{VERSION 5 0 "SUN SPARC SOLARIS" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "Hyperlink" -1 17 "" 0 1 0 128 128 1 2 0 1 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 3" -1 5 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 18 1 {CSTYLE "" -1 -1 "Tim es" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 } {PSTYLE "Author" -1 19 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 8 8 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 4" -1 20 1 {CSTYLE "" -1 -1 "Times" 1 10 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times " 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Normal" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 18 "" 0 "" {TEXT -1 47 "Surfaces of Revolution: Volume and Surface Area" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 256 27 "Calc ulus I Lab -- Fall 2002" }}{PARA 19 "" 0 "" {TEXT -1 11 "prepared by" }}{PARA 19 "" 0 "" {TEXT -1 16 "Douglas B. Meade" }}{PARA 256 "" 0 "" {TEXT -1 25 "Department of Mathematics" }}{PARA 256 "" 0 "" {TEXT -1 28 "University of South Carolina" }}{PARA 256 "" 0 "" {TEXT -1 18 "Col umbia, SC 29208" }}{PARA 256 "" 0 "" {TEXT -1 25 "E-mail: meade@math.s c.edu" }}{PARA 19 "" 0 "" {TEXT -1 16 "27 November 2002" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {SECT 1 {PARA 3 "" 0 "" {TEXT -1 7 "Purpose" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 550 "The purpose of this week's lab is to introduce surfaces \+ of revolution and to show how calculus can be used to determine the su rface area and volume of these surfaces. In particular, both the surfa ce area and the volume of a surface of revolution can be expressed as \+ definite integrals. (Evaluating these definite integrals can be compli cated; you will learn additional techniques for evaluating integrals i n Math 142.) Note that while Maple can evaluate most definite integral s we will encounter, you must determine the appropriate definite integ ral." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 161 " This lab can be prepared in any format -- Word document, Maple workshe et, text file, handwritten (provided it is neat!). All reports should be mailed to Wally (" }{TEXT 17 14 "waltman@sc.edu" }{TEXT -1 66 ") w ith a postmark no later than midnight Friday, December 6, 2002." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 " " {TEXT -1 10 "Background" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 26 "Revi ew: Area and Arclength" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Area" }} {SECT 1 {PARA 20 "" 0 "" {TEXT -1 15 "Vertical Slices" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 " " }{XPPEDIT 18 0 "A" "6#%\"AG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Limit( Sum( `( area of rectangle )`, i=1..n ), n=infinity )" "6#-%&LimitG6$-%$SumG6$%6(~area~of~rectangle~)G/%\"iG; \"\"\"%\"nG/F.%)infinityG" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 11 " = " }{XPPEDIT 18 0 "Limit( Sum( `( height )`[i] * Delt a*x[i], i=1..n ), n=infinity )" "6#-%&LimitG6$-%$SumG6$*(&%+(~height~) G6#%\"iG\"\"\"%&DeltaGF.&%\"xG6#F-F./F-;F.%\"nG/F5%)infinityG" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 11 " = " }{XPPEDIT 18 0 " Int( `( top - bottom )`, x=a..b )" "6#-%$IntG6$%1(~top~-~bottom~)G/%\" xG;%\"aG%\"bG" }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 20 "" 0 "" {TEXT -1 17 "Horizontal Slices" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 " " }{XPPEDIT 18 0 "A" "6#%\" AG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Limit( Sum( `( area of rectangle )`, i=1..n ), n=infinity )" "6#-%&LimitG6$-%$SumG6$%6(~area~of~rectan gle~)G/%\"iG;\"\"\"%\"nG/F.%)infinityG" }{TEXT -1 14 " \n = \+ " }{XPPEDIT 18 0 "Limit( Sum( `( width )`[i] * Delta*y[i], i=1..n ), n =infinity )" "6#-%&LimitG6$-%$SumG6$*(&%*(~width~)G6#%\"iG\"\"\"%&Delt aGF.&%\"yG6#F-F./F-;F.