Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://orion.math.iastate.edu/hentzel/class.166.05 Textbook: Calculus by Varberg, Purcell, Rigdon, eight edition. January 14 6.3 Shells p292:1,2,5,6,9,(10),(14),17,18,(20),22 Main Idea: Think about making a log into a long sheet of veneer. Key Words: Shell Method Goal: Learn the Shell method to find volumes -------------------------------------------------------- Previous assignment: Page 285 Problem 14 Find the volume of the sold generaed by revolving R about the y axis. 2/3 x = y y = 27 x = 0 Horizontal Slice _ | /| | \ \_____/ | |-------------. (9,27) | . |____. |___| | . -------------.-------------------- | Disk method. y=27 2 INT Pi x dy y=0 y=27 4/3 Pi INT y dy y=0 7 | 7/3 | y=27 3 6561 Pi Pi | y | = Pi ---------- = --------- | ------- | 7/3 7 | 7/3 | y=0 -------------------------------------------------- Vertical Slice _ | /|| \ \___/ |-------------. (9,27) | | | . | | | | | | . -------------.-------------------- | x=9 INT 2 Pi x (27-y) dx x=0 x=9 3/2 INT 2 Pi x (27-x ) dx x=0 x=9 5/2 2 Pi INT 27x -x dx x=0 | 2 7/2 | x = 9 2 Pi | 27 x /2 - x / (7/2) | | | x = 0 7 2 Pi ( 27 81/2 - 3 /(7/2)) = 6561 Pi/7 -------------------------------------------- -------------------------------------------------------------------------- a = ParametricPlot3D[ { z^(2/3) Cos[t],z^(2/3) Sin[t],z},{z,0,27},{t,0,2 Pi}]; b = Show[a,PlotLabel->"P285p14 x = y^(2/3) "]; Display["14.ps",b]; -------------------------------------------------------------------------- Page 285 Problem 20 Find the volume of the solid generated by revolving about the x-axis the region in the first quadrant bounded by the circle x^2 + y^2 = r^2 , the x-axis, and the line x = r-h, 0 < h < r and thus find the volume of a spherical segment of height h of a sphere of radius r. The thing is a sphere with the top filed off flat. ... | X `| . | X | . | X | . | X | . /\ | X | . | ---------------+------------------| _ ----- dx <--h--> \/| x=r-h 2 INT Pi y dx x=0 x=r-h 2 2 INT Pi (r -x ) dx x=0 | 2 3 | x = r-h Pi | r x - x /3 | | | x = 0 2 3 Pi ( r (r-h) - (r-h) /3 ) 3 3 h Pi 2 2 Pi r ----- - h Pi r + ------- 3 3 3 When h = 0, we should get 1/2 of a sphere = 1/2 ( 4/3 Pi r ) -------------------------------------------------------------- We use horizontal slices to figure the end cap. Integrate[ 2 Pi y (Sqrt[r^2 - y^2] - (r-h) ),{y,0,Sqrt[r^2 - (r-h)^2]}] 2 2 Sqrt[r - (r-h) ] 2 2 INT 2 Pi y ( Sqrt[r - y ] - (r-h) ) dy y=0 2 2 | 2 2 3/2 2 | y = Sqrt[r - (r-h) ] | (r - y ) y | 2 Pi | --------------- -(r-h)----- | | 3/2 (-2) 2 | | | y = 0 2 2 | 2 2 3/2 2 | y = Sqrt[r - (r-h) ] | -(r - y ) -(r-h) y | 2 Pi | --------------- + ---------- | | 3 2 | | | y = 0 2 2 | 2 2 3/2 | y = Sqrt[r - (r-h) ] 2 Pi |-2(r - y ) -3(r-h) y^2 | ---- | | 6 | | | | y = 0 | | 2 Pi | 2 3/2 2 2 3 | ---- |-2( (r-h) ) -3(r-h)( r -(r-h) ) + 2 r | 6 | | | | | 3 2 3 3 2 Pi |-2(r-h) -3(r-h)r + 3 (r-h) ) + 2 r | ---- | | 6 | | | 3 2 3 2 Pi | (r-h) -3(r-h)r + + 2 r | ---- | | 6 | | | | 2 Pi | r^3 - 3 r^2 h + 3 r h^2 - h^3 | ---- | -3r^3 + 3 r^2 h | 6 | 2r^3 | | 2 3 | 2 Pi | 3r h - h | ---- | | 6 | | 2 Pi h ------- (3 r - h ) 3 This plus the previous should equal the volume of the 1/2 sphere. ============================================================================== a = ParametricPlot3D[ {x,Sqrt[1-x^2] Cos[t], Sqrt[1-x^2] Sin[t]},{t,0,2 Pi},{x,0,0.8}]; b = Show[a,PlotLabel->"P285p20 frustrum of a Sphere"]; Display["20.ps",b]; -------------------------------------------------------- Page 285 