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 13   5.4  Length       

p300:18,20,21,28
 

Main Idea:  Approximate a curve with tiny little straight lines. 

Key Words:  ds, Surface area = INT 2 Pi f(x) ds 

          _           _            _           _     
         |      /dy\ 2 |          |  /dx\ 2     |       
ds = Sqrt| 1 + (----)  | dx = Sqrt| (----)  + 1 | dy  
         |_     \dx/  _|          |_ \dy/      _|         

          _                _              _         _
         |  /dx\ 2   /dy\ 2 |            |   2    2  |
   = Sqrt| (----) + (----)  | dt  =  Sqrt| dx + dy   |
         |_ \dt/     \dt/  _|            |_         _|


Goal:  Find arc length and surface area. 

--------------------------------------------------------
Previous assignment:           p293:(13),(14),(16),(19)

Page 293 Problem 13

Consider the region R.  Set up an integral for the

volume of the solid obtained when R is revolved about the

given line using the indicated method.

                     f(x)
        |             .....     
        |        .-'''     ''-.. 
        |      |'               `|
        |      |                 |
        |      |`..          ...'|
        |      |   ''-....-''    |
        |      |       g(x)      |
        |      |                 |
   -----|------------------------------------
        |      a                 b

(a)  The x-axis (washers)

        x=b          2           2
        INT   Pi f(x)   - Pi g(x)    dx
        x=a

(b)   The y-axis (shells)

      x=b
      INT 2 Pi x (f(x)-g(x)) dx 
      x=a      







(c)   The line x=a (shells)

      x=b
      INT  2 Pi (x-a) (f(x)-g(x)) dx
      x=a

(d)   The line x=b (shells)

      x=b
      INT  2 Pi (b-x) (f(x)-g(x)) dx
      x=a

Page 293 Problem 14

A region R is shown in Figure 9.  Set up an integral

for the volume of the solid obtained when R is revolved 

about each line.  Use the indicated method.

      3 |-------------------------------------------
        |
        |
       d|____________.______________________.________
        |              `\\\\\\\\\\\///////'
        |                `\\\\\\\\/////'
        |        x = g(y)  `\\\\\\////  x = f(y)
        |                    ://///\\\.
        |                  .///////\\\\\.
      c |________________./////////\\\\\\\.____ 
        |                                   
--------+----------------------------------------------   
        |                                   

(a)  The y-axis (washers)

 y=d         2           2
 INT  Pi f(y)   - Pi g(y)    dy
 y=c

(b)  The x-axis (shells)

y=d
INT  2 Pi y (f(y)-g(y)) dy
y=c

(c)  The line y = 3 shells.

y=d
INT  2 Pi (3-y) (f(y)-g(y)) dy
y=c

Page 293 Problem 16
                                    3
Sketch the region R bounded by y = x   + 1 and y = 0

between x=0 and x=2.

Set up, but do not evaluate integrals for each of the following.

(a)  Area of R

(b)  Volume of the solid obtained when R is revolved about the y-axis

(c)  Volume of the solid obtained when R is revolved about y = -1

(d)  Volume of the solid obtained when R is revolved about  x = 4.


                 |      '       3             
                 |     .|  y = x  + 1                  
                 |    . |                    
                 |   .  |                    
                 |_.'   |                    
              x=0|      |x=2                   
-----------------+-----------------------------------
                 | y=0  2                      
(a)  Area of R