%\"nG/F5%)infinityG" }{TEXT -1 14 " \n \+ = " }{XPPEDIT 18 0 "Int( `( right - left )`, y=c..d )" "6#-%$IntG6$%1( ~right~-~left~)G/%\"yG;%\"cG%\"dG" }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 9 "A rclength" }}{SECT 1 {PARA 20 "" 0 "" {TEXT -1 23 "Horizontal Subdivisi ons" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 " " }{XPPEDIT 18 0 "L" " 6#%\"LG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Limit( Sum( `( length of li ne segment )`, i=1..n ), n=infinity )" "6#-%&LimitG6$-%$SumG6$%;(~leng th~of~line~segment~)G/%\"iG;\"\"\"%\"nG/F.%)infinityG" }{TEXT -1 2 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 11 " = " }{XPPEDIT 18 0 "Limit( S um( Delta*s[i], i=1..n ), n=infinity )" "6#-%&LimitG6$-%$SumG6$*&%&Del taG\"\"\"&%\"sG6#%\"iGF+/F/;F+%\"nG/F2%)infinityG" }{TEXT -1 2 " " }} {PARA 0 "" 0 "" {TEXT -1 11 " = " }{XPPEDIT 18 0 "Limit(Sum(sq rt(1+`f'`(c[i])^2)*Delta*x[i],i = 1 .. n),n = infinity);" "6#-%&LimitG 6$-%$SumG6$*(-%%sqrtG6#,&\"\"\"F.*$-%#f'G6#&%\"cG6#%\"iG\"\"#F.F.%&Del taGF.&%\"xG6#F6F./F6;F.%\"nG/F>%)infinityG" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 11 " = " }{XPPEDIT 18 0 "Int( sqrt( 1 + Dif f(f(t), t )^2 ), t=a..b )" "6#-%$IntG6$-%%sqrtG6#,&\"\"\"F**$-%%DiffG6 $-%\"fG6#%\"tGF2\"\"#F*/F2;%\"aG%\"bG" }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 20 "" 0 "" {TEXT -1 21 "Vertical Subdivisions" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 " " }{XPPEDIT 18 0 "L" "6#%\"LG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Limit( Sum( `( length of line segment )`, i=1..n ), n=infinity ) " "6#-%&LimitG6$-%$SumG6$%;(~length~of~line~segment~)G/%\"iG;\"\"\"%\" nG/F.%)infinityG" }{TEXT -1 14 " \n = " }{XPPEDIT 18 0 "Limit ( Sum( Delta*s[i], i=1..n ), n=infinity )" "6#-%&LimitG6$-%$SumG6$*&%& DeltaG\"\"\"&%\"sG6#%\"iGF+/F/;F+%\"nG/F2%)infinityG" }{TEXT -1 14 " \+ \n = " }{XPPEDIT 18 0 "Limit(Sum(sqrt(`g'`(c[i])^2+1)*Delta*y[ i],i = 1 .. n),n = infinity);" "6#-%&LimitG6$-%$SumG6$*(-%%sqrtG6#,&*$ -%#g'G6#&%\"cG6#%\"iG\"\"#\"\"\"F7F7F7%&DeltaGF7&%\"yG6#F5F7/F5;F7%\"n G/F>%)infinityG" }{TEXT -1 14 " \n = " }{XPPEDIT 18 0 "Int( s qrt( Diff(g(t), t )^2 + 1), t=c..d )" "6#-%$IntG6$-%%sqrtG6#,&*$-%%Dif fG6$-%\"gG6#%\"tGF1\"\"#\"\"\"F3F3/F1;%\"cG%\"dG" }{TEXT -1 2 " " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 20 "" 0 " " {TEXT -1 18 "Arclength Function" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "The arclength function is" }}{PARA 0 "" 0 "" {TEXT -1 6 " " } {XPPEDIT 18 0 "s(x)" "6#-%\"sG6#%\"xG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int(``,s = a .. x);" "6#-%$IntG6$%!G/%\"sG;%\"aG%\"xG" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 15 " = " }{XPPEDIT 18 0 "I nt(sqrt(1+`f'`(t)^2),t = a .. x);" "6#-%$IntG6$-%%sqrtG6#,&\"\"\"F**$- %#f'G6#%\"tG\"\"#F*/F/;%\"aG%\"xG" }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 23 "Volume and Surface Area" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 47 "General Ideas for slice, approximate, integrate" }} {SECT 1 {PARA 20 "" 0 "" {TEXT -1 6 "Volume" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 " " }{XPPEDIT 18 0 "V = Limit(Sum(`( cross sectional a rea )`,i = 1 .. n),n = infinity);" "6#/%\"VG-%&LimitG6$-%$SumG6$%9(~cr oss~sectional~area~)G/%\"iG;\"\"\"%\"nG/F0%)infinityG" }{TEXT -1 2 " \+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 20 " " 0 "" {TEXT -1 12 "Surface Area" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 " " }{XPPEDIT 18 0 "S = Limit(Sum(`( cross sectional surface area )`,i = 1 .. n),n = infinity)" "6#/%\"SG-%&LimitG6$-%$SumG6$%A(~cross~ sectional~surface~area~)G/%\"iG;\"\"\"%\"nG/F0%)infinityG" }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 151 " Ttypical cross-sections are triangles, squares, rectangles, circles, a nd washers. Circles and washers are typically obtained from solids of \+ revolution." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 31 "Volume of a solid of revolution" }} {SECT 1 {PARA 20 "" 0 "" {TEXT -1 15 "Horizontal axis" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "For a solid obtained by revolving about a horiz ontal axis (x-axis or y=c) " }}{PARA 0 "" 0 "" {TEXT -1 6 " " } {XPPEDIT 18 0 "V = Limit(Sum(`( cross sectional area )`,i = 1 .. n),n \+ = infinity)" "6#/%\"VG-%&LimitG6$-%$SumG6$%9(~cross~sectional~area~)G/ %\"iG;\"\"\"%\"nG/F0%)infinityG" }{TEXT -1 12 "\n = " } {XPPEDIT 18 0 "Limit( Sum( ( Pi*`( outer radius )`[i]^2 - Pi*`( inner \+ radius )`[i]^2 )*Delta*x[i], i=1..n ), n=infinity )" "6#-%&LimitG6$-%$ SumG6$*(,&*&%#PiG\"\"\"*$&%1(~outer~radius~)G6#%\"iG\"\"#F-F-*&F,F-*$& %1(~inner~radius~)G6#F2F3F-!\"\"F-%&DeltaGF-&%\"xG6#F2F-/F2;F-%\"nG/F@ %)infinityG" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 11 " \+ = " }{XPPEDIT 18 0 "Int(Pi*(f[outer](x)-c)^2-Pi*(f[inner](x)-c)^2,x = \+ a .. b);" "6#-%$IntG6$,&*&%#PiG\"\"\"*$,&-&%\"fG6#%&outerG6#%\"xGF)%\" cG!\"\"\"\"#F)F)*&F(F)*$,&-&F.6#%&innerG6#F2F)F3F4F5F)F4/F2;%\"aG%\"bG " }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {SECT 1 {PARA 20 "" 0 "" {TEXT -1 13 "Vertical axis" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "For a solid obtained by revolving about a vertica l axis (y-axis or x=a)" }}{PARA 0 "" 0 "" {TEXT -1 10 " V = " } {XPPEDIT 18 0 "Limit(Sum(`( cross sectional area )`,i = 1 .. n),n = in finity)" "6#-%&LimitG6$-%$SumG6$%9(~cross~sectional~area~)G/%\"iG;\"\" \"%\"nG/F.%)infinityG" }{TEXT -1 12 "\n = " }{XPPEDIT 18 0 "Li mit( Sum( ( Pi*`( outer radius )`[i]^2 - Pi*`( inner radius )`[i]^2 )* Delta*y[i], i=1..n ), n=infinity )" "6#-%&LimitG6$-%$SumG6$*(,&*&%#PiG \"\"\"*$&%1(~outer~radius~)G6#%\"iG\"\"#F-F-*&F,F-*$&%1(~inner~radius~ )G6#F2F3F-!\"\"F-%&DeltaGF-&%\"yG6#F2F-/F2;F-%\"nG/F@%)infinityG" } {TEXT -1 14 " \n = " }{XPPEDIT 18 0 "Int(Pi*(g[outer](y)-a)^2 -Pi*(g[inner](y)-a)^2,y = c .. d);" "6#-%$IntG6$,&*&%#PiG\"\"\"*$,&-&% \"gG6#%&outerG6#%\"yGF)%\"aG!\"\"\"\"#F)F)*&F(F)*$,&-&F.6#%&innerG6#F2 F)F3F4F5F)F4/F2;%\"cG%\"dG" }{TEXT -1 3 " \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 39 "Surfa ce area of a surface of revolution" }}{SECT 1 {PARA 20 "" 0 "" {TEXT -1 15 "Horizontal axis" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "For a su rface obtained by revolving about a horizontal axis (x-axis or y=c) " }}{PARA 0 "" 0 "" {TEXT -1 6 " " }{XPPEDIT 18 0 "S" "6#%\"SG" } {TEXT -1 3 " = " }{XPPEDIT 18 0 "Limit(Sum(`( cross sectional surface \+ area )`, i = 1 .. n),n = infinity)" "6#-%&LimitG6$-%$SumG6$%A(~cross~s ectional~surface~area~)G/%\"iG;\"\"\"%\"nG/F.%)infinityG" }{TEXT -1 14 " \n = " }{XPPEDIT 18 0 "Limit(Sum(`( circumference )`[i] \+ * Delta*s[i], i = 1 .. n),n = infinity)" "6#-%&LimitG6$-%$SumG6$*(&%2( ~circumference~)G6#%\"iG\"\"\"%&DeltaGF.&%\"sG6#F-F./F-;F.%\"nG/F5%)in finityG" }{TEXT -1 14 " \n = " }{XPPEDIT 18 0 "Limit(Sum( 2*P i*`( radius )`[i]*Delta*s[i],i = 1 .. n),n = infinity)" "6#-%&LimitG6$ -%$SumG6$*,\"\"#\"\"\"%#PiGF+&%+(~radius~)G6#%\"iGF+%&DeltaGF+&%\"sG6# F0F+/F0;F+%\"nG/F7%)infinityG" }{TEXT -1 14 " \n = " } {XPPEDIT 18 0 "Int(2*Pi*(f(x)-c)*sqrt(1+Diff(f(x),x)^2),x = a .. b);" "6#-%$IntG6$**\"\"#\"\"\"%#PiGF(,&-%\"fG6#%\"xGF(%\"cG!\"\"F(-%%sqrtG6 #,&F(F(*$-%%DiffG6$-F,6#F.F.F'F(F(/F.;%\"aG%\"bG" }{TEXT -1 1 " " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 20 "" 0 " " {TEXT -1 13 "Vertical axis" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "Fo r a surface obtained by revolving about a vertical axis (y-axis or x=a ) " }}{PARA 0 "" 0 "" {TEXT -1 10 " S = " }{XPPEDIT 18 0 "Limit(S um(`( cross sectional surface area )`, i = 1 .. n),n = infinity)" "6#- %&LimitG6$-%$SumG6$%A(~cross~sectional~surface~area~)G/%\"iG;\"\"\"%\" nG/F.%)infinityG" }{TEXT -1 14 " \n = " }{XPPEDIT 18 0 "Limit (Sum(`( circumference )`[i] * Delta*s[i], i = 1 .. n),n = infinity)" " 6#-%&LimitG6$-%$SumG6$*(&%2(~circumference~)G6#%\"iG\"\"\"%&DeltaGF.&% \"sG6#F-F./F-;F.