View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/17_Fundamental_Theorems_of_Vector_Analysis/17.1_Green's_Theorem/17.1.9.pg

    1 ## DBsubject('Calculus')
    2 ## DBchapter('Fundamental Theorems of Vector Analysis')
    3 ## DBsection('Greens Theorem')
    4 ## KEYWORDS('calculus')
    5 ## TitleText1('Calculus: Early Transcendentals')
    6 ## EditionText1('2')
    7 ## AuthorText1('Rogawski')
    8 ## Section1('17.1')
    9 ## Problem1('9')
   10 ## Author('JustAsk - Kobi Fonarov')
   11 ## Institution('W.H.Freeman')
   12 ## UsesAuxiliaryFiles('image_17_1_9.png')
   13 
   14 DOCUMENT();
   15 
   16 loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl");
   17 loadMacros("Parser.pl");
   18 loadMacros("freemanMacros.pl");
   19 loadMacros("PGauxiliaryFunctions.pl");
   20 loadMacros("PGgraphmacros.pl");
   21 loadMacros("PGchoicemacros.pl");
   22 
   23 TEXT(beginproblem());
   24 
   25 $a=random(2,9);
   26 $b=random(2,9);
   27 $b2=$b**2;
   28 $fxdeg=random(2,9);
   29 
   30 
   31 $GreenResult=Real($a*$b2);
   32 $answer=-Real($a*$b2);
   33 
   34 
   35 $boundary="\partial\mathcal{D}";
   36 $curve="\mathcal{C}";
   37 $domain="\mathcal{D}";
   38 $FF="\mathbf{F}";
   39 
   40 TEXT('<SCRIPT>jsMath.Extension.Require("AMSmath");</SCRIPT>')
   41        if $displayMode eq 'HTML_jsMath';
   42 
   43 Context()->texStrings;
   44 
   45 BEGIN_TEXT
   46 \{ textbook_ref_exact("Rogawski ET 2e", "17.1","9") \}
   47 $PAR
   48 Use Green's Theorem to evaluate the
   49 line integral of \($FF = \left< x^{$fxdeg}, $a x\right>\)  $PAR around the
   50 boundary of the parallelogram in the following figure
   51 (note the orientation). $PAR
   52 $BR \{image("image_17_1_9.png", width=>234, height=>157)\} With \(x_0=$b\) and \(y_0=$b\).
   53 $PAR
   54 \(\int_{$curve} x^{$fxdeg} \,dx+$a x \,dy =\) \{ans_rule()\}
   55 $PAR
   56 
   57 END_TEXT
   58 
   59 ANS($answer->cmp);
   60 
   61 Context()->texStrings;
   62 SOLUTION(EV3(<<'END_SOLUTION'));
   63 $PAR
   64 $SOL First note that the orientation of the boundary curve is clockwise. We will use
   65 Green's Theorem remembering that the boundary curve must be oriented counterclockwise.
   66 We have \(P= x^{$fxdeg}\) and \(Q=$a x\), therefore
   67 \[
   68 \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=$a-0=$a
   69 \]
   70 Hence, Green's Theorem implies
   71 \[\int_{$boundary} x^{$fxdeg} \,dx+$a x \,dy =
   72 \iint_{$domain} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \,dA =\]\[
   73 \iint_{$domain} $a \,dA=$a \iint_{$domain}  \,dA
   74 = $a\, \mathrm{Area}($domain)=$a \cdot $b2=$GreenResult
   75 \]
   76 So now accounting for the orientation,
   77 \[
   78 \[\int_{$curve} x^{$fxdeg} \,dx+$a x \,dy =
   79 -\[\int_{$boundary} x^{$fxdeg} \,dx+$a x \,dy =$answer
   80 \]
   81 
   82 END_SOLUTION
   83 
   84 ENDDOCUMENT();