     x=2   3          |  4        | x=2
     INT  x  +1 dx =  | x  /4 + x |      = 6 
     x=0              |           | x=0
-----------------------------
Integrate[x^3+1,{x,0,2}] = 6
Get["font.math"];
P1 = Table[ Plot[ i/2,{x,0,2}],{i,1,18}];
P2 = Table[ ParametricPlot[{i/2,y},{y,0,9}],{i,0,4}]; 
P3 = Plot[x^3+1,{x,0,2},PlotStyle->{RGBColor[1,0,0],Thickness[0.02]}];
P4 = Plot[0,{x,0,2},PlotStyle->{RGBColor[1,0,0],Thickness[0.02]}];
P5 = ParametricPlot[{2,y},{y,0,9},
     PlotStyle->{RGBColor[1,0,0],Thickness[0.02]}];
P6 = ParametricPlot[{0,y},{y,0,1},
     PlotStyle->{RGBColor[1,0,0],Thickness[0.02]}];
P7 = Show[P1,P2,P3,P4,P5,P6,PlotLabel->
      "P393 p16(a): y = x^3 + 1:Sq=1/4 ",AspectRatio->Automatic];
Display["p16a.ps",P7];
--------------------------
(b)  Volume of the solid obtained when R is revolved 
about the y-axis

      x=2           3                  | 5      2   | x=2
      INT  2 Pi x (x   + 1) dx  = 2 Pi |x /5 + x /2 |     
      x=0                              |            | x=0

      = 2 Pi (32/5 + 2) = 84 Pi/5 
-----------------------------------------------
Integrate[2 Pi x (x^3+1),{x,0,2}] = 84 Pi/5
----------------------------------------------
Get["font.math"];
f[x_] := x^3+1;
P1 = ParametricPlot3D[{ x Cos[t], x Sin[t], x^3+1}, 
         {x,0,2  },  {t,0, 3 Pi/2}];
P2 = ParametricPlot3D[{ 2 Cos[t], 2 Sin[t], y},    
           {y,0,9},{t,0, 3 Pi/2}];
P3 = ParametricPlot3D[{ 2 Cos[t], 2 Sin[t], 0},
           {t,0,2 Pi}];
P4 = ParametricPlot3D[{ 2 Cos[t], 2 Sin[t], 9},
           {t,0,2 Pi}];
P5 = ParametricPlot3D[ {x,0,x^3+1,{RGBColor[1,0,0],Thickness[0.01]}},
           {x,0,2}];
P6 = ParametricPlot3D[ {2,0,z,{RGBColor[1,0,0],Thickness[0.01]}},
           {z,0,9}];
P7 = ParametricPlot3D[ {0,0,z,{RGBColor[1,0,0],Thickness[0.01]}},
           {z,0,1}];
P8 = ParametricPlot3D[ {x,0,0,{RGBColor[1,0,0],Thickness[0.01]}},
           {x,0,2}];
P9 = ParametricPlot3D[ {0,-y,y^3+1,{RGBColor[1,0,0],Thickness[0.01]}},
           {y,0,2}];
P10 = ParametricPlot3D[ {0,-2,z,{RGBColor[1,0,0],Thickness[0.01]}},
           {z,0,9}];
P11 = ParametricPlot3D[ {0,0,z,{RGBColor[1,0,0],Thickness[0.01]}},
           {z,0,1}];
P12 = ParametricPlot3D[ {0,y,0,{RGBColor[1,0,0],Thickness[0.01]}},
           {y,0,-2}];
P13 = Show[P1,P2,P3,P4,P5,P6,P7,P8,P9,P10,P11,P12,
   PlotLabel->"P293 p16(b) y = 1+x^3 about the y axis";
   PlotRange->All,AspectRatio->Automatic];
Display["p16b.ps",P13];

-------------------------------------------------------------------------
(c)  Volume of the solid obtained when R is revolved

about y = -1

x=2     3    2        2                6       3
INT Pi(x  +2)   - Pi 1   dx  = Pi INT x   + 4 x   + 4 -1 
x=0
      |  7        4         | x=2
=  Pi | x /7 + 4 x /4 + 3 x |     = Pi (128/7 + 16 + 6) = 282 Pi/7
      |                     | x=0
----------------------------------------------------
Integrate[Pi (x^3+2)^2 - Pi ,{x,0,2}] = 282 Pi/7

Get["font.math"];

f[x_] := x^3+1;

P1 = ParametricPlot3D[{ x,   Cos[t](x^3+2),-1+Sin[t](x^3+2)}, 
         {x,0,2},{t,5 Pi/4,10 Pi/4}];

