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. 

Friday, January 18   5.3  Shells       



p293:13,14,16,19

Main Idea:  Think about making a log into a long sheet

of veneer. 

Key Words: Shell Method 

Goal: Learn the Shell method to find volumes  

--------------------------------------------------------
Previous assignment:           p286:(12),(18),(24),(26)


Page 286 Problem 12

Find the volume generated by rotating the region 

about the y-axis.

x = 2/y

y = 2

y = 6

x = 0

y=6          2                   y=6     2
INT  Pi (2/y)   dy       =  4 Pi INT  1/y   dy
y=2                              y=2

     |       | y=6
4 Pi |(-1/y) |        =  4 Pi ( -1/6 -(-1/2)) = 4 Pi/3
     |       | y=2

------------------------------------------------------------
Integrate[ Pi (2/y)^2,{y,2,6}] = 4 Pi/3
-------------------------------------------
Get["font.math"];
p1 = Plot[2/x,{x,1/3,1},PlotStyle->{RGBColor[1,0,0],Thickness[0.02]}];
p2 = Plot[2,{x,  0,1},PlotStyle->{RGBColor[1,0,0],Thickness[0.02]}];
p3 = Plot[6,{x,0,1/3},PlotStyle->{RGBColor[1,0,0],Thickness[0.02]}];
p4 = ParametricPlot[{0,y},{y,2,6},
      PlotStyle->{RGBColor[1,0,0],Thickness[0.02]}];
p5 = Show[p1,p2,p3,p4,PlotLabel->"P286 p12 x = 2/y",
      PlotRange->All,AspectRatio->Automatic];

Display["p12x.ps",p5];

p6 = ParametricPlot3D[{2/y Cos[t], 2/y Sin[t], y},{t,0,2 Pi}, {y,2,6}];
p7 = Show[p6,PlotLabel->"P286 P12: x = 2/y",
         AspectRatio->Automatic,Axes->True];
Display["p12y.ps",p7];
=================================================================



Page 286 Problem 18

Find the volume of the solid generated by revolving

about the x-axis the region bounded by the line y = 6x

                        2
and the parabola y = 6 x  

----------------------------------------------------------------
Get["font.math"];
p1 = Plot[6 x,   {x,0 ,1},PlotStyle->{RGBColor[1,0,0],Thickness[0.05]}];
p2 = Plot[6 x^2 ,{x,0 ,1},PlotStyle->{RGBColor[1,0,0],Thickness[0.05]}];

p3 = Show[p1,p2,PlotLabel->"P286 P18 y = 6x and y = 6x^2",
     AspectRatio->Automatic];

Display["p18x.ps",p3];

f[y_] := ParametricPlot3D[{x,y Cos[t],y Sin[t]},{t,-Pi/2,Pi/2},
        {x,y/6,Sqrt[y/6]},
     PlotRange->All,AspectRatio->Automatic]; 

p4 = Show[ f[1],f[ 2],f[ 3],f[ 4],f[ 5],f[ 5.9],
          AspectRatio->Automatic,PlotRange->All,
          PlotLabel->"P286 P18 y = 6x and y =6x^2"];

g[t_] := ParametricPlot3D[ { x, 6x Cos[t],6x Sin[t],
           {RGBColor[1,0,0],Thickness[0.01]}},{x,0,1}];

h[t_] := ParametricPlot3D[ { x, 6x^2 Cos[t],6x^2 Sin[t],
           {RGBColor[1,0,0],Thickness[0.02]}},{x,0,1}];


p5 = Show[ 
           g[-1 Pi/8],h[-1 Pi/8],
           g[-2 Pi/8],h[-2 Pi/8],
           g[-3 Pi/8],h[-3 Pi/8],
           g[-4 Pi/8],h[-4 Pi/8],
           g[ 0 Pi/8],h[ 0 Pi/8],
           g[ 1 Pi/8],h[ 1 Pi/8],
           g[ 2 Pi/8],h[ 2 Pi/8],
           g[ 3 Pi/8],h[ 3 Pi/8],
           g[ 4 Pi/8],h[ 4 Pi/8] 
          ];

p6 = Show[p4,p5,PlotLabel->"P286 P18 y = 6x and y = 6x^2  ##############"]

