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| 1 : | aubreyja | 2584 | ## 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('15') | ||
| 10 : | ## Author('JustAsk - Kobi Fonarov') | ||
| 11 : | ## Institution('W.H.Freeman') | ||
| 12 : | |||
| 13 : | DOCUMENT(); | ||
| 14 : | |||
| 15 : | loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl"); | ||
| 16 : | loadMacros("Parser.pl"); | ||
| 17 : | loadMacros("freemanMacros.pl"); | ||
| 18 : | loadMacros("PGauxiliaryFunctions.pl"); | ||
| 19 : | loadMacros("PGgraphmacros.pl"); | ||
| 20 : | loadMacros("PGchoicemacros.pl"); | ||
| 21 : | |||
| 22 : | TEXT(beginproblem()); | ||
| 23 : | |||
| 24 : | $r=random(2,9); | ||
| 25 : | $r2=$r**2; | ||
| 26 : | |||
| 27 : | $answer=Real($PI*$r2); | ||
| 28 : | |||
| 29 : | $curve="\mathcal{C}"; | ||
| 30 : | $domain="\mathcal{D}"; | ||
| 31 : | $FF="\mathbf{F}"; | ||
| 32 : | |||
| 33 : | Context()->texStrings; | ||
| 34 : | |||
| 35 : | BEGIN_TEXT | ||
| 36 : | \{ textbook_ref_exact("Rogawski ET 2e", "17.1","15") \} | ||
| 37 : | $PAR | ||
| 38 : | Use the equation | ||
| 39 : | \[ A=\frac12 \int_{$curve} x \,dy-y \,dx \] | ||
| 40 : | to calculate the area of the circle of radius \($r\) centered at the origin. $PAR | ||
| 41 : | \(A=\) \{ans_rule()\} | ||
| 42 : | $PAR | ||
| 43 : | |||
| 44 : | END_TEXT | ||
| 45 : | |||
| 46 : | ANS($answer->cmp); | ||
| 47 : | |||
| 48 : | Context()->texStrings; | ||
| 49 : | SOLUTION(EV3(<<'END_SOLUTION')); | ||
| 50 : | $PAR | ||
| 51 : | $SOL We parametrize the circle by \(x=$r \cos \theta \), \(y=$r \sin \theta\), hence, | ||
| 52 : | \[ | ||
| 53 : | x \,dy-y \,dx =\]\[ | ||
| 54 : | $r \cos \theta \cdot $r \cos \theta \,d\theta -$r \sin \theta (-$r \sin \theta) \,d\theta=\]\[ | ||
| 55 : | ($r2 \cos^2\theta +$r2 \sin^2\theta ) \,d\theta =$r2 \,d\theta | ||
| 56 : | \] | ||
| 57 : | Therefore, | ||
| 58 : | \[A=\frac12 \int_{$curve} x \,dy-y \,dx=\]\[ | ||
| 59 : | \frac12 \int_0^{2\pi } $r2 \,d\theta = | ||
| 60 : | \frac{$r2}2\cdot 2\pi =$r2\pi .\] | ||
| 61 : | |||
| 62 : | END_SOLUTION | ||
| 63 : | |||
| 64 : | ENDDOCUMENT(); |
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