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Tue Nov 8 15:17:41 2011 UTC (19 months, 1 week ago) by aubreyja
File size: 2344 byte(s)
Rogawski problems contributed by publisher WHFreeman. These are a subset of the problems available to instructors who use the Rogawski textbook. The remainder can be obtained from the publisher.

    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
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();