P2 = ParametricPlot3D[{ 2,  Cos[t](1+y),  -1+Sin[t](1+y)}, 
         {y,1,9},{t,5 Pi/4,10 Pi/4}];

P3 = ParametricPlot3D[{ x,        Cos[t] 2,-1+Sin[t] 2}, {x,0,2},
                 {t,5 Pi/4,10 Pi/4}];

t=5 Pi/4;
P4 = ParametricPlot3D[{ x,   Cos[t](x^3+2),-1+Sin[t](x^3+2),
      {RGBColor[1,0,0],Thickness[0.01]} }, {x,0,2}];

P5 = ParametricPlot3D[{ 2,  Cos[t](1+y),  -1+Sin[t](1+y), 
      {RGBColor[1,0,0],Thickness[0.01]} }, {y,1,9}];

P6 = ParametricPlot3D[{ x,        Cos[t] 2,-1+Sin[t] 2,
      {RGBColor[1,0,0],Thickness[0.01]} }, {x,0,2}];
            
t=10 Pi/4;
P7 = ParametricPlot3D[{ x,   Cos[t](x^3+2),-1+Sin[t](x^3+2),
      {RGBColor[1,0,0],Thickness[0.01]} }, {x,0,2}];

P8 = ParametricPlot3D[{ 2,  Cos[t](1+y),  -1+Sin[t](1+y), 
      {RGBColor[1,0,0],Thickness[0.01]} }, {y,1,9}];

P9 = ParametricPlot3D[{ x,        Cos[t] 2,-1+Sin[t] 2,
      {RGBColor[1,0,0],Thickness[0.01]} }, {x,0,2}];

P10 = ParametricPlot3D[{y,0,-1,Thickness[0.02]},{y,-1,3}];
            
P11= Show[P1,P2,P3,P4,P5,P6,P7,P8,P9,P10, 
    PlotLabel->"P293 p16(c) y = 1+x^3 about the y=-1 axis",
     AspectRatio->Automatic,PlotRange->All];

Display["p16cx.ps",P11];

P12 = Show[P11,ViewPoint->{0,-2,0}];

Display["p16cy.ps",P12];

-----------------------------------------------------------------------











(d)  Volume of the solid obtained when R is revolved about x=4.

x=2
INT 2 Pi (4-x) (x^3+1) dx = 156 Pi/5 
x=0

     x=2     3        4
2 Pi INT  4 x  + 4 - x - x  dx
     x=0

     |  4           5      2     | x=2
2 Pi | x   + 4 x - x /5 - x  /2  |
     |                           | x=0

     |                    |
2 Pi |  16 + 8 - 32/5 - 2 | = Pi 156/5
     |                    |
----------------------------------------------------------------------
Integrate[2 Pi (4-x) (x^3+1),{x,0,2}] = 156 Pi/5

