Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.08 Textbook: Calculus by Varberg, Purcell, Rigdon, ninth edition. Wednesday, January 16 5.2 Washers p286:12,15,18,24,26 Main Idea: Stacking coins and rolls of paper. Key Words: Volume. Coin Method, Shell method, Derivative of the Volume is the area. Goal: Learn to figure the volumes of objects. =============================================== Previous assignment: p280:(20),(24),(29),(30) Page 280 Problem 20 (a) Sketch the region bounded by the graphs of the given equations: (b) Show a typical slice, approximate its area, (c) Set up an integral, and calculate the area of the region. (d) Make an estimate of the area to confirm your answer. y = Sqrt[x] y = x-4 x = 0 Sqrt[x] = x-4 2 x = x - 8 x + 16 2 x - 9 x + 16 = 0 9 +/- Sqrt[81-64] 9 +/- Sqrt[17] x = ------------------- = ------------------ 2 2 ---------------------------------------------------- 9+Sqrt[17] x = -------------- 2 INT Sqrt[x]-x+4 dx x=0 9 + Sqrt[17] x = -------------- | 3/2 2 2 | x x | | ------- - --- + 4 x | | 3/2 2 | x=0 3/2 2 (9+Sqrt[17]) (9+Sqrt[17]) 2/3 ------------ - --------------- + 18+2 Sqrt[17] 3/2 2 2 2 2 9+Sqrt[17] 1+Sqrt[17] 98 + 18 Sqrt[17] 2/3 ------------ ------------ - ------------------ + 18 + 2 Sqrt[17 2 2 8 26+10 Sqrt[17] -49 - 9 Sqrt[17] +108 + 12 Sqrt[17] --------------- + ------------------ + ------------------ 6 4 6 52+20 Sqrt[17] -147-27 Sqrt[17] +216 + 24 Sqrt[17] -------------------------------------------------------------- 12 121+17 Sqrt[17] ------------------------------------ = 15.9244 12 The area is approximately 1/2*5*(9+Sqrt[17])/2 = 16.4039 ------------------------------------------------------ Integrate[Sqrt[x]-x+4,{x,0,(9+Sqrt[17])/2}] = 15.9244 ----------------------------------------------------- Get["font.math"]; xo = (9+Sqrt[17])/2; yo = (1+Sqrt[17])/2; p1 = Plot[Sqrt[x],{x,0,8},PlotStyle->{Thickness[0.002]}]; p2 = Plot[x-4,{x,0,8},PlotStyle->{Thickness[0.002]}]; p3 = ListPlot[{ {xo,yo} } ,PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; p4 = Show[p1,p2,p3,PlotLabel->"P280 P20, y=Sqrt[x], y=x-4", AspectRatio->Automatic]; Display["p20x.ps",p4]; p5 = ListPlot[{{0,1},{xo,yo},{0,-4},{0,1}}, PlotStyle->{RGBColor[1,0,0],Thickness[0.002]}, PlotJoined->True]; p6 = ListPlot[{{0,1},{xo,yo},{0,-4},{0,1}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.03]}, PlotJoined->False]; p7 = Show[p1,p2,p3,p5,p6,PlotLabel->"P280 P20, y=Sqrt[x], y=x-4", AspectRatio->Automatic]; Display["p20y.ps",p7]; ----------------------------------------------------- Page 280 Problem 24 (a) Sketch the region bounded by the graphs of the given equations: (b) Show a typical slice, approximate