%\"nG/F5%)infinityG" }{TEXT -1 14 " \n = " } {XPPEDIT 18 0 "Limit(Sum( 2*Pi*`( radius )`[i]*Delta*s[i],i = 1 .. n), n = infinity)" "6#-%&LimitG6$-%$SumG6$*,\"\"#\"\"\"%#PiGF+&%+(~radius~ )G6#%\"iGF+%&DeltaGF+&%\"sG6#F0F+/F0;F+%\"nG/F7%)infinityG" }{TEXT -1 14 " \n = " }{XPPEDIT 18 0 "Int(2*Pi*(g(y)-a)*sqrt(Diff(g(y), y)^2+1),y = c..d)" "6#-%$IntG6$**\"\"#\"\"\"%#PiGF(,&-%\"gG6#%\"yGF(% \"aG!\"\"F(-%%sqrtG6#,&*$-%%DiffG6$-F,6#F.F.F'F(F(F(F(/F.;%\"cG%\"dG" }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} }{SECT 1 {PARA 3 "" 0 "" {TEXT -1 8 "Example:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "restart;\nwith( plots ):\nwith( Student[Calculus 1] ):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Let R be the region in the first quadrant bound ed by " }{XPPEDIT 18 0 "y = (4-x^(2/3))^(3/2);" "6#/%\"yG),&\"\"%\"\" \")%\"xG*&\"\"#F(\"\"$!\"\"F.*&F-F(F,F." }{TEXT -1 6 ", the " } {XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 20 "-axis, and the line " } {XPPEDIT 18 0 "x=1" "6#/%\"xG\"\"\"" }{TEXT -1 173 ". We will see how \+ to use Maple to graph the region, find the area of R, the arclength of the curved segment of the boundary of the region, the solid of revolu tion about the " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 7 "-axis (" } {XPPEDIT 18 0 "y=0" "6#/%\"yG\"\"!" }{TEXT -1 73 ") and its volume and the surface area of the curved portion of the solid." }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "Graph" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "Graph the region" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "f : = (4-x^(2/3))^(3/2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "The left \+ endpoint is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "a := 1;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "The right endpoint is where the cu rve crosses the x-axis" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "s olve( f=0, x );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "b := 8;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "plot( f, x=a..b, filled=true, view=[0..10,0..6], scaling=constrained );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 14 "Area of Region" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "To find the area of the region, with ver tical slices, it is necessary to identify the top and bottom of the re gion:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "top := f;\nbottom \+ := 0;" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "area := Int( top-bottom, x=a..b );" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "Note that this definite integral \+ is one that we do not know how to evaluate. But, asking Maple to evalu ate it we find" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "area = va lue( area );\n `` = simplify(value( area ));\n `` = evalf( area );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Arclength" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "To f ind the length of curved segment of boundary we find the differential \+ arclength" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "ds := sqrt( 1 \+ + diff( top, x )^2 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "ds := simplify( ds );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Then, the \+ length of this segment is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "L := Int( ds, x=a..b );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 " L = value( L );\n`` = simplify( value(L) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 30 "Sol id of Revolution about the " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 7 "-axis (" }{XPPEDIT 18 0 "y=0" "6#/%\"yG\"\"!" }{TEXT -1 1 ")" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "To plot the solid of revolution fo rmed by rotation R about the x-axis, i.e., y=0, use the " }{HYPERLNK 17 "VolumeOfRevolution" 2 "Student,Calculus1,VolumeOfRevolution" "" } {TEXT -1 9 " command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 121 "V olumeOfRevolution( f, x=a..b, axis=horizontal, output=plot, axes=box, \+ view=[0..10,-8..8,-8..8], orientation=[-120,45] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "The definite integral for the volume of this solid is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "volume := VolumeOfRevolution( f, x=a..b, ax is=horizontal, output=integral );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "This integral can be evaluated, but only after the integrand is ex panded" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "volume := expand( volume );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "According to Maple, the exact and numeric al values for the volume are:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "volume = value( volume );\n `` = simplify( value(volume) );\n `` = evalf( volume );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 149 "The surface area of curved \+ portion of the solid is a surface of revolution -- the surface of revo lution of the curved segment of R rotated about the " }{XPPEDIT 18 0 " x" "6#%\"xG" }{TEXT -1 11 "-axis. The " }{HYPERLNK 17 "SurfaceOfRevolu tion" 2 "Student,Calculus1,SurfaceOfRevolution" "" }{TEXT -1 75 " comm and can be used to obtain the definite integral for this surface area: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "surfacearea := SurfaceO fRevolution( f, x=a..b, axis=horizontal, output=integral );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "surfacearea := expand( surfa cearea );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 185 "The first term in t his integral can be evaluated with a substitution u=4-x^(2/3). However , the second term cannot be evaluated using the techniques we know. Re gardless, Maple informs us" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "surfacearea = value( surfacearea );\n `` = simplify( valu e(surfacearea) );\n `` = evalf( surfacearea );" }{TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 69 "Exer cise 1 -- OMIT! (solution provided for those who are interested)" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "Consider the solid of revolution f ormed by rotating the region R in the Example around the y-axis." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 " a) Find \+ the definite integrals for the volume and surface area of this solid. " }}{PARA 0 "" 0 "" {TEXT -1 153 " (Note that the surface area is f ound in three pieces - a cylinder, a disk, and a surface of revolution . Be sure to clearly identify all three terms.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 " b) Which of the definite integrals could be evaluated using the methods from this course?" }} {PARA 0 "" 0 "" {TEXT -1 85 " (If you can evaluate the integral, g ive an antiderivative. If not, explain why.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 " c) What are the exact and nume ric values for these integrals?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 208 "To rotate about a vertical axis requires horizontal slic es. For this we need to identify the left and right edges of the regio n (as functions of y) and find the appropriate vertical interval for i ntegration. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "The upper and low er limits of integration will be" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "c := 0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 " d := eval( f, x=a );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "The left \+ edge is not difficult to find:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "left := 1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "To find the right edge (the curved segment), we must convert y=f(x) into x=g(y): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eq1 := isolate( y=f, x \+ );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "right := rhs( eq1 ); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "To test this, plot the region described by these limits:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "reflect := plottools[transform]( (x,y) -> [y,x] ):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "Pright := plot( right, y=c..d ):\n Pleft := plot( left, y=c..d ):\ndisplay( [reflect( Pright ), reflect( \+ Pleft )], view=[0..10,0..6] );" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "Good, this is the same region as before. [Note: I do \+ not expect students to do all this Maple in their solution.]" