View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/15_Multiple_Integration/15.4_Integration_in_Polar_Cylindrical_and_Spherical_Coordinates/15.4.23.pg

    1 ## DBsubject('Calculus')
    2 ## DBchapter('Multiple Integration')
    3 ## DBsection('Integration in Polar, Cylindrical, and Spherical Coordinates')
    4 ## KEYWORDS('calculus', '')
    5 ## TitleText1('Calculus: Early Transcendentals')
    6 ## EditionText1('2')
    7 ## AuthorText1('Rogawski')
    8 ## Section1('15.4')
    9 ## Problem1('23')
   10 ## Author('JustAsk - Kobi Fonarov')
   11 ## Institution('W.H.Freeman')
   12 ## UsesAuxiliaryFiles('image_15_4_27_a.png','image_15_4_27_b.png','image_15_4_27_c.png')
   13 
   14 DOCUMENT();
   15 
   16 loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl");
   17 loadMacros("Parser.pl");
   18 loadMacros("freemanMacros.pl");
   19 
   20 $a=random(1,9);
   21 
   22 
   23 $w="\mathcal{W}";
   24 $d="\mathcal{D}";
   25 $fout=4*$a**2;
   26 $gout=$a**2;
   27 $f="x^2+y^2=$fout";
   28 $g="(x-$a)^2+y^2=$gout";
   29 $rf=$a*2;
   30 $rg=$a;
   31 $sol1=Real((2/3)*$PI*($rf)**3);
   32 $sol2=Real((4/9)*($rf**3));
   33 $answer=Real(($sol1)-($sol2));
   34 
   35 Context()->texStrings;
   36 BEGIN_TEXT
   37 \{ beginproblem() \}
   38 \{ textbook_ref_exact("Rogawski ET 2e", "15.4","23") \}
   39 $PAR
   40 Evaluate \($IINT _{$d}\sqrt{x^2+y^2}\,dA\),
   41 where \($d\) is the domain in Figure 4
   42 $PAR
   43 \{image("image_15_4_27_a.png", width=>228, height=>244)\}
   44 \(\begin{array}{ll} F: $f\quad G: $g\\R_f=$rf \quad R_g=$rg \end{array}\)
   45 $PAR
   46 \($IINT _{$d}\sqrt{x^2+y^2}\,dA=\) \{ans_rule()\}
   47 $PAR
   48 
   49 END_TEXT
   50 
   51 ANS($answer->cmp);
   52 
   53 Context()->texStrings;
   54 
   55 SOLUTION(EV3(<<'END_SOLUTION'));
   56 $PAR
   57 $SOL We denote by \({$d}_1\) and \({$d}_2\) the regions enclosed by the circles \($f\) and \($g\). $BR
   58 Therefore,
   59 \[
   60 $IINT _{$d} \sqrt{x^2+y^2} \,dx \,dy =
   61 $IINT _{{$d}_1} \sqrt{x^2+y^2} \,dx \,dy
   62  - $IINT _{{$d}_2} \sqrt{x^2+y^2} \,dx \,dy \quad (\mathbf{1})
   63 \]
   64 We compute the integrals on the right hand-side.
   65 $PAR
   66 \({$d}_1 \):$PAR
   67 \{image("image_15_4_27_b.png", width=>224, height=>227)\}
   68 \(\begin{array}{ll} F: $f\quad \\R_f=$rf \end{array}\)
   69 $PAR
   70 The circle \($f\) has polar equation \(r=$rf\), therefore \({$d}_1\) is determined by the following inequalities:
   71 \[{$d}_1: \,0\le \theta \le 2\pi,\ 0\le r\le $rf \]