Problem 25 The base of a solid is bounded by one arch of y = Sqrt[Cos[x]] -Pi/2 <= x <= Pi/2 and the x axis. Each cross section perpendicular to the x-axis is a square sitting on this base. Find the volume of the solid. x= Pi/2 2 INT Sqrt[Cos[x]] dx x=-Pi/2 x= Pi/2 INT Cos[x] dx x=-Pi/2 | | x= Pi/2 | Sin[x] | = 1-(-1) = 2 | | x= -Pi/2 -------------------------------------------- f[x_] := Sqrt[Cos[x]]; g[x_] := Graphics3D[ Line[{{x,0,0},{x,f[x],0},{x,f[x],f[x]},{x,0,f[x]},{x,0,0}}]]; n=10 H = Table[g[ -Pi/2 + k Pi/(2 n)],{k,0,2n}]; K = ParametricPlot3D[{x,f[x],f[x]},{x,-Pi/2,Pi/2}]; L = ParametricPlot3D[{x, 0 ,f[x]},{x,-Pi/2,Pi/2}]; M = ParametricPlot3D[{x,f[x], 0 },{x,-Pi/2,Pi/2}]; ans = Show[H,K,L,M,PlotLabel->"P285p25 y = Sqrt[Cos[x]]",AspectRatio->Automatic]; Display["25.ps",ans]; ----------------------------------------------------------------------------- Page 285 Problem 28 Find the volume inside the "+" shown in Figure 16. Assume that both cylinders have radius 2 inches and length 12 inches. Hint: The volume is equal to the volume of the first cylinder plus the volume of the second cylinder minus the region common to both. The volume Common to both is something with horizontal cross sections which are squares of side x^2 + z^2 = 4 y^2 + z^2 = 4 The points on both cylinders have x = y = Sqrt[4-z^2]. z= 2 INT (2 Sqrt[4-z^2])^2 dz z=-2 z= 2 4 INT 4-z^2 dz z=-2 | 3 | z = 2 4 | 4z - z /3 | | | z = -2 4 ( 8 - 8/3 - (-8 + 8/3) ) = 4(16-16/3) = 64(1-1/3) = 128/3 2 Notice that a sphere of radius 2 has volume 4/3 Pi 2 = 33.5103 < 128/3 = 42.6667 The total volume is 2 12 Pi 4 - 128/3 = 96 Pi - 128/3 = 258.926 ---------------------------------------------------------- f[z_] := Sqrt[4-z^2]; g[z_] := Graphics3D[ Line[{{-f[z],-f[z],z},{-f[z],f[z],z},{f[z],f[z],z},{f[z],-f[z],z},{-f[z],-f[z],z} }]]; n = 3; H = Table[g[ -2 + k /n],{k,0,4n}]; K = ParametricPlot3D[{ f[z], f[z],z},{z,-2,2}]; L = ParametricPlot3D[{-f[z], f[z],z},{z,-2,2}]; M = ParametricPlot3D[{ f[z],-f[z],z},{z,-2,2}]; n = ParametricPlot3D[{-f[z],-f[z],z},{z,-2,2}]; ans = Show[H,K,L,M,n,PlotLabel->"P285p28 PLUS",AspectRatio->Automatic]; Display["28.ps",ans]; =================================================================== New Material Page 288 Example 1 The region bounded by y = 1/Sqrt[x] and the x-axis, x=1 and x=4 is rotated about the y axis. Find the volume of the resulting solid. x=4 V = INT 2 Pi x 1/Sqrt[x] dx = 28 Pi/3 x=1 Using washer y=1 2 2 V = INT Pi x - Pi 1 dy y=1/2 y=1 V = Pi INT 1/y^4 - 1 dy y=1/2 | -3 | y=1 V = Pi | y /(-3) - y | | | y = 1/2 V = Pi ( -1/3 -1 - (-8/3 -1/2) ) = 11 Pi/6 2 The base has volume 1/2 Pi 4 - 1/2 Pi = 15 Pi/2 ------------------------------------------------------- f[x_] := 1/Sqrt[x]; a = ParametricPlot3D[ {x Cos[t], x Sin[t], f[x]}, {t, 0, Pi },{x,1,4}]; b = Show[a,PlotLabel->"P288e1 f[x] = 1/Sqrt[x]" ]; Display["e1.ps",b]; ---------------------------------------------------------------- Page 289 Example 2. The region bounded by the line y = (r/h)x the x axis and x=h is revolved about the x axis. Find the volume. y=r V = INT 2 Pi y (h - hy/r) dy y=0 2 V = 2 Pi h INT y(1-y/r) dy = 1/3 Pi r h. OR x=h 2 x=h V = INT Pi y dx = Pi INT (r/h)^2 x^2 dx x=0 x=0 | | x=h Pi(r/h)^2 | x^3/3 | = Pi(r/h)^2 h^3/3 = 1/3 Pi r^2 h | | x=0 --------------------------------------------------------- A 10 dip candle burns 2 hours. A 30 dip candle burns 10 hours. How long will a 50 dip candle burn? The candles are 10 inches long. ----------------------------------------------------