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "To graph the solid of revolution use the VolumeOfRevoluti on command as before" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 145 "so lid := VolumeOfRevolution( left, right, y=c..d, axis=horizontal, outpu t=plot, axes=normal, view=[0..8,-10..10,-10..10], labels=[`y`,`x`,`z`] ):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "display( solid, orie ntation=[0,180] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "displ ay( solid, orientation=[0,90] );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Note the hole in the center of this solid! This is to be expected " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "volume := VolumeOfRevol ution( right, left, y=c..d, axis=horizontal, output=integral );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "volume = value( volume );\n \+ `` = evalf( volume );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "To verify this answer it shou ld be possible to write down the definite integral for the volume:" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "volumeA := Int( Pi*right^2 \+ - Pi*left^2, y=c..d );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "v olumeA = expand(volumeA);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "Fant astic! This is the same integral as found above!!!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "Now for the surface area. First, inner cylinder has radius 1 and height \+ ( " }{XPPEDIT 18 0 "3*sqrt(3) - 0" "6#,&*&\"\"$\"\"\"-%%sqrtG6#F%F&F& \"\"!!\"\"" }{TEXT -1 25 " ) so the surface area is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "surfacearea1 := SurfaceOfRevolution( left , y=c..d, axis=horizontal, output=integral );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "surfacearea1 = value( surfacearea1 );\n \+ `` = evalf( surfacearea1 );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "T he surface area of the curved surface is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "surfacearea2 := SurfaceOfRevolution( right, y=c..d, a xis=horizontal, output=integral );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "surfacearea2 = expand( surfacearea2 );\n `` = value( surfacearea2 );\n `` = evalf( surfacearea2 );" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "The third component of the surfac e area is the area of the washer with outer radius 8 and inner radius \+ 1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "surfacearea3 := Pi*b^ 2 - Pi*a^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "The total surface \+ area of this surface is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 " totalsurfacearea := surfacearea1 + surfacearea2 + surfacearea3;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "totalsurfacearea = value( to talsurfacearea );\n `` = evalf( totalsurfacearea );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 153 "To verify these computations, find alternate approaches \+ to the surface area. For the inner surface, this is simply a cylinder \+ with radius 1 and height ( " }{XPPEDIT 18 0 "sqrt(3)*3 - 0" "6#,&*&-%% sqrtG6#\"\"$\"\"\"F(F)F)\"\"!!