   72 The function in polar coordinates is \(f( x,y )=\sqrt{{x}^2+{y}^2}=r\).
   73 Using change of variables in the integral gives
   74 \[
   75 $IINT _{{$d}_1} \sqrt{{x}^2+{y}^2} \,dx \,dy=
   76 \int_0^{2\pi } \int_0^{$rf} r\cdot r \,dr \,d\theta =
   77 \int_0^{2\pi } \int_0^{$rf} {r}^2 \,dr \,d\theta=
   78 \]\[
   79 \int_0^{2\pi } \frac{{r}^3}{3}\bigg|_{r=0}^{$rf}\,d\theta =
   80 \int_0^{2\pi } \frac{\{$rf**3\}}{3} \,d\theta =
   81 \frac{\{2*($rf**3)\}\pi }{3} \quad (\mathbf{2})
   82 \]
   83 \({$d}_2\) : $PAR
   84 \{image("image_15_4_27_c.png", width=>227, height=>230)\}
   85 \(\begin{array}{lll} G: $g\\R_f=$rf \quad R_g=$rg \\ 0 \le r \le \{2*$rg\}\cos\theta\end{array}\)
   86 $PAR
   87 \({$d}_2\) lies in the angular sector \(-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}\).$BR
   88 We find the polar equation of the circle \($g\):
   89 \[
   90 {(x-$rg)}^2+{y}^2=
   91 {x}^2-\{2*$rg\}x+\{$rg**2\}+{y}^2=
   92 {x}^2+{y}^2-\{2*$rg\}x+\{$rg**2\}=\{$rg**2\}
   93 \]\[
   94 \begin{array}{lll} {x}^2+{y}^2=\{2*$rg\}x\\
   95  {r}^2=\{2*$rg\}r \cos \theta\\
   96  r=\{2*$rg\} \cos \theta \end{array}
   97 \]
   98 Thus, the domain \({$d}_2\) is defined by the following inequalities:
   99 \[{$d}_2:\,-\frac{\pi}{2}\le\theta\le\frac{\pi}{2},\ 0\le r\le \{2*$rg\} \cos \theta\]
  100 We use the change of variables in the integral and integration table to obtain
  101 \[
  102 $IINT _{{$d}_2\,} \sqrt{{x}^2+{y}^2} \,dx \,dy=
  103 \int_{-\pi/2}^{\pi/2} \int_0^{\{2*$rg\} \cos \theta } r\cdot r \,dr \,d\theta=
  104 \int_{-\pi/2}^{\pi/2} \int_0^{\{2*$rg\} \cos \theta } {r}^2 \,dr \,d\theta=
  105 \]\[
  106 \int_{-\pi/2}^{\pi/2} \frac{{r}^3}{3}\bigg|_{r=0}^{\{2*$rg\}\cos \theta}d\theta=
  107 \int_{-\pi/2}^{\pi/2} \frac{\{(2*$rg)**3\}\, \cos^3\theta }{3} \,d\theta=
  108 2 \int_0^{\pi/2} \frac{\{(2*$rg)**3\}\, \cos^3\theta }{3} \,d\theta=
  109 \]\[
  110 \frac{\{2*((2*$rg)**3)\}}{3}\left(\frac{\cos^2\theta \sin \theta }{3}+\frac{2}{3} \sin \theta \right)\bigg|_{\theta =0}^{\pi/2}=
  111 \frac{\{2*((2*$rg)**3)\}}{3}\cdot \frac{2}{3} \sin \frac{\pi }{2}=
  112 \frac{\{4*((2*$rg)**3)\}}{9}\quad (\mathbf{3})
  113 \]
  114 Substituting \((\mathbf{2})\) and \((\mathbf{3})\) in \((\mathbf{1})\), we obtain the following solution:
  115 \[
  116 $IINT _{$d} \sqrt{{x}^2+{y}^2} \,dx \,dy=
  117 \frac{\{2*($rf**3)\}\pi }{3}-\frac{\{4*((2*$rg)**3)\}}{9}
  118 \approx $answer
  119 \]
  120 
  121 END_SOLUTION
  122 
  123 ENDDOCUMENT();