its area, (c) Set up an integral, and calculate the area of the region. (d) Make an estimate of the area to confirm your answer. x = (3-y)(y+1) x = 0 y= 3 INT (3-y)(y+1) dy y=-1 y= 3 2 INT 3+2y-y dy y=-1 | 2 3 | y=3 = | 3y + y -y /3 | | | y=-1 = 9 + 9 -9 -(-3+1+1/3) = 9+2-1/3 = 10 2/3 The area of the triangle is 1/2*4*5 = 10 which is the approximate area. ----------------------------------------------------- Get["font.math"]; p1 = ParametricPlot[{(3-y)(y+1),y},{y,-1,3}]; p2 = Show[p1,PlotLabel->"P280 P24, x = (3-y)(y+1)",AspectRatio->Automatic]; Display["p24x.ps",p2]; p3 = ListPlot[{ { 0,-1},{0,3},{5,1},{ 0,-1} }, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}, PlotJoined->True]; p4 = ListPlot[{ { 0,-1},{0,3},{5,1},{ 0,-1}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.03]}, PlotJoined->False]; p5 = Show[p1,p2,p3,p4,PlotLabel->"P280 P24, y = (3-y)(y+1)", AspectRatio->Automatic]; Display["p24y.ps",p5]; ----------------------------------------------------- Page 280 Problem 29 Sketch the region R bounded by 3 y = x+6, y = x , and 2y+x = 0. Then find its area. 3 x = x+6 3 x - x - 6 = 0 2 (x-2)(x + 2x + 3 ) = 0 x = 2 x=2 3 INT x+6 - x dx x=0 | 2 4 | x=2 | (1/2) x + 6 x - 1/4 x | | | x=0 2 + 12 - 4 = 10 Area of triangle = 1/2 4 6 = 12 Total area is 22. The approximate area is the sum of three triangles: 12 + 1/2*2*6 +1/2*1*8= 22. -------------------------------------------------------- Get["font.math"]; p1 = Plot[x+6,{x,-4,4}]; p2 = Plot[x^3,{x,-1,2.1}]; p3 = Plot[-x/2,{x,-4,1}]; p4 = ListPlot[ { {2,8},{0,0},{-4,2} }, PlotStyle->{RGBColor[1,0,0],PointSize[0.03]}]; p5 = Show[p1,p2,p3,p4,PlotLabel->"P280 P29 y=x+6,y=x^3,y=-x/2", AspectRatio->Automatic]; Display["p29x.ps",p5]; p6 = ListPlot[{ {0,0},{2,8},{-4,2},{0,0} },PlotJoined->True, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; p7 = ListPlot[{ {0,0},{1,0},{ 2,8},{0,0} },PlotJoined->True, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; p8 = ListPlot[{ {0,0},{0,6}},PlotJoined->True, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; p9 = Show[p1,p2,p3,p4,p6,p7,p8,PlotLabel->"P280 P29 y=x+6,y=x^3,y=-x/2", AspectRatio->Automatic]; Display["p29y.ps",p9]; -------------------------------------------------------- Page 280 Problem 30 Find the area of the triangle with vertices at (-1,4), (2,-2), and (5,1) by integration. --------------------------------------------------------- Get["font.math"]; a = ListPlot[{{-1,4},{2,-2},{5,1},{-1,4}},PlotJoined->True] b = ParametricPlot[{ 2,y},{y,-2,5/2}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; c = Show[a,b, PlotLabel->"P280 P30, {-1,4},{2,-2},{5,1} ", AspectRatio->Automatic]; Display["p30.ps",c]; -------------------------------------------------------------- y-4 = (-3/6)(x+1) (-1,4) . y = -1/2 x + 7/2 \ ` . \ ` . \ ` . (5,1) y-4 = (6/-3)(x+1) \ . ' y = -2x + 2 \ . ' \ . ' (2,-2) ' y+2 = 3/3(x-2) y = x-4 x= 2 x= 5 INT (-1/2 x + 7/2)-(-2x+2) dx + INT (-1/2 x + 7/2) - (x-4) dx x=-1 x= 2 x= 2 x= 5 INT 3/2 x + 3/2 + INT -3/2 x + 15/2 dx x=-1 x= 2 | 2 | x= 2 | 2 | x=5 | 3/4 x + 3/2 x | + | -3/4 x + 15/2 x | | | x=-1 | | x=2 3+3-(3/4-3/2) + -75/4 + 75/2 -( -3 + 15) 6 +3/4 +75/4 - 12 24+3+75-48 ---------- = 54/4 = 27/2 4 ------------------------------------------------------------------------ | -1 4 1 | 2+20+2 - (-10-1+8) 24 -(-3) 27 1/2 | 2 -2 1 | = --------------------- = -------- = ----- | 5 1 1 | 2 2 2 ------------------------------------------------------ New Material: Page 283 Example 1. Find the volume of the solid of revolution obtained by revolving the plane region R bounded by y = Sqrt[x], the x-axis and the line x = 4 about the x axis. x=4 2 x=4 | 2 | x=4 INT Pi y dx = Pi INT x dx = Pi | x /2 | = 8 Pi x=0 x=0 | | x=0 Ans 8 Pi. ------------------------------------------------------- Get["font.math"]; p1 = ParametricPlot3D[ {x, Sqrt[x] Cos[t], Sqrt[x] Sin[t]}, {x,0,2},{t,-Pi/2, Pi/2}]; p2 = Show[p1,PlotLabel->"P283 ex 1 y=Sqrt[x]",AspectRatio->Automatic]; Display["ex1x.ps",p2]; f[r_] := ParametricPlot3D[ {r,Sqrt[r 0.99] Cos[t],Sqrt[r 0.99] Sin[t], {RGBColor[1,0,0],Thickness[0.01]}}, {t,-Pi/2, Pi/2}]; g[r_,s_] := Graphics3D[ {RGBColor[1,0,0],Thickness[0.01], Line[{ {r,Sqrt[s] Cos[-Pi/2], Sqrt[s] Sin[-Pi/2]}, {s,Sqrt[s] Cos[-Pi/2], Sqrt[s] Sin[-Pi/2]}, {s,Sqrt[s] Cos[+Pi/2], Sqrt[s] Sin[+Pi/2]}, {r,Sqrt[s] Cos[+Pi/2], Sqrt[s] Sin[+Pi/2]}, {r,Sqrt[s] Cos[-Pi/2], Sqrt[s] Sin[-Pi/2]} }]}]; h[r_] := ParametricPlot3D[ {r,Sqrt[r+6/25] Cos[t],Sqrt[r+6/25] Sin[t], {RGBColor[1,0,0],Thickness[0.01]}}, {t,-Pi/2, Pi/2}]; p3 = Table[ f[(7r+1)/25],{r,1,7,3 }]; p4 = Table[ g[(7r-6)/25,(7r+1)/25],{r,1,7,3}]; p5 = Table[ h[(7r-6)/25],{r,1,7,3}]; p6 = Show[p5,p1,p3,p4,PlotLabel->" P283 Ex 1: y = Sqrt[x] "]; Display["ex1y.ps",p6]; ------------------------------------------------------------------------------ Page 283 Example 2. Find the volume of the solid generated by revolving the region bounded 3 by the curve y = x , the y-axis, and the line y = 3 about the y axis. 1/3 Ans 1/5 9 9 -------------------------------------------------------------- Get["font.math"]; f[x_] := x^3; p1 = ParametricPlot3D[ { x Cos[t], x Sin[t], f[x] },{x,0,3^(1/3)},{t,0, Pi}, PlotLabel->"P283 Ex 2: y = x^3"]; Display["ex2x.ps",p1]; g[r_] := ParametricPlot3D[ {0.99 r Cos[t], 0.99 r Sin[t], f[r+2 c/50], {RGBColor[1,0,0],Thickness[0.01]}}, {t, 0 , Pi }]; h[r_,s_] := Graphics3D[ {RGBColor[1,0,0],Thickness[0.01], Line[{ {r Cos[ 0 ], r Sin[ 0 ], f[r]}, {r Cos[ 0 ], r Sin[ 0 ], f[s]}, {r Cos[+Pi ], r Sin[+Pi ], f[s]}, {r Cos[+Pi ], r Sin[+Pi ], f[r]}, {r Cos[ 0 ], r Sin[ 0 ], f[r]} }]}]; k[r_] := ParametricPlot3D[ {0.99 r Cos[t], 0.99 r Sin[t], f[r], {RGBColor[1,0,0],Thickness[0.01]}}, {t, 0 , Pi }]; c = 3^(1/3); p3 = Table[ g[(7r-1)c/50],{r,1,7,1.5 }]; p4 = Table[ h[(7r-1)c/50,(7r+1)c/50],{r,1,7,1.5}]; p5 = Table[ k[(7r-1)c/50],{r,1,7,1.5}]; p6 = Show[p5,p1,p3,p4,PlotLabel->" P283 Ex 2: y = Sqrt[x] "]; Display["ex2y.ps",p6]; p7 = ParametricPlot3D[{3^(1/3) Cos[t],3^(1/3) Sin[t],y},{y,0,3},{t,0,Pi}]; p8 = Show[p6,p7,PlotLabel->" P283 Ex 2: y = x^3 "]; Display["ex2z.ps",p8]; ---------------------------------------------------------------------- 5/3 y=3 2 y=3 2/3 | y | y = 3 5/3 INT Pi x dy = Pi INT y dy = Pi | ----- | = 3/5 Pi 3 y=0 y=0 | 5/3 | y = 0 = 11.7626 Approximation is Pi 3^(2/3) 3 = 19.6043 -------------------------------------------------------------------------- Page 284 Example 3 Find the volume of the solid generated by revolving the region bounded 2 2 by the parabolas y = x and y = 8x about the x-axis. --------------------------------------------------- Get["font.math"]; p1 = Plot[x^2,{x,0,2.5}]; p2 = ParametricPlot[ {y^2/8, y}, {y,-1,Sqrt[20]}]; p3 = Show[p1,p2,PlotLabel->"P282 Ex 3: y = x^2 and y^2 = 8x", AspectRatio->Automatic,PlotRange->{{0,4.1},{0,4.1}}]; Display["ex3x.ps",p3]; p4 = ParametricPlot3D[ {r,r^2 Cos[t], r^2 Sin[t] }, {t, -Pi/2, Pi/2}, {r,0,2}]; p5 = ParametricPlot3D[ {r,Sqrt[8r] Cos[t], Sqrt[8r] Sin[t]}, {t,-Pi/2, Pi/2}, {r,0,2}]; p6 = Show[p4,p5,PlotLabel->"P282 Ex 3: y = x^2 and y^2 = 8x about X-axis"]; Display["ex3y.ps",p6]; ---------------------------------------------------------------------- x=2 2 x=2 2 INT Pi y dx - INT Pi y dx x=0 1 x=0 2 x=2 x=2 4 INT Pi 8x dx - INT Pi x dx x=0 x=0 | 2 | x=2 | 5 | x=2 Pi |8 x /2 | - Pi | x /5 | | | x=0 | | x=0 16 Pi - Pi 32/5 = Pi ( 16 - 32/5 ) = Pi (80-32)/5 = 48 Pi/5 <== answer ------------------------------------------------------------------------ Page 284 Example 4. 2 The semicircular region bounded by the curve x = Sqrt[4-y ] and the y axis is revolved about the line x= -1. Set up the integral representing the volume. y= 2 2 2 2 V = Pi INT ( 1 + Sqrt[4-y ]) - 1 dy y=-2 -------------------------------------------------------------------------- Get["font.math"]; p1 = ParametricPlot[ { Sqrt[4-y^2],y},{y,-2,2},PlotStyle->{Thickness[0.01]}]; p2 = ParametricPlot[ {0,y},{y,-2,2},PlotStyle->{Thickness[0.01]}]; p3 = ParametricPlot[ {-1,y},{y,-2,2},PlotStyle->{RGBColor[1,0,0]}]; p4 = Show[p1,p2,p3,PlotLabel->"P284 