\"\"" }{TEXT -1 3 " ):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "surfacearea1A := 2*Pi*left*(d-c);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "For the curved surface, the defini te integral can be written directly as" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "g := right;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "ds := sqrt( diff(g,y)^2 + 1 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "surfacearea2A := Int( 2*Pi*right*ds, y=c..d );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 129 "surfacarea2A = expand( surf acearea2A );\n `` = simplify( value( surfacearea2A ) );\n \+ `` = evalf( surfacearea2A );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "There is no need to cross-verify the third component of the sur face area." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "surfacearea3A := surfacearea3;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "The total su rface area by these methods is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "totalsurfaceareaA := surfacearea1A + surfacearea2A + surfacear ea3A;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "totalsurfaceareaA \+ = value( totalsurfaceareaA );\n `` = evalf( totalsurface areaA );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Great news - everythi ng checks!!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 10 "Exercise 2" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "Let R be the region defined in the Example (and Exercise 1) and consider the surface formed by rotating the region R about " } {XPPEDIT 18 0 "y=-2" "6#/%\"yG,$\"\"#!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 " a) Graph this so lid of revolution --- OMIT! (solution provided for anyone who is inter ested)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 " b) Determine definite integrals for the volume and the surface area. " }}{PARA 0 "" 0 "" {TEXT -1 74 " (How are these integrals differe nt from the integrals in Exercise 2?)" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 37 " c) Evaluate these definite integrals " }}{PARA 0 "" 0 "" {TEXT -1 141 " (Be sure to identify, by giving an antiderivative, all terms in these integrals that can be evaluated using the methods in this course.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {SECT 1 {PARA 4 "" 0 "" {TEXT -1 11 "a) Solution" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Here is a plot of the surface of revolution. The " } {HYPERLNK 17 "VolumeOfRevolution" 2 "Student,Calculus1,VolumeOfRevolut ion" "" }{TEXT -1 36 " command allows rotations about the " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 5 "- or " }{XPPEDIT 18 0 "y" "6#%\"yG" } {TEXT -1 134 "-axis only. To force this problem into this structure si mply shift the entire region up by 2 units. (In general, for revolutio n about " }{XPPEDIT 18 0 "y=c" "6#/%\"yG%\"cG" }{TEXT -1 25 ", shift t he region up by " }{XPPEDIT 18 0 "-c" "6#,$%\"cG!\"\"" }{TEXT -1 26 " \+ units.) To implement this" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "c := -2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "solid := Volum eOfRevolution( top-c, bottom-c, x=a..b, axis=horizontal, output=plot ) :" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display( solid );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "Here are two additional views of the soli d. The first is looking down the " }{XPPEDIT 18 0 "y" "6#%\"yG" } {TEXT -1 57 "-axis (towards the origin). The second is looking in the \+ " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 27 "-axis (towards the origin )." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "display( solid, orien tation=[-90,90] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "displ ay( solid, orientation=[0,90] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "volume2 := VolumeOfRevolution( top-c, bottom-c, x=a..b, axis=horizontal, outp ut=integral );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "volume2 = expand( volume2 );\n `` = value( volume2 );\n `` = evalf( vol ume2 );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 99 "This value should agre e with the value that you find for the integrals you write down in b) \+ and c)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }