6.4: SHELL METHOD 6.5: WORK AND ENERGY NAME: SOLUTIONS Math 1910 September 26, 2017
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1 6.4: SHELL METHOD 6.5: WORK AND ENERGY NAME: SOLUTIONS Mt 9 September 26, 27 ONE-PAGE REVIEW Sell Metod: Wen you rotte te region between two grps round n xis, te segments prllel to te xis generte cylindricl sells. Te volume V of te solid of revolution is te integrl of te surfce res of te sells. V = rdiuseigt of sell dr. 2 Wt is te volume of te region between fx, fx, nd te x-xis for x [, b], rotted round te y-xis? V = xfx dx. 2 b te region between fx nd gx, fx gx, for x [, b] rotted round te y-xis? V = fx gx dx. 3 c te region between fx, fx nd te x-xis for x [, b], rotted round te line c? If c,v = x cfx dx. 4 If c,v = c xfx dx. 5 3 Te work W performed to move n object from to b long te x-xis by pplying force of mgnitude Fx is W = Fx dx. 4 To compute work ginst grvity, first decompose n object into N lyers of equl tickness y, nd ten express te work performed on tin lyer s Ly y, were Ly = g density re of y distnce lifted. Te totl work performed to lift te object from eigt to eigt b is W = Ly dy 6.
2 PROBLEMS Sketc te solid obtined by rotting te region undernet te grp of f over te intervl bout te given xis, nd clculte its volume using te sell metod. fx = x 3, x [, ], bout x = 2. SOLUTION: Ec sell s rdius 2 x nd eigt x 3, so te volume of tis solid is 2 xx 3 dx = x 2x 3 x 4 4 dx = 2 x5 = 3π 2 5. b fx = x 3, x [, ], bout x = 2. SOLUTION: Ec sell s rdius x 2 = x + 2 nd eigt x 3, so te volume of tis solid is 2 + xx 3 dx = x 2x 3 + x 4 4 dx = 2 + x5 = 7π
![c fx =, x [, 2], bout x =.](/docs-images/86/93213342/images/3-0.jpg "x 2 + SOLUTION: Ec sell s rdius x, nd eigt, so te volume of te solid is x 2 + 2 x x 2 + dx = x 2 + 2 = 5.")
3 c fx =, x [, 2], bout x =. x 2 + SOLUTION: Ec sell s rdius x, nd eigt, so te volume of te solid is x x x 2 + dx = x = 5. 2 Use te most convenient metod disk/wser or sell to find te given volume of rottion. 3
4 Region between x = y5 y nd x =, rotted bout te y-xis. SOLUTION: Exmine te picture below, wic sows te region in question. If te indicted region is sliced verticlly, ten te top of te slice lies long one brnc of te prbol x = y5 y nd te bottom lies long te oter brnc. On te oter nd, if te region is sliced orizonlly, ten te rigt endpoint of slice is lwys long te prbol nd te left endpoint is on te y-xis. So it s esier to do orizontl slices. Now suppose te region is rotted bout te y-xis. Becuse orizontl slice is perpendiculr to te y-xis, we will clculte te volume of te resulting solid using te disk metod. Ec cross section is disk of rdius R = y5 y, so te volume is π y 2 5 y 2 dy = π 25y 2 y 3 + y 4 dy = π 25 3 y3 5 2 y y5 = 625π 6. 4
5 b Region between x = y5 y nd x =, rotted round te x-xis. SOLUTION: Exmine te?gure below, wic sows te region bounded by x = y5?y nd x =. If te indicted region is sliced verticlly, ten te top of te slice lies long one brnc of te prbol x = y5 y nd te bottom lies long te oter brnc. On te oter nd, if te region is sliced orizontlly, ten te rigt endpoint of te slice lwys lies long te prbol nd left endpoint lwys lies long te y-xis. Clerly, it will be esier to slice te region orizontlly. Now, suppose te region is rotted bout te x-xis. Becuse orizontl slice is prllel to te x-xis, we will clculte te volume of te resulting solid using te sell metod. Ec sell s rdius of y nd eigt of y5 y, so te volume is y 2 5 y dy = 5y 2 y 3 dy = 5 3 y3 5 4 y4 = 625π 6. 5
6 c Region between y = x 2 nd x = y 2, rotted bout x = 3. SOLUTION: Exmine te figure below, wic sows te region bounded by te y = x 2 nd x = y 2. If te indicted region is sliced verticlly, ten te top of te slice lies long x = y 2 nd te bottom lies long y = x 2. On te oter nd, if te region is sliced orizontlly, ten te rigt endpoint of te slice lwys lies long y = x 2 nd left endpoint lwys lies long x = y 2. Tus, for tis region, eiter coice of slice will be convenient. To proceed, let s coose verticl slice. Now rotte te region bout x = 3. Becuse verticl slice is prllel to x = 3, we will clculte te volume of te resulting solid using te sell metod. Ec sell s rdius of 3 x nd eigt of x x 2, so te volume is 3 x x x 2 dx = 3x /2 x 3/2 3x 2 + x 3 dx = 2x 3/2 25 x5/2 x x4 = 7π. 6
7 3 Clculte te work in Joules required to pump ll of te wter out of full tnk wit te spe described. Distnces re in meters, nd te density of wter is kg/m 3. A rectngulr tnk, wit wter exiting from smll ole in te top. SOLUTION: Plce te origin t te top of te box, nd let te positive y-xis point downwrd. Te volume of one lyer of wter is 32 y cubic meters, so te force needed to lift it is y = 336 y Newtons. Ec lyer must be lifted y meters, so te totl work needed to empty te tnk is 336y dy = 568y 3 5 = Joules. b A troug s in te picture, were te wter exits by pouring over te sides. SOLUTION: Plce te origin long te bottom edge of te troug, nd let te positive y-xis point upwrd. From similr tringles, te widt of lyer of wter t eigt y meters is so te volume of ec lyer is wc y = c Tus, te force needed to lift lyer is 98c w = yb meters, yb y meters 3. yb y Newtons. Ec lyer must be lifted y meters, so te totl work needed to empty te tnk is yb 3 98 yc + dy = 98c 3 + b2 Joules. 6 7
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