Get["font.math"];
f[x_] := x^3+1;
P1 = ParametricPlot3D[{4+(4-x) Cos[t], Sin[t](4-x),x^3+1}, 
      {x,0,2},{t, 0, 3 Pi/2}];
P2 = ParametricPlot3D[{4+(4-x) Cos[t], (4-x) Sin[t] ,  0}, 
      {x,0,2}, {t, 0, 3 Pi/2 }];
P3 = ParametricPlot3D[{ 4 + 2 Cos[t], 2 Sin[t], y}, 
      {y,0,9},{t,0, 3 Pi/2}];
P4 = ParametricPlot3D[{4 + 4 Cos[t], 4 Sin[t], y},{y,0,1}, {t,0, 3 Pi/2}];
t = 0;
P5 = ParametricPlot3D[{4+(4-x) Cos[t], Sin[t](4-x),x^3+1,
     {RGBColor[1,0,0],Thickness[0.01]}  }, {x,0,2}];
P6 = ParametricPlot3D[{4+(4-x) Cos[t], (4-x) Sin[t] ,  0, 
     {RGBColor[1,0,0],Thickness[0.01]}  }, {x,0,2}];
P7 = ParametricPlot3D[{ 4 + 2 Cos[t], 2 Sin[t], y, 
     {RGBColor[1,0,0],Thickness[0.01]}  }, {y,0,9}];
P8 = ParametricPlot3D[{4 + 4 Cos[t], 4 Sin[t], y,
     {RGBColor[1,0,0],Thickness[0.01]}  }, {y,0,1}];
t = 3 Pi/2;
P9 = ParametricPlot3D[{4+(4-x) Cos[t], Sin[t](4-x),x^3+1,
     {RGBColor[1,0,0],Thickness[0.01]}  }, {x,0,2}];
P10 = ParametricPlot3D[{4+(4-x) Cos[t], (4-x) Sin[t] ,  0, 
     {RGBColor[1,0,0],Thickness[0.01]}  }, {x,0,2}];
P11 = ParametricPlot3D[{ 4 + 2 Cos[t], 2 Sin[t], y, 
     {RGBColor[1,0,0],Thickness[0.01]}  }, {y,0,9}];
P12 = ParametricPlot3D[{4 + 4 Cos[t], 4 Sin[t], y,
     {RGBColor[1,0,0],Thickness[0.01]}  }, {y,0,1}];
P13 = Show[P1,P2,P3,P4,P5,P6,P7,P8,P9,P10,P11,P12,
    PlotLabel->"P293 p16(d)  y = 1+x^3 about the y=-1 axis"];
Display["p16d.ps",P13];
------------------------------------------------------------------------------


















Page 293 Problem 19

A round hole of radius a is drilled through the

center of a solid sphere of radius b (assume that b > a.  Find

the volume of the solid that remains.

                        .......       2     2     2
                    . |    |    |` . x   + y   = b  
                   '  |    .    |   .
                 .    |    .    |    .
                      |    .    |
                .     |    .    |     . 
                .<----b--->|<-a->     .  
                      |    .    |
                 .    |    .    |    . 
                      |    .    |   . 
                  ` . |         |  .
                      . ...|... . '



         x=b                  2    2
Volume = INT (2 Pi x) 2 Sqrt[b  - x  ] dx
         x=a

               |    2   2  3/2  | x=b
               |  (b  -x  )     |
        = 4 Pi | -----------    |
               |  3/2 (-2)      | x=a

                             2  2  3/2
        =  - 4 Pi/3  ( 0 - (b -a  )    )


                   2   2  3/2          3
            4 Pi (b  -a  )       4 Pi h  
        =    ---------------  =  -------  where h is 1/2 length of hole
                   3               3
--------------------------------------------------------------------

                                                  2    2 3/2
                                             4 ( b  - a )    Pi
Integrate[ 4 Pi x Sqrt[b^2-x^2],{x,a,b}]  = ------------------
                                                  3
Get["font.math"];
a = 1/9; b = 1; f[x_] := Sqrt[b^2-x^2];
P1 = ParametricPlot3D[{ x Cos[t], f[x], x Sin[t]}, {x,a,b},{t, 2 Pi/4, 8 Pi/4}];
P2 = ParametricPlot3D[{ x Cos[t],-f[x], x Sin[t]}, {x,a,b},{t, 2 Pi/4, 8 Pi/4}];
P3 = ParametricPlot3D[{ a Cos[t], y, a Sin[t]}, 
      {y,-f[a],f[a]}, {t, 2 Pi/4, 8 Pi/4}];
t = 2 Pi/4;
P4 = ParametricPlot3D[{ x Cos[t], f[x], x Sin[t],
        {RGBColor[1,0,0],Thickness[0.01]}}, {x,a,b}];
P5 = ParametricPlot3D[{ x Cos[t],-f[x], x Sin[t],
        {RGBColor[1,0,0],Thickness[0.01]}}, {x,a,b}];
P6 = ParametricPlot3D[{ a Cos[t], y, a Sin[t],
        {RGBColor[1,0,0],Thickness[0.01]}},{y,-f[a],f[a]}];
t = 8 Pi/4;
P7 = ParametricPlot3D[{ x Cos[t], f[x], x Sin[t],
        {RGBColor[1,0,0],Thickness[0.01]}}, {x,a,b}];
P8 = ParametricPlot3D[{ x Cos[t],-f[x], x Sin[t],
        {RGBColor[1,0,0],Thickness[0.01]}}, {x,a,b}];
P9 = ParametricPlot3D[{ a Cos[t], y, a Sin[t],
        {RGBColor[1,0,0],Thickness[0.01]}},{y,-f[a],f[a]}];
P10 = Show[P1,P2,P3,P4,P5,P6,P7,P8,P9, 
         PlotLabel->"Page 293 Problem 19 hole in a sphere"];
Display["p19.ps",P10];


