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# Annotation of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/7_Techniques_of_Integration/7.3_Trigonometric_Substitution/7.3.58.pg

 1 : aubreyja 2584 ## 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(); 14 : loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl"); 15 : loadMacros("PGchoicemacros.pl"); 16 : loadMacros("Parser.pl"); 17 : loadMacros("freemanMacros.pl"); 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 :