume V = rhr Fin work one using the formul W = Fx, noting tht one common instnce of force is weight Prctice Problems For ll problems, show correct, lbele igrm n complete setup of the problem in terms of single vrible. Use correct units where pplicble. This is esigne to be one with clcultor. Remember, when giving pproximte nswers, to give three eciml plces. Let R be the she region boune by the grphs of y= x n y = e x n the verticl line x =, s shown in the figure t right. Fin the re of R. b Fin the volume of the soli generte when R is revolve bout the horizontl line y =. c The region R is the bse of soli. For this soli, ech crosssection perpeniculr to the xxis is rectngle whose height is 5 times the length of its bse in region R. Fin the volume of this soli. A continer is in the shpe of regulr squre pyrmi. The height of the pyrmi is ft n the sies of the squre bse re ft long. The tnk is full of liqui with weight ensity Fin the work one in pumping the liqui to point ft bove the top of the tnk. 8 lb ft. Let R be the region in the first qurnt boune by the grph of y = x x n the xxis. Fin the volume of the soli generte when R is revolve bout the ( xxis n (b yxis R If the force F, in ft lb, prticle from F x cting on prticle on the xxis is given by ( x = ft to ft ft lb b x = is equl to ft lb c ft lb ft lb e 7 =, then the work one in moving the x ft lb 5 The bse of soli is circle of rius, n every plne crosssection perpeniculr to one specific imeter is squre. The soli hs volume 8 b c e 8 The region whose bounries re = n y x x y = is revolve bout the xxis. The resulting soli hs volume ( x + x x b ( x x x c ( y yy e y y x x x
2 7 The re of the region enclose by the grphs of y = x n y = x is b c 5 e 8 When the region enclose by the grphs of y = x n y x x soli generte is given by ( ( x x x b ( x xx x c ( ( x x x e ( x x x = is revolve bout the yxis, the volume of the x x x Wht is the volume of the soli generte by rotting bout the xxis the region enclose by the grph of y = sec x n the lines x =, y =, n x =? b c 8 e ln( + If the region in the first qurnt boune between the yxis n the grph of ( the xxis, the volume of the soli generte is given by ( ( y y y b ( y ( y y 8 e ( ( 8 x ( x x x 8 c x( x x x x x = y y is revolve bout x Fin the re enclose by the grphs of b = + n y = x. 7 c y x x x e 85
3 Answers. b. c.55 8 ft lb 8. b.88 5 b 5 b 7 8 e c Solutions First we nee to fin the beginning of the intervl over which to integrte, which is the point of intersection of y= x with x y = e. Therefore we solve x x = e ; this cnnot be solve for x exctly, but n pproximtion cn be foun:.. Since the top function is y= x, the bottom function is x y = e, n the upper limit is, x we integrte ( x e x x. This cn be evlute s e + x. or just plugge into clcultor; the. nswer is.. b We fin this object s volume using isks centere roun the line y =. Ech isk hs inner rius x n ( x outer rius e x, so ech one hs re = ( ( = (( ( x V e x x. To fin the totl volume, we integrte ( x ( A e x n, with thickness x, volume ( e x x.. Don t bother fining n ntierivtive for the integrn; it s relly ugly n you ll nee to pproximte the nswer nywy. Your clcultor will give you the pproximtion V =.. x e x x n height 5 ( x e. They ech hve re A 5 ( x e x, thickness is x, ech volume is = ( The totl volume is given by ( c Ech rectngle hs with = n if their x V 5 x e x. 5 x V = x e x.. An pproximtion to this is V =.55. Consier squre horizontl slb of liqui t height h below the pyrmi s pex n with sie length x. A resulting sie view of hlf the pyrmi is shown t right. Clerly, the two tringles re h similr, so we cn set up the proportion =, mening x = h. Thus slb locte h below h x the pex hs sie length, h re ( A = h = h, n if it hs thickness h, volume x 7 V = h h. This mens tht ech slb s weight is 8 ( h h = h h. Ech hs to be lifte istnce h to get to the pex n then further to the esire point, for totl istnce of h +. Therefore the 7 7 work one to lift ech slb is W = h h( + h, n the totl work is ( h + h h= 8 ft lb. A igrm of the region is shown t right. The volume cn be foun by isks centere roun the xxis; ech isk hs rius y = x x n thickness x, V = x x x. The object s totl volume is then given by 8 V = ( x x x=.. 5 for volume of ( R
4 b This requires the metho of cylinricl shells, which shoul be centere roun the yxis. Ech shell hs rius x, thickness x, n height y x x =. So ech cyliner hs volume V = x( x x x. The totl volume is given by V = x( x x x= Since W = Fx n the prticle is moving from x = ft to x = ft uner force of F = lb, the totl work x ft ft one is x = = ft ft lb. This is choice b. x x ft 5 The circle is given by x + y =, so x = y. Ech squre hs bse x n height x for n re of ( x = x. Since the squres re prllel to the xxis, they hve thickness y, n ech slb hs volume V = x y. Fortuntely, since we know x = y, we cn substitute tht in to fin = ( Then the totl volume is V 8 8 = ( y y = ( y y = ( =, choice. V y y. The region is shown t right; its left bounry is t x = n its right bounry is t x =. The volume of the soli escribe is foun by isks centere roun the x xis; ech isk hs rius x x. If the isks hve thickness x, ech one s volume is V = ( x x x, so the totl volume is given by V = ( xx x. This is choice b. 7 The region is shown t left; its right bounry is t x = ; the top function is y = x, n the bottom function is y = x. Therefore, the region s re is ( ( ( ( ( A= x x x= x x = =, choice. 8 The region is shown t right; its left bounry is t x = n its right bounry is t x =. We cn fin the volume of the soli escribe with cylinricl shells centere roun the yxis. Ech shll hs rius x, thickness x, n height ( x x x= x x. Therefore the volume of the soli is ( or ( x x x, choice e. x x xx, The region is shown t left. We cn fin the volume of the soli with isks centere roun the xxis. Ech isk hs rius y = sec x n thickness x, so its volume is V = sec xx. The totl volume is thus sec xx = tnx] =, choice c.
5 The region is shown t right. Its lower bounry is y = n its upper bounry is y =. Fining the volume escribe is tricky; we nee to use cylinricl shells centere roun the xxis. Ech shell hs rius y, thickness y, n height = (. The volume of ech shell is given by ( x y y the totl volume is ( ( ( V = y y y y = y y y, ( V = y y y y, so choice. The grphs with the two regions in question shown re t right. The left bounry of the left region is x =, the curves intersect t x =, n the right bounry of the right region is x =. Since in the left region the top function is y= x + x x while in the right region the top function is y = x, the ( + = ( + ( + = ( + + left region s re is ( n the right s is ( The first integrl is evlute s x x x x x x x x x x x x x x x x x x. x + x x x =. The totl re is thus x x + x + x =, n the secon s 7 + =, choice.
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