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 30 5.8 Chapter Review p322:1-18 Assignment page 324: (4),(12),(20),(22) -------------------------------------------------------- Previous assignment: p313:(4),(16),(22),(30) Page 313 Problem 4 3 A straight wire 7 units long has density d(x) = 1+x at a point x units from one end. Find the distance from this end to the center of mass. x=7 3 | 4 | x=7 mass = INT (1+x ) dx = | x + x /4 | = 7+2401/4 = 2429/4 x=0 | | x=0 x=7 3 x=7 4 | 2 5 | x=7 Moment = INT x(1+x )dx = INT x+x dx = | x /2 + x /5 | x=0 x=0 | | x=0 = 49/2 + 16807/5 = 33859/10 _ x = (33859/10)/(2429/4) = 9674/1735 = 5.57579 ------------------------------------------------------------ Integrate[x(1+x^3),{x,0,7}]/ Integrate[1+x^3,{x,0,7}]; Get["font.math"]; f[x_] := 1+x^3; r[x_] := Sqrt[f[x]/Pi]; xbar = 5.57579; a = ParametricPlot3D[ {x, r[x] Sin[t], r[x] Cos[t]},{t,0,2 Pi},{x,0,7}]; b = ParametricPlot3D[ {xbar,y,z },{y,-1-r[xbar],1+r[xbar]}, {z,-1-r[xbar],1+r[xbar]}]; c = Show[a,b,PlotLabel->"P313 P4 Cross section = density y=1+x^3", PlotRange->All, AspectRatio->Automatic]; d = Show[c,ViewPoint->{0,-7,0}]; Display["p4x.ps",c]; Display["p4y.ps",d]; --------------------------------------------------------------- Page 313 Problem 16 Find the centroid of the region bounded by the given curves. 2 x = y - 3 y - 4 x = -y -y = y^2 - 3 y - 4 0 = y^2 - 2 y - 4 2 (+/-)Sqrt[4+16] y = ------------------- = 1 (+/-) Sqrt[5] 2 2 Area = Integrate[-y -(y -3y-4),{y, 1-Sqrt[5],1+Sqrt[5]}] = 20 Sqrt[5]/3 2 2 My = Integrate[1/2 (-y +(y -3y-4))(-y -(y -3y-4)),{y, 1-Sqrt[5],1+Sqrt[5]}] = -20 Sqrt[5] 2 Mx = Integrate[y(-y -(y -3y-4)),{y, 1-Sqrt[5],1+Sqrt[5]}] = 20 Sqrt[5]/3 _ x = My/Area = -3 _ y = Mx/Area = 1 ------------------------------------------------------- Get["font.math"]; f[y_] := y^2 - 3 y - 4; ymin = 1-Sqrt[5]; ymax = 1+Sqrt[5]; a = ParametricPlot[{y^2-3y-4,y},{y,1-Sqrt[5],1+Sqrt[5]}]; b = ParametricPlot[{-y,y},{y,1-Sqrt[5],1+Sqrt[5]}]; c = ListPlot[{{-3,1}},PlotStyle->{RGBColor[1,0,0],PointSize[0.03]}]; d = Show[a,b,c,AspectRatio->Automatic,PlotLabel->"P313 P16: x=y^2-3y-r, x=-y"]; Display["p16x.ps",d]; e = ListPlot[ { { f[3/2],3/2}}, PlotStyle->{RGBColor[0,1,0],PointSize[0.02]}]; a = 0; g = ListPlot[ { { a+f[3/2], 3/2}, { f[ymin],ymin}, { f[ymax],ymax}, { a+f[3/2], 3/2}}, PlotStyle->{ RGBColor[0,1,0], Thickness[0.01] }, PlotJoined->True]; h = ListPlot[{ {f[3/2]+f[ymin]+f[ymax], 3/2+ymin+ymax}/3}, PlotStyle->{RGBColor[0,1,0],PointSize[0.01]}]; i = Show[d,e,g,h]; Display["p16y.ps",i]; ------------------------------------------------------------------- Page 313 Problem 22 (-3,4)______________________________(1.4) | | | | | | | | | (-1,3/2) | | 4x5 | | | | | | ___________ | | (-2,-1) (0,-1) | | | | | | | | (1/2,-3/2) | | (-5/2,-2)| |____1x1_____| | 1x2 | {0,-2} (1,-2) |___________| (-3,-3) (-2-3) 20 {-1,3/2} + 2 {-5/2,-2} + {1/2,-3/2} = {-49/2,49/2} {-49/2,49/2}/23 = {-1.06522, 1.06522} ------------------------------------------------------------------- a = ListPlot[{ {-3,4},{1,4},{1,-2},{0,-2},{0,-1},{-2,-1},{-2,-3},{-3,-3},{-3,4}}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]},PlotJoined->True]; b = ListPlot[{{-1.06522, 1.06522}},PlotStyle->{RGBColor[1,0,0],PointSize[0.05]}]; c = Show[a,b,AspectRatio->Automatic, PlotLabel->"P313 p22 {-3,4},{1,4},{1,-2},{0,-2},{0,-1},{-2,-1},{-2,-3},{-3,-3}"]; Display["p22.ps",c]; ------------------------------------------------------------------------- Page 313 Problem 30 y=k --------------------------------------------------- h /\ / \ / \ / \ / \ /__________\ 0 b _ (a) Show that y = h/3. (b) Find the volume of the solid obtained when T is revolved around y = k, A = 1/2 b h y=h y=h | 2 3 | y=h Mx = INT y b(1-y/h) = b INT y-y^2/h = b | y /2 - y /(3h) | y=0 y=0 | | y=0 2 3 2 Mx = b( h /2 - h /(3h) ) = b h /6 _ 2 y = (b h /6)/(1/2 b h) = h/3 The new and different way. About y = k, The distance traveled by the center of mass is 2 Pi (k-h/3). The volume is -h 1/2 b h 2 Pi (k-h/3) = b h (k - --- ) Pi 3 ......................................... The same old way y=h Volume = INT 2 Pi (k-y) b(1-y/h) dy y=0 y=h Volume = 2 Pi k INT b(1-y/h) dy -2 Pi INT y b(1-y/h) dy y=0 = 2 Pi k Area - 2 Pi ybar Area = 2 Pi (k-ybar) Area So volume is the area times the distance moved by the center of mass. Turn to page 322 and work problems 1 through 18. 1. False. The integral will be zero because Cos[x] is negative on [Pi/2,Pi] 2. True. 3. False. The curves f(x) and g(x) may cross each other and then it is false. 4. True. 5. True. 6. False. It will be doubled 7. False. Shells is easier. 8. -(1-x)^2 + (1-x) = x - x^2 True. They have the same volume. 9. False. Think of a spiral, or even pick a square and bounce up and down an infinite number of times. 10. False. It will be four times. 11. False. The cone will be easier if it is point down. It will be harder if it is point up. 12. False. The deeper windows have a lot more pressure. 13. True. That is the whole point about the center of mass. 14. True. It is symmetrical about that point. 15. True. The centroid is at x=Pi/2 by symmetry. 16. True. 17. True. By symmetry. 18. True.