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, Feb 5   7.1  Area       

p427: 1,(2),7,8,(10),14,18,25,28,31,33,35,36,(39),42,43,47,48,53,55

Main Idea:  Write down the cross section and integrate it. 

Key Words: Cross section. 

Goal: Learn to find areas between curves. 
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Previous Assignment 
Monday, Feb  3  6.5 Substitution

Page 407 Problem 54

x=-4
 INT Sqrt[1-2x] dx
 x=0
          x=-4
   (-1/2) INT Sqrt[1-2x] (-2 dx)
          x=0

          (1-2x)^(3/2)  | x=-4
    -1/2  ------------  |
             (3/2)      | x=0


     -1/3( 9^(3/2) - 1) = -26/3

a = Plot[Sqrt[1-2x],{x,-10,1/2}];
b = Show[a,PlotLabel->"Page 407 Problem 54"];
Display["54.ps",b];

Estimate of area  1/2 4 2 + 4  = 8 compared to 8.66.
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Page 407 Problem 68

x=a
INT  x Sqrt[a^2-x^2] dx
x=0
         x=a
  (-1/2) INT   Sqrt[a^2-x^2] (-2x dx)
         x=0

          (a^2-x^2)^(3/2)   | x=a
  (-1/2)  ---------------   |
              (3/2)         | x=0

    -1/3 (0 - |a|^3) = 1/3 |a|^3

a = Plot[x Sqrt[16-x^2],{x,-4,4}];
b = Show[a,PlotLabel->"Page 407 Problem 68"];
Display["68.ps",b];
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Page 407 Problem 78
          1
Evaluate INT x Sqrt[1-x^4] dx  by making a substitution and 
          0
interpreting the resulting integral in terms of an area

 u = x^2
du = 2x dx

u=1
INT Sqrt[1-u^2] du/2   = 1/2 the area of a quarter circle of radius 1
u=0

or Pi/8.

a = Plot[x Sqrt[1-x^4],{x,-2,2}];
b = Show[a,PlotLabel->"Page 407 Problem 78"];
Display["78.ps",b];

c = Plot[Sqrt[1-u^2],{u,-1,1}];
d = Show[c,PlotLabel->"Page 407 Problem 78 Sqrt[1-u^2]",AspectRatio->True];
Display["78x.ps",d];
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More on changing the variables.

Page 407 Problem 82

                        9                     3
If f is continuous and INT f(x) dx = 4, find INT x f(x^2) dx
                        0                     0

Let  u = x^2
    du = 2x dx

      x=3                        u=9
(1/2) INT f(x^2)  (2x) dx  = 1/2 INT f(u) du = 1/2 4 = 2.
      x=0                        u=0

Of course, we believe the mathematical gobblydigook, but let's

see how it works out with one simple example.

Suppose f(x) = 4/9 as one example of a function satisfying 

x=9                            3                            x=3
INT f(x) dx = 4.        Then  INT (4/9) x dx = 4/9 (x^2)/2 |  = 2
x=0                            0                            x=0
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New Material:

Page 423  Example 1.

Find the area bounded by the parabolas y = x^2 and y = 2x - x^2.

a = Plot[{x^2,2x-x^2},{x,-1,5}];
b = Show[a,PlotLabel->"y=x^2, y=2x-x^2"];
 Display["1.ps",b]; 
ans = 1/3.
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Page 424 Example 2  Find the area bounded by the curves y = Sin[x], 

y = Cos[x], and x=0, and x = Pi/2.

a = Plot[{Sin[x],Cos[x]},{x,0,2 Pi}];
b = Show[a,PlotLabel->"y=Sin[x], y=Cos[x]",AspectRatio->Automatic];
Display["2.ps",%]
Ans = 2(Sqrt[2]-1)
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Page 424 Example 4.   Find the area of the region bounded by the
line y = x-1 and the parabola y^2 = 2x+6.

a = Plot[{x-1, Sqrt[2x+6], -Sqrt[2x+6]},{x,-3,10}];
b = Show[%,PlotLabel->"y = x-1, y^2 = 2x+6",AspectRatio->Automatic];
Display["4.ps",b];
ans = 18