Display["p18y.ps",p6];
----------------------------------------------------------------


x=1         2          2 2
INT  Pi (6x)   - Pi (6x )   dx
x=0

      x=1   2     4
36 Pi INT  x   - x  dx
      x=0

      |  3      5    | x=1
36 Pi | x /3 - x /5  |
      |              | x=0

36 Pi (1/3-1/5) = 72 Pi/15  = 15.0796
---------------------------------------------------------------
Integrate[ Pi (6x)^2 - Pi (6x^2)^2,{x,0,1}] = 24 Pi/5
Integrate[ 6x-6x^2,{x,0,1}] = 1 so original area is one.
---------------------------------------------------------


Get["font.math"];
p1 = ParametricPlot3D[{x, 6x Cos[t], 6x Sin[t]},{t,-Pi/2   , Pi/2 },{x,0,1}];
p2 = ParametricPlot3D[{x, 6x^2 Cos[t], 6x^2 Sin[t]},{t,-Pi/2,Pi/2 },{x,0,1}];
p3 = Show[p1,p2,PlotLabel->"P286 P18 y=6x, y = 6x^2 ##########"];
Display["p18x.ps",p3];
------------------------------------------------------------------

Page 286 Problem 24

The base of a solid is the region inside the circle

 2    2
x  + y   = 4.  Find the volume of the solid if every

cross section by a plane perpendicular to the x-axis

is an isosceles triangle with base on the xy-plane 

and altitude 4.

x= 2
INT    1/2 base*height dx
x=-2

x= 2                 2
INT    1/2 2 Sqrt[4-x ]*4 dx
x=-2

   x=2             2
4  INT     Sqrt[4-x ]   dx    The integral is the area of 1/2 
   x=-2                       of a circle of radius 2 or 
                              1/2 2^2 Pi = 2 Pi.
4 (1/2) Pi 2^2 = 8 Pi

------------------------------------------------------------------
Integrate[ 1/2 2 Sqrt[4-x^2] 4,{x,-2,2}]
-----------------------------------------------
Get["font.math"];
f[x_] :=   Sqrt[4-x^2];
p[x_] := Graphics3D[ Line[{{x,-f[x],0},{x,f[x],0},{x,0,4},{x,-f[x],0}}]]; 
p1 = Table[p[2 x/20],{x,-20,20}];
p2 = ParametricPlot3D[{2 Cos[t],2 Sin[t],0},{t,0,2 Pi}];
p3 = ParametricPlot3D[{x,0,4,RGBColor[1,0,0]},{x,-2,2}];
p4 = Show[p1,p2,p3,PlotLabel->"Page 286 Problem 24"];

Display["p24.ps",c];
--------------------------------------------------------


Page 286 Problem 26

The base of a solid is the region bounded by

       2            4
y = 1-x  and y = 1-x .  Cross sections of the

solid that are perpendicular to the x-axis are

squares.  Find the volume of the solid.

---------------------------------------------------------------------------
Get["font.math"];
p1 = Plot[{1-x^2,1-x^4},{x,-1,1}]
p2 = Show[p1,PlotLabel->"Page 286 Problem 26 y = 1-x^2, y=1-x^4",
     PlotRange->All,AspectRatio->Automatic];
Display["p26.ps",p2];
---------------------------------------------------------------------------


We will find the volume in the first quadrant and then double it.

x=1      4      2  2
INT  (1-x  - 1+x  )   dx
x=0

x=1     2  4 2
INT   (x -x )   dx
x=0

x=1        4       6     8
INT       x   - 2 x   + x   dx
x=0
     |    5       7      9   | x=1
     |   x /5 -2 x /7 + x /9 |
     |                       | x=0