Ex 4: x = Sqrt[4-y^2] about x = -1", AspectRatio->Automatic]; Display["ex4x.ps",p4]; r[y_] := 1+ Sqrt[4-y^2]; p5 = ParametricPlot3D[ {r[y] Cos[t]-1, r[y] Sin[t], y},{y, -2,2},{t,0,Pi}] p6 = ParametricPlot3D[ { -1+Cos[t], Sin[t], z},{z,-2,2},{t,0,Pi}] p7 = Show[p5,p6,PlotLabel->" P284 Ex 4: x = Sqrt[4-y^2] about x=-1"]; Display["ex4y.ps",p7]; ------------------------------------------------------------------------ Page 285 Example 5 Let the base of a solid be the first quadrant plane region bounded by 2 y = 1-x /4, the x-axis and the y-axis. Suppose that the cross sections perpendicular to the x-axis are squares. Find the volume of the solid. --------------------------------------------- Get["font.math"]; f[x_] := 1-x^2/4; p1 = ParametricPlot3D[ {x,f[x],0},{x,0,2} ]; p2 = ParametricPlot3D[ {x,0,0},{x,0,2}]; p3 = ParametricPlot3D[ {0,y,0},{y,0,1}]; p4 = ParametricPlot3D[ {x, 1-x^2/4,1-x^2/4,{RGBColor[1,0,0],Thickness[0.01]}}, {x,0,2}]; p5 = ParametricPlot3D[ {x, 0,1-x^2/4,{RGBColor[1,0,0],Thickness[0.01]}}, {x,0,2}]; p[x_] := Graphics3D[{Thickness[0.01], Line[{ {x,0,f[x]},{x,f[x],f[x]},{x,f[x],0},{x,0,0},{x,0,f[x]}}]} ]; p6 = Table[ p[x/10],{x,0,20}]; p7 = Show[p1,p2,p3,p4,p5,p6,PlotLabel->"P283 Ex 5: f[x] = 1-x^2/4"]; Display["ex5.ps",p7]; ------------------------------------------------------------------ x=2 2 2 Volume = INT (1-x /4) dx x=0 x=2 2 4 = INT 1 - 2 x /4 + x /16 dx x=0 | 3 5 | x=2 = | x -1/2 x /3 + 1/16 x /5 | | | x=0 = 2 - 4/3 + 2/5 = 16/15 <== answer -------------------------------------------------------------------- Page 284 Example 6. The base of a solid is the region between one arch of y = Sin[x] and the x-axis. Each cross section perpendicular to the x-axis is an equilateral triangle sitting on this base. Find the volume of the solid. --------------------------------------------- Get["font.math"]; f[x_] := Sin[x]; p1 = ParametricPlot3D[ {x,f[x],0,{RGBColor[1,0,0],Thickness[0.01]}},{x,0,Pi} ]; p2 = ParametricPlot3D[ {x,0,0,{RGBColor[1,0,0],Thickness[0.01]}},{x,0,Pi}]; p3 = ParametricPlot3D[ {x, f[x]/2, f[x] Sqrt[3]/2, {RGBColor[1,0,0],Thickness[0.01]}},{x,0,Pi}]; p[x_] := Graphics3D[ {Thickness[0.01], Line[{ {x,0,0}, {x,f[x],0},{x,f[x]/2,f[x] Sqrt[3]/2}, {x,0,0}} ]} ]; p4 = Table[ p[x Pi/15],{x,0,15}]; p5 = Show[p1,p2,p3,p4,PlotLabel->"P284 Ex 6: f[x] = Sin[x] "]; Display["ex6.ps",p5]; ------------------------------------------------------------------ x=Pi 2 INT s Sqrt[3]/4 dx x=0 x=Pi 2 INT Sin[x] Sqrt[3]/4 dx x=0 x=Pi 1-Cos[2 x] Sqrt[3]/4 INT -------------- dx x=0 2 x=Pi Sqrt[3]/8 INT 1-Cos[2 x] dx x=0 | | x=Pi Sqrt[3]/8 | x - Sin[2 x]/2 | | | x=0 Sqrt[3]/8 (Pi - 0 - (0-0) ) = Pi Sqrt[3]/8 <=== answer