------------------------------------------------------------------------------

New Material

The length of a curve is defined by the combined lengths 

of approximating straight line segments.  This idea

fits nicely into the style of calculus.  It also makes

practical sense because when we follow a curve, we 

actually approximate it by a path of line segments.  The

distance we will experience will be the limit of these

types of paths. 
                       ..
                ds . '   `.
                .'    . ' |
            . '   . '     |
         .'   . '  ds     |  
       .' . '             |  dy 
      ._'________dx_______| 
               
From the Pythagorian theorem

                       2      2
          ds = Sqrt[ dx   + dy   ]

The basis formula for arc length is given in three ways

and the one you use depends on how the curve is specified.

(a)  parametric   R = (f[t], g[t])    


(b)  y = f(x)


(c)  x = g(y)

               _     2          2  _ 
      t=b     |  /dx\       /dy\    |
(a)   INT Sqrt| (----)  +  (----)   | dt
      t=a     |_ \dt/       \dt/   _|

               _          2 _
      t=b     |      / dy\   | 
(b)   INT Sqrt| 1 + (-----)  | dx
      t=a     |_     \ dx/  _|

               _     2    _
      t=b     |  /dx\      |
(c)   INT Sqrt| (----) + 1 | dy
      t=a     |_ \dy/     _|

Each of these equations expresses the same idea.  We

approximate the curve with line segments and add the

lengths of these segments.


The introduction of the derivative is based on the  

mean value theorem that says



          f(b)-f(a)      /
         ------------ = f (c) for some c where a < c < b.
             b-a                          
                                             _     _      _ 2 _
                                            |     |  /\    |   |
                   /\   2    /\   2         |     | /__\ y |   |  /\
Thus   ds = Sqrt[ /__\ x  + /__\ y  ] = Sqrt| 1 + |-----   |   | /__\ x 
                                            |     |  /\    |   | 
                                            |_    |_/__\ x_|  _|

                 _        2  _ 
                |     /dy\    |  /\
       ds = Sqrt| 1+ (----)   | /__\ x
                |_    \dx/   _|


Page 297 Example 3.  Find the circumference of the 

         2     2     2
circle  x   + y   = a  

The path is given by   (a Cos[t], a Sin[t])

              t=2 Pi                 2              2
The length =   INT  Sqrt[ (-a Sin[t])   + (a Cos[t])   ] dt
               t=0


              t=2 Pi       2         2          2
           =   INT  Sqrt[ a   (Sin[t]   + Cos[t]  )  ] dt
               t=0
           
              t=2 Pi
           =   INT   a dt  = 2 Pi a
               t=0


Page 297 Example 4.  Find the length of the line segment from

(0,1) to (5,13).   

                           2     2
Answer:   Length = Sqrt[ 12   + 5   ] = 13.


Calculus should give us the same answer.


      y-1 = 12/5(x-0)

      y = 12/5 x + 1

check:

     when x = 0   y = 1.

     when x = 5. y = 13.

x=5               2     x=5              2
INT Sqrt[1+(dy/dx)  ] = INT Sqrt[1+(12/5)  ] dx 
x=0                     x=0

= Sqrt[1+144/25] 5 = Sqrt[169] = 13

Thank goodness, the answers are the same.