     1/5 - 2/7 + 1/9 = 8/315   So the total volume is 16/315.
-------------------------------------------
Integrate[ (1-x^4 -1+x^2)^2,{x,0,1}]
------------------------------------------------------
Get["font.math"];
f[x_] := 1-x^2;

g[x_] := 1-x^4;

d[x_] := Abs[f[x]-g[x]];

p[x_] := Graphics3D[ {Thickness[0.005],
Line[{  {x,f[x],0},
        {x,g[x],0},
        {x,g[x],d[x]},
        {x,f[x],d[x]},
        {x,f[x],0}}]}];

p1 = Table[p[x/20],{x,-19,19}];

p2 = ParametricPlot3D[{x,f[x],0},{x,-1,1}];

p3 = ParametricPlot3D[{x,g[x],0},{x,-1,1}];

p4 = ParametricPlot3D[{x,f[x],g[x]-f[x],
              {RGBColor[1,0,0],Thickness[0.005]}}, {x,-1,1}];

p5 = ParametricPlot3D[{x,g[x],g[x]-f[x],
              {RGBColor[1,0,0],Thickness[0.005]}}, {x,-1,1}];

p6 = Show[p1,p2,p3,p4,p5,PlotLabel->"Page 286 Problem 26"];

Display["p26.ps",p6];
===================================================================
New Material

Page 288 Example 1

The region bounded by y = 1/Sqrt[x] and the x-axis, x=1 and x=4

is rotated about the y axis.  Find the volume of the resulting

solid.

      x=4
V  =  INT  2 Pi x 1/Sqrt[x] dx   = 28 Pi/3
      x=1
-----------------------------------------------------------------





Get["font.math"];
f[x_] := 1/Sqrt[x];
q1 = Plot[f[x],{x,1,4}];
q2 = ParametricPlot[{1,y},{y,0,f[1]}];
q3 = ParametricPlot[{4,y},{y,0,f[4]}];
q4 = Plot[0,{x,1,4}];
q5 = ParametricPlot[{0,y},{y,0,f[1]},
     PlotStyle->{RGBColor[1,0,0],Thickness[0.02]}];
q6 = Show[q1,q2,q3,q4,q5,PlotLabel->"P288 Ex 1 y = 1/Sqrt[x]",
     AspectRatio->Automatic,PlotRange->All,Axes->True];
Display["ex1x.ps",q6];
p[t_] := Graphics3D[ Line[{ 
         { 1 Cos[t], 1 Sin[t],0},
         { 4 Cos[t], 4 Sin[t],0},
         { 4 Cos[t], 4 Sin[t],f[4]},
         { 3 Cos[t], 3 Sin[t],f[3]},
         { 2 Cos[t], 2 Sin[t],f[2]},
         { 1 Cos[t], 1 Sin[t],f[1]},
         { 1 Cos[t], 1 Sin[t], 0}     }]];
A = Table[ p[ 2 Pi t/20],{t,0,20}];
p1 = ParametricPlot3D[ {Cos[t],Sin[t],0},{t,0,2 Pi}];
p2 = ParametricPlot3D[ 4 {Cos[t],Sin[t],0},{t,0,2 Pi}];
p3 = ParametricPlot3D[ {Cos[t],Sin[t],f[1]},{t,0,2 Pi}];
p4 = ParametricPlot3D[ {4 Cos[t],4 Sin[t], f[4]},{t,0,2 Pi}];
p5 = ParametricPlot3D[ {x,0,0,{RGBColor[1,0,0],Thickness[0.01]}},{x,1,4}];
p6 = ParametricPlot3D[ {4,0,z,{RGBColor[1,0,0],Thickness[0.01]}},{z,0,1/2}];
p7 = ParametricPlot3D[ {1,0,z,{RGBColor[1,0,0],Thickness[0.01]}},{z,0,1  }];
p8 = ParametricPlot3D[ {x,0,1/Sqrt[x],{RGBColor[1,0,0],Thickness[0.01]}},
                                 {x,1,4  }];
f[x_] :=  ParametricPlot3D[ {x Cos[t], x Sin[t],z},
            {t,0,3 Pi/2},{z,0,1/Sqrt[x]}];
p9 = Show[p1,p2,p3,p4,p5,p6,p7,p8,A,f[1],f[2],f[3],f[4],
          AspectRatio->Automatic,PlotRange->All,
     PlotLabel->"P288 ex1: f[x] = 1/Sqrt[x]"];
Display["ex1y.ps",p9];
----------------------------------------------------------------
Page 289 Example 2.  The region bounded by the line y = (r/h)x

the x axis and x=h is revolved about the x axis.  Find the

volume.

