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Tue Nov 8 15:17:41 2011 UTC (18 months, 2 weeks ago) by aubreyja
File size: 3590 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('Techniques of Integration')
3 ## DBsection('Trigonometric Substitution')
4 ## KEYWORDS('calculus', 'integration', 'integral', 'trigonometric substitution', 'substitution', 'trigonometry', 'trigonometric', 'trig')
5 ## TitleText1('Calculus: Early Transcendentals')
6 ## EditionText1('2')
7 ## AuthorText1('Rogawski')
8 ## Section1('7.3')
9 ## Problem1('58')
10 ## Author('Christopher Sira')
11 ## Institution('W.H.Freeman')
12
13 DOCUMENT();
18 $context = Context(); 19 20$a = 0;
21 # Introduced more variation and kept same level of difficulty
22 $c = random(2, 4, 1); 23$d = random(5, 9, 1);
24 $b =$c*$c; 25$d2 = 2*$d; 26 #$dc4 = $d *($c**4)/4;
27 $dc4 =$d *($c**4)/8; 28$ans = Formula(" $dc4 * (pi ** 2) ")->reduce(); 29 30 Context()->texStrings; 31 BEGIN_TEXT 32 \{ beginproblem() \} 33 \{ textbook_ref_exact("Rogawski ET 2e", "7.3","58") \} 34$PAR
35 Find the volume of the solid obtained by revolving the graph of $$y = d x\sqrt{b - x^2}$$
36 over [$a,$c] about the y-axis.
37 $PAR 38 \{ans_rule()\} 39$PAR
40 END_TEXT
41 Context()->normalStrings;
42
43 ANS($ans->cmp); 44 45 Context()->texStrings; 46 SOLUTION(EV3(<<'END_SOLUTION')); 47$PAR
48 $SOL 49$PAR
50 Using the method of cylindrical shells, the volume is given by
51 $PAR 52 $$V = 2\pi \int^{c}_{a} x (d x\sqrt{b - x^2}) \, dx = d2 \pi \int^{c}_{a} x^2 \sqrt{b - x^2} \, dx$$. 53$PAR
54 To evaluate this integral, let $$x = c \sin \theta$$.  Then $$dx = c \cos \theta \, d\theta$$,
55 $PAR 56 $$b - x^2 = b (1 - \sin^2 \theta) = b \cos^2 \theta$$, 57$PAR
58 and
59 $60 \begin{array}{ll} 61 I & = \int x^2\sqrt{b - x^2} \, dx \cr 62 & = \int (c^2 \sin^2 \theta )\,( c \cos \theta )\, (c \cos \theta) \, d\theta \cr 63 & = c^{4} \int (1 - \cos^2 \theta) \cos^2 \theta \, d\theta. \cr 64 & \cr 65 \frac{1}{c^{4}} I &= \int \cos^2 \theta \, d\theta - \int \cos^4 \theta \, d\theta . 66 \end{array} 67$
68
69 Now we use the reduction formula for $$\int \cos^4 \theta \, d\theta$$:
70 $71 \begin{array}{ll} 72 \frac{1}{c^{4}} I & = \int \cos^2 \theta \, d\theta - \left[ \frac{cos^3 \theta \, \sin \theta}{4} + \frac{3}{4}\int \cos^2 \theta \, d\theta \right] \cr 73 & = -\frac{1}{4}\cos^3 \theta \, \sin \theta + \frac{1}{4} \int \cos^2 \theta \, d\theta \cr 74 & = -\frac{1}{4}\cos^3 \theta \, \sin \theta + \frac{1}{4} \left[ \frac{1}{2} \theta + \frac{1}{2} \sin \theta \, \cos \theta \right] + C \cr 75 & = -\frac{1}{4}\cos^3 \theta \, \sin \theta + \frac{1}{8} \theta + \frac{1}{8} \sin \theta \, \cos \theta + C. 76 \end{array} 77$
78
79 $PAR 80 Since $$\sin \theta = \frac{x}{c}$$, we know that $$\cos \theta = \frac{\sqrt{b - x^2} }{c}$$. 81 Then we have 82$PAR
83 $$\frac{1}{c^{4}} I = -\frac{1}{4}\left(\frac{\sqrt{b - x^2} }{c}\right)^3 \, \left(\frac{x}{c}\right) + \frac{1}{8}\sin^{-1} \left(\frac{x}{c}\right) 84 + \frac{1}{8}\left(\frac{x}{c}\right) \left(\frac{\sqrt{b - x^2} }{c}\right) + C$$.
85 $PAR 86 Now we can complete the volume calculation: 87 $88 \begin{array}{ll} 89 V & = \left . d2 \pi I \right |^{c}_{a} \cr 90 & = d2 \cdot c^{4} \pi \left [ -\frac{1}{4}\left(\frac{\sqrt{b - x^2} }{c}\right)^3 \, \left(\frac{x}{c}\right) + \frac{1}{8}\sin^{-1} \left(\frac{x}{c}\right) \right .\cr 91 & \left . \quad + \frac{1}{8}\left(\frac{x}{c}\right) \left(\frac{\sqrt{b - x^2} }{c}\right) \right ]^{c}_{a} \cr 92 & = d2 \cdot c^{4} \pi \frac{1}{8} \frac{\pi}{2} = d \cdot c^{4} \pi^2 \frac{1}{4} \cr 93 & = ans . 94 \end{array} 95$ 96$PAR
97 END_SOLUTION
98
99 ENDDOCUMENT();
100
101