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Tue Aug 24 14:40:25 2010 UTC (2 years, 8 months ago) by apizer
File size: 3073 byte(s)
Results of running convert_fun_in_dir.sh to clean up problems

    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
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();