     y=r
V =  INT  2 Pi y (h - hy/r) dy 
     y=0

           y=r
V = 2 Pi h INT y(1-y/r) dy  
           y=0

           y=r      2
V = 2 Pi h INT y - y /r) dy  
           y=0

            |   2      3       | y=r
V = 2 Pi h  |  y /2 - y /(3r)  | 
            |                  | y=0


              2      3                   2
V = 2 Pi h ( r /2 - r /(3r) )  = 2 Pi h r /6 

          2
= 1/3 Pi r  h.   The volume of a cone is 1/3 base x height.
---------------------------------------------------------





Get["font.math"];
r = 4;
h = 8;
f[x_] := r/h x

q1 = Plot[f[x],{x,0,h}];
q2 = ParametricPlot[{h,y},{y,0,r}];
q3 = Plot[0,{x,0,h}];
q4 = Show[q1,q2,q3,PlotLabel->"P289 Ex2 y = r/h x",PlotRange->All,
     AspectRatio->Automatic];
Display["ex2x.ps",q4];

g[y_] := h/r y;
p[y_] := ParametricPlot3D[ { x, y Cos[t], y Sin[t]},{t,- Pi/2,1 Pi/4},
        {x,g[y],h}];
A = Table[ p[ r y/10],{y,1, 9}];
B = Graphics3D[{RGBColor[1,0,0],Thickness[0.02],Line[{{0,0,0},{h,0,0}}]}];
a = Show[A,B,PlotLabel->"P289 ex2: y = r/h x",AspectRatio->Automatic];
Display["ex2y.ps",a];

p1 = ParametricPlot3D[{x,r/h x Cos[t],r/h x Sin[t]},{t,0,2 Pi},{x,0,h}];
p2 = Show[p1,PlotLabel->"P289 Ex2 y = r/h x",PlotRange->All];
Display["ex2z.ps",p2];
----------------------------------------------------------------------------
Now rework the first two homework problems for the assignment

due today using the shell method.











Previous assignment:

Page 286 Problem 12

Find the volume generated by rotating the region

about the y-axis.

x = 2/y

y = 2     x=1

y = 6     x=1/3

x = 0

         x= 1                                 2  
Volume = INT   2 Pi x (2/x - 2 ) dx + Pi (1/3)  4
         x=1/3

              x=1
Volume = 2 Pi INT   2 - 2 x dx  + Pi (1/3)^2 4
             x=1/3

              |           2    | x=1
Volume = 2 Pi | 2x -     x     |           + 4 Pi/9
              |                | x=1/3



Volume = 2 Pi ( 2 -1   - (2/3 -1/9 )) + 4 Pi/9  = 4 Pi/3 

..................................................................



Get["font.math"];
q1 = Plot[2/x,{x,1/3,1}];
q2 = Plot[2,{x,  0,1}];
q3 = Plot[6,{x,0,1/3}];
q4 = ParametricPlot[{0,y},{y,2,6},
      PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}];
q5 = Show[q1,q2,q3,q4,q5,PlotLabel->"P286 p12 x = 2/y",
      PlotRange->All,AspectRatio->Automatic];
Display["p12s.ps",q5];
a = ParametricPlot3D[{2/y Cos[t], 2/y Sin[t], y},{t,0,2 Pi}, {y,2,6}];
b = Show[a,PlotLabel->"P286 P12: x = 2/y",AspectRatio->Automatic];
Display["p12t.ps",b];
---------------------------------------------------------
Page 286 Problem 18

Find the volume of the solid generated by revolving

about the x-axis the region bounded by the line y = 6x

                        2
and the parabola y = 6 x  


    y=6
V = INT  2 Pi y ( Sqrt[y/6] - y/6 ) dy
    y=0 



          y=6    3/2           2
V = 2 Pi  INT   y  /Sqrt[6] - y /6 dy 
          y=0 


         |  5/2                  3     | y=6
V = 2 Pi | y   /(5/2 Sqrt[6]) - y /18  |
         |                             | y=0


                 5/2    
V = 2 Pi (  2/5 6   /Sqrt[6] - 216/18 ) = 24 Pi/5  
       
--------------------------------------
A 10 dip candle burns  2 hours.
A 30 dip candle burns 10 hours.
How long will a 50 dip candle burn?
The candles are 10 inches long.
----------------------------------------------------