View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/7_Techniques_of_Integration/7.3_Trigonometric_Substitution/7.3.58.pg

    1 ## DBsubject('Calculus')
    2 ## DBchapter('Techniques of Integration')
    3 ## DBsection('Trigonometric Substitution')
    4 ## KEYWORDS('calculus', 'integration', 'integral', 'trigonometric substitution', 'substitution', 'trigonometry', 'trigonometric', 'trig')
    5 ## TitleText1('Calculus: Early Transcendentals')
    6 ## EditionText1('2')
    7 ## AuthorText1('Rogawski')
    8 ## Section1('7.3')
    9 ## Problem1('58')
   10 ## Author('Christopher Sira')
   11 ## Institution('W.H.Freeman')
   12 
   13 DOCUMENT();
   14 loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl");
   15 loadMacros("PGchoicemacros.pl");
   16 loadMacros("Parser.pl");
   17 loadMacros("freemanMacros.pl");
   18 $context = Context();
   19 
   20 $a = 0;
   21 # Introduced more variation and kept same level of difficulty
   22 $c = random(2, 4, 1);
   23 $d = random(5, 9, 1);
   24 $b = $c*$c;
   25 $d2 = 2*$d;
   26 #$dc4 = $d *($c**4)/4;
   27 $dc4 = $d *($c**4)/8;
   28 $ans = Formula(" $dc4 * (pi ** 2) ")->reduce();
   29 
   30 Context()->texStrings;
   31 BEGIN_TEXT
   32 \{ beginproblem() \}
   33 \{ textbook_ref_exact("Rogawski ET 2e", "7.3","58") \}
   34 $PAR
   35 Find the volume of the solid obtained by revolving the graph of \( y = $d x\sqrt{$b - x^2} \)
   36 over [$a,$c] about the y-axis.
   37 $PAR
   38 \{ans_rule()\}
   39 $PAR
   40 END_TEXT
   41 Context()->normalStrings;
   42 
   43 ANS($ans->cmp);
   44 
   45 Context()->texStrings;
   46 SOLUTION(EV3(<<'END_SOLUTION'));
   47 $PAR
   48 $SOL
   49 $PAR
   50 Using the method of cylindrical shells, the volume is given by
   51 $PAR
   52 \( V = 2\pi \int^{$c}_{$a} x ($d x\sqrt{$b - x^2}) \, dx = $d2 \pi \int^{$c}_{$a} x^2 \sqrt{$b - x^2} \, dx \).
   53 $PAR
   54 To evaluate this integral, let \( x = $c \sin \theta \).  Then \( dx = $c \cos \theta \, d\theta \),
   55 $PAR
   56 \( $b - x^2 = $b (1 - \sin^2 \theta)  = $b \cos^2 \theta \),
   57 $PAR
   58 and
   59 \[
   60 \begin{array}{ll}
   61 I & = \int x^2\sqrt{$b - x^2} \, dx \cr
   62 & = \int ($c^2 \sin^2 \theta )\,( $c \cos \theta )\, ($c \cos \theta) \, d\theta \cr
   63 & = $c^{4} \int (1 - \cos^2 \theta) \cos^2 \theta \, d\theta. \cr
   64 & \cr
   65 \frac{1}{$c^{4}} I &=  \int \cos^2 \theta \, d\theta - \int \cos^4 \theta \, d\theta .
   66 \end{array}
   67 \]
   68 
   69 Now we use the reduction formula for \( \int \cos^4 \theta \, d\theta \):
   70 \[
   71 \begin{array}{ll}
   72 \frac{1}{$c^{4}} I & = \int \cos^2 \theta \, d\theta - \left[ \frac{cos^3 \theta \, \sin \theta}{4} + \frac{3}{4}\int \cos^2 \theta \, d\theta \right] \cr
   73 & = -\frac{1}{4}\cos^3 \theta \, \sin \theta + \frac{1}{4} \int \cos^2 \theta \, d\theta \cr
   74 & = -\frac{1}{4}\cos^3 \theta \, \sin \theta + \frac{1}{4} \left[ \frac{1}{2} \theta + \frac{1}{2} \sin \theta \, \cos \theta \right] + C \cr
   75 & = -\frac{1}{4}\cos^3 \theta \, \sin \theta + \frac{1}{8} \theta + \frac{1}{8} \sin \theta \, \cos \theta + C.
   76 \end{array}
   77 \]
   78 
   79 $PAR
   80 Since \( \sin \theta = \frac{x}{$c} \), we know that \( \cos \theta = \frac{\sqrt{$b - x^2} }{$c}\).
   81 Then we have
   82 $PAR
   83 \(\frac{1}{$c^{4}} I = -\frac{1}{4}\left(\frac{\sqrt{$b - x^2} }{$c}\right)^3 \, \left(\frac{x}{$c}\right) + \frac{1}{8}\sin^{-1} \left(\frac{x}{$c}\right)
   84 + \frac{1}{8}\left(\frac{x}{$c}\right) \left(\frac{\sqrt{$b - x^2} }{$c}\right) + C \).
   85 $PAR
   86 Now we can complete the volume calculation:
   87 \[
   88 \begin{array}{ll}
   89  V & = \left . $d2 \pi I \right |^{$c}_{$a} \cr
   90 & = $d2 \cdot $c^{4} \pi  \left [ -\frac{1}{4}\left(\frac{\sqrt{$b - x^2} }{$c}\right)^3 \, \left(\frac{x}{$c}\right) + \frac{1}{8}\sin^{-1} \left(\frac{x}{$c}\right) \right .\cr
   91 & \left . \quad + \frac{1}{8}\left(\frac{x}{$c}\right) \left(\frac{\sqrt{$b - x^2} }{$c}\right) \right ]^{$c}_{$a} \cr
   92 & = $d2 \cdot $c^{4} \pi \frac{1}{8} \frac{\pi}{2} = $d \cdot $c^{4} \pi^2 \frac{1}{4}  \cr
   93 & = $ans .
   94 \end{array}
   95 \]
   96 $PAR
   97 END_SOLUTION
   98 
   99 ENDDOCUMENT();
  100 
  101