View of /trunk/NationalProblemLibrary/Michigan/Chap16Sec4/Q21.pg

    1 # DESCRIPTION
    2 # Problem from Calculus, multi-variable, Hughes-Hallett et al.,
    3 # originally from 5ed (with updates)
    4 # WeBWorK problem written by Gavin LaRose, <glarose@umich.edu>
    5 # ENDDESCRIPTION
    6 
    7 ## KEYWORDS('polar coordinates', 'integral', 'calculus')
    8 ## Tagged by glr 04/29/10
    9 
   10 
   11 ## DBsubject('Calculus')
   12 ## DBchapter('Multiple Integrals')
   13 ## DBsection('Double Integrals in Polar Coordinates')
   14 ## Date('')
   15 ## Author('Gavin LaRose')
   16 ## Institution('University of Michigan')
   17 ## TitleText1('Calculus')
   18 ## EditionText1('5')
   19 ## AuthorText1('Hughes-Hallett')
   20 ## Section1('16.4')
   21 ## Problem1('21')
   22 
   23 ## Textbook tags
   24 ## HHChapter1('Integrating Functions of Several Variables')
   25 ## HHSection1('Double Integrals in Polar Coordinates')
   26 
   27 DOCUMENT();
   28 
   29 loadMacros(
   30 "PGstandard.pl",
   31 "PGchoicemacros.pl",
   32 "MathObjects.pl",
   33 # "parserNumberWithUnits.pl",
   34 # "parserFormulaWithUnits.pl",
   35 # "parserFormulaUpToConstant.pl",
   36 # "PGcourse.pl",
   37 );
   38 
   39 Context("Numeric");
   40 Context()->variables->are( r=>'Real', t=>'Real' );
   41 $showPartialCorrectAnswers = 1;
   42 
   43 $ao2 = random(1,5,1);
   44 $a = 2*$ao2;
   45 $t0 = Compute("-pi/4");
   46 $t1 = Compute("pi/4");
   47 $r0 = Compute("0");
   48 $r1 = Compute("sqrt($a)/cos(t)");
   49 
   50 $rvar = Compute("r");
   51 
   52 $integr = Compute( "$ao2/cos(t)^2" );
   53 $antide = Compute( "$ao2*tan(t)" );
   54 $result = Compute( "$a" );
   55 
   56 Context()->texStrings;
   57 TEXT(beginproblem());
   58 BEGIN_TEXT
   59 
   60 Convert the integral
   61 \[ \int_0^{\sqrt{$a}}\int_{-x}^x\,dy\,dx \]
   62 to polar coordinates and
   63 evaluate it (use \(t\) for \(\theta\)):
   64 $BR
   65 With
   66 \(a = \) \{ans_rule(10)\}, \(b = \) \{ans_rule(10)\},
   67 \(c = \) \{ans_rule(10)\} and \(d = \) \{ans_rule(10)\},
   68 $BR
   69 \( \int_0^{\sqrt{$a}}\int_{-x}^x\,dy\,dx =
   70    \int_{a}^{b}\int_{c}^{d} \) \{ ans_rule(5) \}\(dr\,dt\)
   71 $BR
   72 $BCENTER
   73 \( = \int_{a}^{b}\) \{ ans_rule(25) \} \(\,dt\) $BR
   74 \( = \) \{ ans_rule(25) \}\(\bigg|_{a}^{b} \) $BR
   75 \( = \) \{ ans_rule(25) \}.
   76 $ECENTER
   77 
   78 END_TEXT
   79 Context()->normalStrings;
   80 
   81 ANS($t0->cmp() );
   82 ANS($t1->cmp() );
   83 ANS($r0->cmp() );
   84 ANS($r1->cmp() );
   85 ANS($rvar->cmp() );
   86 ANS($integr->cmp() );
   87 ANS($antide->cmp() );
   88 ANS($result->cmp() );
   89 
   90 Context()->texStrings;
   91 SOLUTION(EV3(<<'END_SOLUTION'));
   92 $PAR SOLUTION $PAR
   93 
   94 In polar coordinates, \(-\pi/4\le\theta\le\pi/4\).  Also,
   95 the outer (right) boundary of the region is
   96 \(\sqrt{$a} = x = r\cos\theta\). Hence,
   97 \(0\le r\le \sqrt{$a}/\cos\theta\). The integral
   98 becomes
   99 \[
  100 \int_0^{\sqrt{$a}}\int_{-x}^x\,dy\,dx
  101   =\int_{-\pi/4}^{\pi/4}\int_0^{\sqrt{$a}/\!\cos\theta}\,r\,dr\,d\theta.
  102 \]
  103 Then
  104 \[
  105 \int_{-\pi/4}^{\pi/4}\int_0^{\sqrt{$a}/\!\cos\theta}\,r\,dr\,d\theta
  106   = \int_{-\pi/4}^{\pi/4}\left(\frac{r^2}{2}\bigg|_0^{\sqrt{$a}/\!\cos\theta}\right)\,d\theta
  107   = \int_{-\pi/4}^{\pi/4}\frac{$ao2}{\cos^2\theta}\,d\theta.
  108 \]
  109 And
  110 \[
  111 \int_{-\pi/4}^{\pi/4}\frac{$ao2}{\cos^2\theta}\,d\theta
  112   =$ao2 \tan\theta\bigg|_{-\pi/4}^{\pi/4} = $ao2\cdot(1 - (-1)) = $a.
  113 \]
  114 Notice that we can check this answer because the integral gives the
  115 area of the shaded triangular region which is \(\frac{1}{2} \cdot \sqrt{$a}
  116 \cdot (2\sqrt{$a}) = $a\).
  117 
  118 END_SOLUTION
  119 Context()->normalStrings;
  120 
  121 
  122 COMMENT('MathObject version');
  123 ENDDOCUMENT();