Try working the problem using the formula x = 5/12(y-1).



Page 297 Example 6.

Find the length of the curve (2 Cos[t], 4 Sin[t]) from t=0 to t=Pi.

t=Pi             2              2
INT   Sqrt[ 4 Sin [t]   + 16 Cos [t]  ] dt
t=0


t=Pi                     2
INT   Sqrt[ 4 + 12 Cos[t]   ] dt
t=0


t=Pi                      2
INT   2 Sqrt[ 1 + 3 Cos[t]   ] dt   Now we have to approximate this
t=0


Integral.  Why not just approximate the length to begin with.

----------------------------------------------------------------
Get["font.math"];
f[t_] := {2 Cos[t],4 Sin[t]};
P1 = ParametricPlot[f[t],{t,0,Pi},
     PlotStyle->{RGBColor[0,0,1],Thickness[0.003]}];
A  = Table[{2 Cos[Pi t/10], 4 Sin[Pi t/10]},{t,0,10}];
P2 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],Thickness[0.003]},PlotJoined->True];
P3 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],Thickness[0.010]},PlotJoined->False];
P4 = Show[P1,P2,P3,
     PlotLabel->"P297 ex6: (2 Cos[t], 4 Sin[t]) from t=0 to t=Pi",
        AspectRatio->Automatic];

Display["ex6.ps",P4];

g[n_] :=   Sum[   Sqrt[(f[ Pi i/n] - f[ Pi (i-1)/n] ).
                       (f[ Pi i/n] - f[ Pi (i-1)/n] )],{i,1,n}];
Do[ Print[n," ",N[g[n]]],{n,10,200,10}];
.............................................................................
  n approximation
 10 9.64866
 20 9.67849
 30 9.68402
 40 9.68596
 50 9.68685
 60 9.68734
 70 9.68764
 80 9.68783
 90 9.68796
100 9.68805
110 9.68812
120 9.68817
130 9.68821
140 9.68824
150 9.68827
160 9.68829
170 9.68831
180 9.68833
190 9.68834
200 9.68835

Integrate[ 2 Sqrt[ 1 + 3 Cos[t]^2 ],{t,0,Pi}] = 9.688448221
......................................................................


Page 297 Example 5.
                                  3/2
Find the length of the curve y = x    from (1,1) to (4,8).


    x=4                       |         3/2   | x=4           3/2       3/2
L = INT Sqrt[ 1+ 9/4 x] dx  = | (1+9/4x)      |     = 8/27 (10 -  (13/4)   )
    x=1                       | ----------    |
                              | (3/2)(9/4)    | x=1

-------------------------------------------
Integrate[Sqrt[1+9/4 x],{x,1,4}] = 7.63371
Straightline = Sqrt[7^2 + 3^2]   = 7.61577 

Get["font.math"];
P1 = Plot[x^(3/2),{x,1,4}];
P2 = Plot[ 1+7/3(x-1),{x,1,4},PlotStyle->{RGBColor[1,0,0]}];
P3 = Show[P1,P2,PlotLabel->"P297 Ex 5: y = x^(3/2)",
PlotRange->All,AspectRatio->Automatic];
Display["ex5.ps",P3];
......................................................................


Page 299 Example 7.  Find the area of the surface generated by

revolving y = Sqrt[x] from x=0 to x=4 about the x axis.

                                        _         _
x=4                    x=4             |        -1 |
INT  2 Pi y ds = 2 Pi  INT Sqrt[x] Sqrt| 1+1/4 x   |   dx 
x=0                    x=0             |_         _|

                                |        3/2| x=4
     x=4                        | (x+1/4)   |                   3/2       3/2
2 Pi INT Sqrt[x + 1/4] dx = 2 Pi|---------  |   = 4 Pi/3 ( (17/4)  - (1/4))
     x=0                        |   3/2     | x=0


             3/2
      Pi ( 17   -1)
     ---------------
            6

------------------------------------------------
Get["font.math"];
f[x_] := Sqrt[x];
P1 = ParametricPlot3D[ {x, f[x] Cos[t], f[x] Sin[t]},
       {x,0,4},{t,-4.5 Pi/4, 1.1 Pi/2}, AspectRatio->Automatic];

t= -4.5 Pi/4;
P2 = ParametricPlot3D[ {x, f[x] Cos[t], f[x] Sin[t],
       {RGBColor[1,0,0],Thickness[0.01]} }, {x,0,4}, AspectRatio->Automatic];
t= 1.1 Pi/2
P3 = ParametricPlot3D[ {x, f[x] Cos[t], f[x] Sin[t],
       {RGBColor[1,0,0],Thickness[0.01]} }, {x,0,4}, AspectRatio->Automatic];

P4 = Show[P1,P2,P3,PlotLabel->"P299 ex7: y = Sqrt[x] about x axis"];

Display["ex7.ps",P4];
-----------------------------------------------------------

---------------------------------------------------------

Write the parametric equation of a cycloid.





...........................................................
Get["font.math"];
a = 1;
P1 = ParametricPlot[ {t-Sin[t],1-Cos[t]},{t,0,4 Pi}];
f[w_] := ParametricPlot[ {w,a} + a{Cos[t],Sin[t]},{t,0,2 Pi},
      PlotStyle->{RGBColor[1,0,0]}];
g[w_] := ListPlot[{ {w-Sin[w],1-Cos[w]} },PlotStyle->{RGBColor[1,0,0],
         PointSize[0.01]}];

P2 = Show[P1,f[1.5], f[3.0], f[2 Pi], f[2.5 Pi], f[3.0 Pi], f[3.5 Pi],
             f[4.0 Pi], g[1.5], g[3.0], g[2 Pi], g[2.5 Pi], g[3.0 Pi],
             g[3.5 Pi], g[4.0 Pi], PlotLabel->"P295 Ex 2",PlotRange->All,
     AspectRatio->Automatic];
Display["Ex2.ps",P2];
.............................................................

                                   2     2     2     2
Find the surface area of a sphere x   + y   + z   = r   

between the planes z = 0 to z = h.
.............................................................
Get["font.math"];
h = 2; R = 3;
P1 = ParametricPlot3D[{r Cos[t],r Sin[t], Sqrt[R^2-r^2]},
        {r,Sqrt[5],3},{t,0,2 Pi}];
r=Sqrt[5];
P2 = ParametricPlot3D[{r Cos[t],r Sin[t], Sqrt[R^2-r^2],
        {RGBColor[1,0,0],Thickness[0.01]} }, {t,0,2 Pi}];
r=3;
P3 = ParametricPlot3D[{r Cos[t],r Sin[t], Sqrt[R^2-r^2]-0.03,
        {RGBColor[1,0,0],Thickness[0.01]} }, {t,0,2 Pi}];
P4 = Show[P1,P2,P3,PlotLabel->"Surface Area of Sphere",PlotRange->All,
     AspectRatio->Automatic];
Display["ex.ps",P4];
-------------------------------------------------------------------------

 x = Sqrt[r^2-y^2]
                                            -y
 dx/dy = 1/2 (r^2-y^2)^(-1/2) (-2y)  = -------------
                                       Sqrt[r^2-y^2]
y=h
INT  2 Pi x ds 
y=0
                              _             _
                             |          2    |
y=h             2   2        |         y     |
INT  2 Pi Sqrt[r  -y  ]  Sqrt|  1 + ------   | dy 
y=0                          |        2   2  |
                             |_      r - y  _|

                             _         _
                            |      2    |
y=h             2   2       |     r     |
INT  2 Pi Sqrt[r  -y  ] Sqrt|  ------   | dy 
y=0                         |    2   2  |
                            |_  r - y  _|

                             _         _
y=h       
INT  2 Pi r dy 
y=0    

           |   | y=h
   2 Pi r  | y |
           |   | y=0


   2 Pi r h    So equal height slices have equal amounts of surface area.