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.ICN Textbook: Calculus Early Vectors Preliminary Edition By Stewart. Wednesday, February 12, 7.3 Shells p444: (1),5,(10),12,16,19,(22),26,27,30,33, 38,(46) Main Idea: You can eat carrot coins or carrot shavings. Key Words: Thin Hollow Cylindrical Shells Goal: Learn to rotate the object around the correct axis and learn to set up volume integrals as either coins or shells. ------------------------------------------------------------- Previous Assignment Page 438 Problem 51 Cap of a sphere with radius r and height h. y=r INT Pi (Sqrt[r^2-y^2])^2 dy y=r-h y=r Pi INT r^2 - y^2 dy y=r-h y=r Pi | r^2 y - y^3/3 | y=r-h y=r Pi/3 | 3 r^2 y - y^3 | y=r-h Pi/3 ( 3r^3 - r^3 - ( 3 r^2 (r-h) - (r-h)^3 ) Pi/3 ( 2 r^3 - (3 r^3 - 3 r^2 h - (r-h)^3 ). Pi/3 ( 2 r^3 - 3 r^3 + 3 r^2 h + (r-h)^3 ) Pi/3 ( -r^3 + 3 r^2 h + r^3 - 3 r^2 h + 3 r h^2 - h^3 ) Pi/3 (3 r h^2 - h^3) ----------------------------------------------------------- Page 438 Problem 56 The base of S is a circular disk with radius r. Parallel cross-sections perpendicular to the base are squares. x=r INT (2 Sqrt[r^2 - x^2] )^2 dx x=-r INT 4 (r^2 - x^2) dx x=r 4 | r^2 x - x^3/3 | x=-r 4 ( r^3 - r^3/3 - (-r^3 + r^3/3) ) 4 (2 r^3/3 + 2r^3/3) = 16 r^3/3. f[t_] := Graphics3D[ Line[ { { Cos[t],Sin[t],0}, { Cos[t],Sin[t],2 Abs[Sin[t]]}, { Cos[t],-Sin[t],2 Abs[Sin[t]]}, { Cos[t],-Sin[t],0}, { Cos[t],Sin[t],0} } ] ]; a = Table[f[2 Pi i/50],{i,1,50}]; Show[a]; b = ParametricPlot3D[{Cos[t],Sin[t],0,RGBColor[1,0,0]},{t,0,2 Pi}]; c = ParametricPlot3D[{Cos[t],Sin[t],2 Abs[Sin[t]],RGBColor[0,1,0]},{t,0,2 Pi}]; d = Show[a,b,c,PlotLabel->"Page 438 Problem 56"] Display["56.ps",d]; e = Show[d,ViewPoint->{3,0,0},PlotLabel->"ViewPoint[3,0,0]"]; f = Show[d,ViewPoint->{0,3,0},PlotLabel->"ViewPoint[0,3,0]"]; g = Show[d,ViewPoint->{0,0,3},PlotLabel->"ViewPoint[0,0,3]"]; Display["e.ps",e]; Display["f.ps",f]; Display["g.ps",g]; ------------------------------------------------------------------- Page 438 Problem 66 A bowl is shaped like a hemisphere with diameter 30 cm. A ball with diameter 10 cm is placed in the bowl and water is poured into the bowl to a depth of h centimeters. Find the volume of water in the bowl. Go back to Problem 51. Volume of endcap of height h in a sphere of radius r is: V[r_,h_] := Pi/3 (3 r h^2 - h^3) Till depth is 5 cm: V[15,h]-V[5,h]; = 10 h^2 Pi 5 cm till 10 cm: V[15,h] - 4/3 Pi 5^3 + V[5,10-h]; = 10 h^2 Pi 10 cm to 15 cm: V[15,h]-4/3 Pi 5^3; = (45 h^2 - h^3-500) Pi/3 f[h_] := 10 h^2 Pi k[h_] := (45 h^2 - h^3 -500) Pi/3; a = Plot[f[h],{h,0,10},PlotStyle->{RGBColor[1,0,0]}]; b = Plot[k[h],{h,10,15}]; c = Show[a,b,PlotLabel->"Page 438 Problem 66"]; Display["66.ps",c]; ======================================================================= Problem: A person takes a peel of width 2b all the way around a spherical apple of radius a. What is the amount of apple removed. . ''' . . /| . / |/// ' a/ |//// c = Sqrt[a^2-b^2] / b|///// ' /____|///// c |///// ' |///// |//// ' |///' |// ' . .| ' ''' Work this three different ways. (1) Use thin hollow cylindrical shells. Manchester and Cedar Rapids. ---------------------------------------------------- x=a INT 2 Pi x 2 Sqrt[a^2-x^2] dx x=Sqrt[a^2-b^2] (a^2-x^2)^(3/2) | x=a -2Pi ------------ | 3/2 | x=Sqrt[a^2-b^2] -4/3 Pi ( 0 - (b^2)^(3/2) = 4/3 Pi b^3. (2) Use Coin method. Ames and Eldridge ---------------------------------------------------- y=b INT [ Pi Sqrt[a^2-y^2]^2 - Pi(Sqrt[a^2-b^2])^2 ] dy y=-b y=b Pi INT a^2 - y^2 - (a^2-b^2) dy y=-b | y^3 |y=b Pi | b^2 y - ---- | | 3 |y=-b | 2b^3 | Pi | 2 b^3 y - ----- | = 4/3 Pi b^3 | 3 | (3) Subtract the 2 end caps and the cylinder from the volume of the whole sphere. End-cap Anamosa and First Row -------------------------------------------------------- y=a INT Pi (Sqrt[a^2-y^2])^2 dy y=b y=a INT Pi (a^2-y^2 ) dy y=b | | y=a Pi|(a^2 y - y^3/3) | | | y=b Pi ( a^3 -a^3/3 -(a^2 b - b^3/3) Pi ( (2/3) a^3 - a^2 b + b^3/3) Cylinder Abraham Lincoln and Back two rows. ------------------------------------------------------ cylinder = Pi Sqrt[a^2-b^2]^2 2b cylinder = 2 Pi b(a^2-b^2) Volume of the ring is Volume(sphere) - 2*end-caps - cylinder 4/3 Pi a^3 -2 Pi ( (2/3) a^3 - a^2 b + b^3/3) - 2 Pi b(a^2-b^2) ========== ================================= =============== | | Pi | 4/3 a^3 -4/3 a^3 + 2 a^2 b - 2 b^3/3 -2 a^2 b + 2 b^3 | | | | | Pi | (4/3) b^3 | | | The volume of the ring does not depend on the radius of the sphere. No matter how big the apple was, the amount removed depends only on the width of the peel. ================================================ Find the volume of a lopsided cone with base B and height h. y=h Volume = INT B(y/h)^2 dh y=0 | |y=h = | B/h^2 y^3/3 | = (1/3) Bh. | |y=0 f[h_] := ParametricPlot3D[ { h(3+Cos[2 t])Cos[t]/10, h(2+Sin[9 t])Sin[t]/10,h,RGBColor[1,0,0]},{t,0,2 Pi}]; g[t_] := ParametricPlot3D[ { h(3+Cos[2 t])Cos[t]/10, h(2+Sin[9 t])Sin[t]/10,h,RGBColor[0,1,0]},{h,0,20}]; A = Table[f[5 h],{h,0,4}]; B = Table[g[2 t Pi/10],{t,0,10}];