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 12 6.2 Washers p285:1-6, 9-11,(14),15,(20),21,22,(25),(28) 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: Page 278 Problem 22 (a) Sketch the region (b) Show a typical slice (c) Approximate its area (of the slice) (d) Set up an integral (e) Calculate the area of the region (f) Make an estimate of the area to confirm your answer. 2 2 y = x - 9 y = (2x-1)(x+3) = 2x + 5 x - 3 . | . | * | * . | . * | * . | . * * | * * | . . | * ---|-----@-----|-----|-----+--.--|-----|-----*-----|--- * | * .* |. .* -| * * .| (a) '@ . | * . . -. * | * * -| * (0,-9) | | Find the points of intersection 2 2 x - 9 = 2x + 5x - 3 2 0 = x + 5x + 6 0 = (x+2)(x+3) (-2,-5) and (-3,0) The area bounded is between x = -2 and x = -3. (a) -3 -2.8 -2.6 -2.4 -2.2 -2 @----_----|---------|---------|---------|---------|-- - - |_| - - - - - - - - - - - - - - - - - - - - - - - ----------o--_--------------------------------------- - - - - - * | | - - - - - - - - - - - - - - - - - - - ------------|_|-----o---___-------------------------- - - - - - - - - - - * - | | - - - - - - - - - - - - - ------------------------|_|---o---------------------- - - - - - - - - - - - - - - - * - _ - - - - - - - - - ---------------------------------|_|----o------------ - - - - - - - - - - - - - - - - - - - - * - - - - - - --------------------------------------------------@-- (b)-(c) the area of the slice is pretty small. x=-2 (d) INT x^2 - 9 - (2x-1)(x+3) x=-3 x=-2 | | x=-2 (d) INT -x^2 - 5 x - 6 = | -x^3/3 -5 x^2/2 - 6 x | x=-3 | | x=-3 = 8/3 -10 +12 - (9-45/2 + 18) = 1/6 (e) The maximum distance between the curves is when -2x - 5 = 0 or at 5/2. -25/4 + 25/2 - 6 = 1/4 so an upper bound on the area is 1/4. ---------------------------------------------- f[x_] := x^2-9; g[x_] := (2x-1)(x+3); e = 0.01; h[x_] := ListPlot[{ {x,g[x+e]}, {x+e,g[x+e]}, {x+e,f[x]},{x,f[x]},{x,g[x+e]}}, PlotStyle->{RGBColor[1,0,0]},PlotJoined->True]; a = Plot[f[x],{x,-5,5}]; b = Plot[g[x],{x,-5,5}]; c = Show[a,b,PlotLabel->"P278p22 y = x^2 - 9 y = (2x-1)(x+3)" ]; Display["22.ps",c]; ap = Plot[f[x],{x,-3,-2}]; bp = Plot[g[x],{x,-3,-2}]; cp = Table[h[-3+i/100],{i,0,99}]; dp = Show[ap,bp,cp,PlotLabel->"P278p22 y = x^2 - 9 y = (2x-1)(x+3)" ]; Display["22x.ps",dp]; ----------------------------------------- Page 278 Problem 26 2 +/- Sqrt[4+4x] x = y^2 - 2 y y = ----------------- = 1 +/- Sqrt[1+x] 2 x-y-4 = 0 The points of intersection: y^2 - 2 y = y+4 y^2 - 3 y - 4 = 0 (y+1)(y-4) = 0 y = -1 y = 4 (3,-1) and (8,4) y=5 INT y+4 - (y^2 - 2 y) dy y=-1 y=5 INT - y^2 + 3 y + 4 dy y=-1 | | y=4 | -y^3/3 + 3 y^2/2 + 4 y | | | y = -1 -64/3 + 48/2 + 16 - ( 1/3 + 3/2 - 4) = 125/6 = 20.833 For the approximation The maximal width is where -2y + 3 = 0 so y = 3/2 and x goes from -3/4 Making two triangles Det[{{-3/4,3/2,1},{3,-1,1},{8,4,1}}]; |-3/4 3/2 1 | 1/2| 3 -1 1 | = 125/8 = 15.625 | 8 4 1 | (8,4) /| / | / | / | / | / | / | / | / | / | / | / | (-3/4, 3/2) | \ 15.625 | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \| ( 3,-1) The approximation of 15.625 should be too small. x = y^2 - 2 y dx/dy = 2y - 2 dx/dy at (3,-1) = -4 dy/dx at (3,-1) = -1/4 dx/dy at (8, 4) = 8 dy/dx at (8. 4) = 1/8 y+1 = -1/4 (x-3) y-4 = 1/8 (x-8) 5 = -1/4 (x-3) - 1/8 (x-8) 5 = -1/4 x + 3/4 - 1/8 x + 1 40 = -2 x + 6 - x + 8 3x = -26 x = -26/3 {-26/3,23/12} | -26/3 23/12 1 | 1/2 | 3 -1 1 | = 875/24 = 36.4583 | 8 4 1 | Should be too big. ----------------------------------------------- a = ParametricPlot[ {y^2-2y,y},{y,-1.5,4.5}]; b = ParametricPlot[ {y+4,y},{y,-1.5,4.5}]; c = Show[a,b,PlotLabel->"P278p26 x=y^2-2y, x-y-4=0"]; Display["26.ps",c]; f[y_] := y^2 - 2 y; g[y_] := y+4; p[y_] := Min[g[y],g[y+e]]; q[y_] := Max[f[y],f[y+e]]; e = 0.10; h[y_] := ListPlot[ { {q[y],y},{p[y],y},{p[y],y+e},{q[y],y+e},{q[y],y} }, PlotStyle->{RGBColor[1,0,0]},PlotJoined->True]; d = Table[h[-1+i 1/10],{i,1,48}]; r = Show[a,b,c,d,PlotLabel->"P278p26 x=y^2-2y, x-y-4 = 0"]; Display["26x.ps",r]; u = ListPlot[{ {-3/4,3/2},{3,-1},{8,4},{-3/4,3/2}}, PlotJoined->True,PlotStyle->{RGBColor[1,0,0]}]; v = Show[a,u, PlotLabel->"P278p26 x=y^2-2y, x-y-4 = 0"] Display["26y.ps",v]; uu = ListPlot[{{3,-1},{8,4},{-26/3,23/12},{3,-1}}, PlotStyle->{RGBColor[1,0,0]},PlotJoined->True]; vv = Show[a,uu,PlotLabel->"P278p26 x=y^2-2y, x-y-4 = 0"]; Display["26z.ps",vv]; --------------------------------------------- Page 278 Problem 28 4 x = 4 y 4 x = 8 - 4 y Find the points of intersection 4 4 4 y = 8 - 4 y 4 8 y = 8 4 y = 1 y = -1,1 {4,1},{4,-1} y=1 INT 8-4y^4 - 4y^4 dy y=-1 y=1 INT 8-8y^4 dy y=-1 | | y = 1 | 8y - 8 y^5/5 | | | y = -1 8-8/5 - (-8 + 8/5) = 64/5 = 12.8 2*8 = 16 should be too big. 1/2(2*8) = 8 should be too small. -------------------------------------------------------------- a = ParametricPlot[ {4y^4,y},{y, -1,1}]; b = ParametricPlot[ {8 - 4y^4,y},{y, -1,1}]; c = Show[a,b,PlotLabel->"P278p28 x = 4 y^4, x = 8-4 y^4" ]; Display["28.ps",c]; uu = ListPlot[{{ 0,-1},{8,-1},{ 8,1},{ 0,1},{ 0,-1}},PlotJoined->True, PlotStyle->{RGBColor[1,0,0]}]; vv = Show[a,b,uu,PlotLabel->"P278p28 x = 4 y^4, x = 8-4 y^4"]; Display["28x.ps",vv]; uuu = ListPlot[{{0,0},{4,-1},{8,0},{4,1},{0,0}},PlotJoined->True, PlotStyle->{RGBColor[1,0,0]}]; vvv = Show[a,b,uuu,PlotLabel->"P278p28 x = 4 y^4, x = 8-4 y^4"]; Display["28y.ps",vvv]; -------------------------------------------- Page 278 Problem 30 Find the area of the triangle with vertices at (-1,4), (2,-2) and (5,1) by integration. (-1,4) * _ . . . |_ . m = 3/-6 = -1/2 . | |. y-1 = -1/2 (x-5) .|_ | . y = -1/2 x +7/2 . | . m=6/-3 = -2 |. | * (5,1) y-4=-2(x+1) | . | . y = -2x+2 | . | . ----------|----+---.|---||----|--.-|----|------- | . | . |_ . | . m = 3/3 = 1 | . | . y+2 = x-2 |_ .|. y = x-4 | * (2,-2) | | | | 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 dx + INT -3/2 x + 15/2 dx x=-1 x=2 | | x=2 | | x=5 | 3/4 x^2 + 3/2 x | + | -3/4 x^2 + 15/2 x | | | x=-1 | | x=2 3 + 3 -( 3/4-3/2) + -75/4 + 75/2 - (-3 + 15) 27/4 + 27/4 = 27/2 ------------------------------------- 1/2 Det[{{2,-2,1},{-1,4,1},{5,1,1}}]; -------------------------------------- Page 278 Problem 32 v(t) = 1/2 + Sin[2 t] and the interval is 0 <= t <= 3 Pi/2. Find the displacement and the total distance traveled. s(t) = 1/2 t -1/2 Cos[2 t] The displacement is s(3 Pi/2) - s(0) = 1 + 3 Pi/4 = 3 Pi/4 - 0 - (-1/2 = 3 Pi/4 + 1/2. The total distance traveled is the velocity is zero when 0 = 1/2 + Sin[2t] Sin[2t] = -1/2 2t = 7 Pi/6 or 11 Pi/6 t = 7 Pi/12 or 11 Pi/ 12 Abs[ s[7 Pi/12] - s[0]] + Abs[s[11 Pi/12] - s[7 Pi/12] ] + Abs[ s[3 Pi/2] - s[11 Pi/12] ] I II III 1 Sqrt[3] 7 Pi I - + ------- + ---- 2 4 24 Sqrt[3] Pi II -------- - -- 2 6 1 Sqrt[3] 7 Pi III - + ------- + ---- 2 4 24 ------------------------------ 5 Pi 1 + Sqrt[3] + ---- <=== Total distance. 12 -------------------------------------------------------- v[t_] := 1/2 + Sin[2t]; s[t_] = 1/2 t -1/2 Cos[2 t]+1/2; delta = (3 Pi/2)/100 f[n_] := delta * Sum[v[i delta],{i,1,n}]; A = Table[{n*delta,f[n]},{n,1,100}]; a = ListPlot[A]; b = Plot[s[t],{t,0,3 Pi/2},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->"P278p32, v[t_] := 1/2 + Sin[2t]",PlotRange->All]; Display["32.ps",c]; -------------------------------------------------------- ----------------------------------------------------------------------- New Material Page 281 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. Ans 8 Pi. Page 281 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 Page 282 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. ans 48 Pi/5. Page 283 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 Page 283 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. Ans. 16/15. 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. Ans = Sqrt[3]/8 Pi.