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 Office Hours 9:15-10:15 MTWThF Textbook: Calculus by Varberg, Purcell, Rigdon, ninth edition. Wednesday, February 20 7.7 Chapter Review Page 419: 7.7 Chapter Review: Previous Assignment: page 421: 44, 46, 48, 50 Page 421 Problem 44 Find the volume of the solid generated by revolving the region under the graph of 1 y = --------------- ______ / 2 \/ 3x-x from x=1 to x=2 about (a) The x-axis. (b) The y-axis x=2 2 x=2 1 (a) INT Pi y dx = Pi INT --------- dx x=1 x=1 2 3x-x -1 -1 ---- ----- -1 -3 3 -------- = ------- + ------ x(x-3) x x-3 / g (x) = 2x-3 1 1/3 1/3 ------- = ------ + ----- x(3-x) x 3-x | |x=2 = Pi | 1/3 Log[x] - 1/3 Log[3-x] | | |x=1 = Pi/3 ( Log[2] +Log[2]) = 2 Log[2]/3 (b) x=2 1 INT Pi x ----------- dx x=1 2 Sqrt[3x-x ] x=2 x x=2 Pi INT ------------ dx + Pi INT --------------- x=1 2 x=1 2 Sqrt[3x-x ] Sqrt[3x-x ] x=2 -1/2(-2x+3) x=2 + 3/2 Pi INT ------------ dx + Pi INT ------------- dx x=1 2 x=1 2 Sqrt[3x-x ] Sqrt[3x-x ] 2 1/2 | x=2 (3x-x ) | x=2 dx -1/2 Pi ------------ | 3/2 Pi INT ---------------------- 1/2 | x=1 x=1 2 Sqrt[(9/4 - (x-3/2) ] | x=2 -Pi(Sqrt[2]-Sqrt[2]) + 3/2 ArcSin[ (x-3/2)/(3/2) | | x=1 3/2 Pi ArcSin[1/3] - 3/2 Pi ArcSin[-1/3] 3 Pi ArcSin[1/3] = 3.20289 <========= ------------------------------------ Get["font.math"]; f[x_] := 1/Sqrt[3x-x^2] P1 = Plot[f[x],{x,1,2}] P2 = Plot[0,{x,0,2}]; P3 = Show[P1,P2,PlotLabel->"P421 P44 y = 1/(3x-x^2)", AspectRatio->Automatic]; Display["44x.ps",P3]; P4 = ParametricPlot3D[ {x,f[x] Cos[t],f[x] Sin[t]},{x,1,2}, {t,-4.5Pi/4,3 1Pi/4}]; Display["44y.ps",P4]; P5 = ParametricPlot3D[ {x Cos[t], x Sin[t], f[x]},{x,1,2},{t,0,2 Pi}]; P6 = ParametricPlot3D[ {2 Cos[t], 2 Sin[t], y },{y,0,f[2]},{t,0,2 Pi}]; P7 = ParametricPlot3D[ {1 Cos[t], 1 Sin[t], y },{y,0,f[1]},{t,0,2 Pi}]; P8 = Show[P5,P6,P7,PlotLabel->"P421 P44 y = 1/(3x-x^2)", AspectRatio->Automatic]; Display["44z.ps",P8]; ---------------------------------------------------------------------- Page 421 Problem 46 The region under the curve 1 y = ---------- 2 x + 5 x + 6 from x=0 to x=3 is rotated about the x-axis. Compute the volume of the solid that is generated. x=3 1 INT Pi --------------- dx x=0 2 2 (x +5x+6) 1 A B C D -------------- = ---------- + -------- + ---------- + ----- 2 2 2 2 (x+2) (x+3) (x+2) (x+2) (x+3) (x+3) 1 -2 1 2 1 -------------- = ---------- + -------- + ---------- + ----- 2 2 2 2 (x+2) (x+3) (x+2) (x+2) (x+3) (x+3) x=3 1 | |x=3 INT Pi ------------ dx = Pi | -2 Log[x+2] -1/(x+2) + 2 Log[x+3]-1/(x+3) | x=0 2 2 | |x=0 (x +5x+6) Pi( (-2 Log[5] -1/5 +2 Log[6]-1/6)-(-2Log[2]-1/2+2Log[3]-1/3) ) = 0.0640243 ======================================== Get["font.math"]; f[x_] = 1/(x^2+5x+6); P1 = Plot[f[x],{x,0,3}]; P2 = Plot[0,{x,0,3}]; P3 = Show[P1,P2,PlotLabel->"P421 P46 y = 1/(x^2+5x+6)", AspectRatio->Automatic]; Display["46x.ps",P3]; P4 = ParametricPlot3D[{x,f[x] Cos[t],f[x] Sin[t]},{x,0,3},{t,0,2 Pi}, AspectRatio->Automatic,PlotLabel->"P421 P46 y = 1/(x^2+5x+6"]; Display["46y.ps",P4]; ============================================ Page 421 Problem 48 Find the volume of the solid created by revolving the region bounded by the x-axis and the curve y = 4 x Sqrt[2-x] about the y axis. x=2 INT 2 Pi x 4 x Sqrt[2-x] dx x=0 x=2 2 8 Pi INT x Sqrt[2-x] dx x=0 u = Sqrt[2-x] u^2 = 2-x x = 2-u^2 x=2 8 Pi INT (2-u^2)^2 u (-2u du) x=0 x=2 8 Pi INT (4-4 u^2 + u^4) (-2u^2 du) x=0 x=2 -16 Pi INT 4 u^2 -4 u^4 + u^6 du x=0 u = 0 | | x=2 -16 Pi | 4 u^3/3 -4 u^5/5 + u^7/7 | | | x=0 u=Sqrt[2] 16 Pi Sqrt[2](8/3-16/5+8/7) = 43.3287 <=================== ============================================ Get["font.math"]; f[x_] = 4 x Sqrt[2-x]; P1 = Plot[f[x],{x,0,2},AspectRatio->Automatic, PlotLabel->"P421 P48 y = 4 x Sqrt[2-x]" ]; Display["p48x.ps",P1]; P2 = ParametricPlot3D[{ x Cos[t],x Sin[t],f[x]}, {x,0,2},{t,-Pi/4, 3/2 Pi}]; P3 = Show[P2,PlotLabel->"P421 P48 y = 4 x Sqrt[2-x]", AspectRatio->Automatic]; Display["p48y.ps",P3]; Page 421 Problem 50 Find the area of the region bounded by the x-axis 18 the curve y = --------------- 2 2 x Sqrt[x + 9] and the lines x=Sqrt[3] and x = 3 Sqrt[3] x = 3 Sqrt[3] 18 INT -------------- x = Sqrt[3] 2 2 x Sqrt[ x + 9 ] | /| |2 / | Sqrt[x + 9]/ | | / | | / | 3 | / | | / | | / | |/t | ----------+------------------------------------ | x x = 3 Ctn[t] 2 dx = -3 Csc [t] dt 2 Sqrt[x + 9] = 3 Csc[t] 18 2 INT -------------------- (-3 Csc [t] dt 2 9 Ctn [t] 3 Csc[t] 2 -2 Sin [t] 1 INT -------------------- --------- dt 2 Cos [t] Sin[t] Sin[t] -2 INT ----------- dt 2 Cos [t] t = Pi/6 t = ArcTan[1/Sqrt[3]] -1 | x = 3 Sqrt[3] -2 ( -1/Cos[t] ) | | x = Sqrt[3] t = ArcTan[Sqrt[3]] = Pi/3 2(2-2/Sqrt[3]) = 1.6906 Get["font.math"]; f[x_] := 18/(x^2 Sqrt[x^2+9]); P1 = Plot[f[x],{x,Sqrt[3],3 Sqrt[3]}]; P2 = Show[P1,PlotLabel->"P421 P50 y = 18/(x^2 Sqrt[x^2+9]", AspectRatio->Automatic]; Display["50.ps",P2]; Answer short answer questions 1-28. 1.___ 2.___ 3.___ 4.___ 5.___ 6.___ 7.___ 8.___ 9.___ 10.___ 11.___ 12.___ 13.___ 14.___ 15.___ 16.___ 17.___ 18.___ 19.___ 20.___ 21.___ 22.___ 23.___ 24.___ 25.___ 26.___ 27.___ 28.___ Find the reduction formula for; n a) INT Sin [x] dx n b) INT Cos [x] dx n c) INT Tan [x] dx n d) INT Sec [x] dx ......................................................................... n-1 Sin [x] -Cos[x] n-2 (n-1) Sin [x] Cos[x] dx Sin[x] dx n n-1 n-2 2 a) INT Sin [x] = -Sin [x] Cos[x] + (n-1) INT Sin [x] Cos [x] dx n n-1 n-2 2 INT Sin [x] = -Sin [x] Cos[x] + (n-1) INT Sin [x] (1-Sin [x]) dx n n-1 n-2 INT Sin [x] = -Sin [x] Cos[x] + (n-1) INT Sin [x] dx n -(n-1) INT Sin [x] dx n n-1 n-2 n INT Sin [x] = -Sin [x] Cos[x] + (n-1) INT Sin [x] dx n-1 n -Sin [x] Cos[x] (n-1) n-2 INT Sin [x] = ---------------- + ----- INT Sin [x] dx n n ......................................................................... n-1 Cos [x] Sin[x] n-2 (n-1) Cos [x](-Sin[x]) dx Cos[x] dx n n-1 n-2 2 b) INT Cos [x] dx = Cos [x] Sin[x] + (n-1) INT Cos [x] Sin [x] dx n n-1 n-2 2 INT Cos [x] dx = Cos [x] Sin[x] + (n-1) INT Cos [x] (1-Cos [x]) dx n n-1 n-2 INT Cos [x] dx = Cos [x] Sin[x] + (n-1) INT Cos [x] dx n -(n-1) INT Cos [x] dx n n-1 n-2 n INT Cos [x] dx = Cos [x] Sin[x] + (n-1) INT Cos [x] dx n-1 n Cos [x] Sin[x] n-1 n-2 INT Cos [x] dx = ------------------ + ----- INT Cos [x] dx n n ......................................................................... n n-2 2 c) INT Tan [x] dx = INT Tan [x] (Sec [x] - 1 ) dx n-2 2 n-2 = INT Tan [x] Sec [x] dx - INT Tan [x] dx n-1 Tan [x] n-2 = ---------- - INT Tan [x] dx n-1 ......................................................................... n-2 Sec [x] dx Tan[x] n-3 2 (n-2) Sec [x] Sec[x] Tan[x] dx Sec [x] dx n n-2 n-2 2 INT Sec [x] dx = Sec [x] Tan[x] - (n-2) INT Sec [x] Tan [x] dx n n-2 n-2 2 INT Sec [x] dx = Sec [x] Tan[x] - (n-2) INT Sec [x] (Sec [x]-1) dx n n-2 n INT Sec [x] dx = Sec [x] Tan[x] - (n-2) INT Sec [x] dx n-2 +(n-2) INT Sec [x] dx n n-2 n-2 (n-1) INT Sec [x] dx = Sec [x] Tan[x] + (n-2) INT Sec [x] dx n-2 n Sec [x] Tan[x] n-2 n-2 INT Sec [x] dx = ------------------ + --- INT Sec [x] dx n-1 n-1 .........................................................................