Math252Sols4F08

Math252Sols4F08 - QUIZ 4 (CHAPTER 17) SOLUTIONS MATH 252...

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QUIZ 4 (CHAPTER 17) SOLUTIONS MATH 252 – FALL 2008 – KUNIYUKI SCORED OUT OF 125 POINTS ± MULTIPLIED BY 0.84 ± 105% POSSIBLE 1) Reverse the order of integration, and evaluate the resulting double integral: y 113 + x 9 dx dy y 4 2 ± 0 16 ± . Give a simplified exact answer; do not approximate. Sketch the region of integration; you may use different scales for the x - and y - axes. (20 points) Sketch the region of integration, R . It may help to label the limits of integration: Let I = y 113 + x 9 dx dy x = y 4 x = 2 ± y = 0 y = 16 ± We aim / fix y , shoot x , and slide y . Reverse the order of integration. Solve x = y 4 for y : y = x 4 x ± 0 ()
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We aim / fix x , shoot y , and slide x . I = y 113 + x 9 dy dx y = 0 y = x 4 ± x = 0 x = 2 ± = 1 113 + x 9 ² y ³ ´ µ · ¸ dy dx y = 0 y = x 4 ± x = 0 x = 2 ± = 1 113 + x 9 ² y 2 2 ¹ º » ¼ ½ ¾ y = 0 y = x 4 ³ ´ µ µ · ¸ ¸ dx x = 0 x = 2 ± = 1 113 + x 9 ² x 4 () 2 2 ¹ º » » » ¼ ½ ¾ ¾ ¾ ¿ 0 ¹ º ¼ ½ ³ ´ µ µ · ¸ ¸ dx 0 2 ± = x 8 2 113 + x 9 dx 0 2 ± We now perform a standard u -substitution. Let u = 113 + x 9 du = 9 x 8 dx ± x 8 dx = 1 9 du You can also “Compensate”: x 8 2 113 + x 9 dx 0 2 ± = 1 9 9 x 8 2 113 + x 9 dx 0 2 ± Change the limits of integration: x = 0 ± u = 113 + 0 9 ± u = 113 x = 2 ± u = 113 + 2 9 ± u = 625
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= 1 9 du 2 u u = 113 u = 625 ± = 1 18 u ² 1/2 du 113 625 ± = 1 18 u ³ ´ µ · ¸ 113 625 = 1 18 2 u ³ ´ · 113 625 = 1 9 u ³ ´ · 113 625 = 1 9 625 ² 113 () = 25 ² 113 9 Note: This is about 1.59665. 2) Evaluate e 3 x 2 + 3 y 2 dy dx 0 ± x 2 ² ± 2 2 ² . Your answer must be exact and simplified. You do not have to approximate it. (16 points) Let’s call this double integral “ I .” It may help to label the limits of integration: I = e 3 x 2 + 3 y 2 dy dx y = 0 y = 4 ± x 2 ² x = ± 2 x = 2 ² Sketch the region of integration, R . If we aim / fix any x in ± 2, 2 ² ³ ´ µ , then y is shot from y = 0 to y = ± x 2 . The graph of y = 4 ± x 2 is a semicircle, the upper half of the circle of radius 2 centered at the origin. Since the x -interval is ± 2, 2 ² ³ ´ µ , we do graph the entire semicircle. Observe that the integrand, e 3 x 2 + 3 y 2 , is continuous throughout R .
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Convert to polar coordinates. Why? R is well suited for that. • The integrand is well suited for that. • It looks like using Cartesian coordinates will be difficult. I = e 3 x 2 + 3 y 2 dy dx y = 0 y = 4 ± x 2 ² x = ± 2 x = 2 ² = e 3 x 2 + y 2 () dA R ²² = e 3 r 2 dA R ²² = e 3 r 2 ³ rdrd ´ r = 0 r = 2 ² = 0 = µ ² At this point, we can separate the double integral into a product of single integrals. This is because the limits of integration are all constants, and the integrand is separable. = d ± = 0 = ² ³ ´ µ · ¸ ¹ re 3 r 2 dr r = 0 r = 2 ³ ´ µ · ¸ ¹ Evaluate the first integral: d = 0 = ³ = ´ µ · = 0 = = To evaluate the second integral, re 3 r 2 dr r = 0 r = 2 ± , we perform a standard u -substitution: Let u = 3 r 2 du = 6 rdr ± = 1 6 du You can also “Compensate”: re 3 r 2 dr r = 0 r = 2 ± = 1 6 6 re 3 r 2 dr r = 0 r = 2 ± Change the limits of integration: r = 0 ± u = 30 2 ± u = 0 r = 2 ± u = 32 2 ± u = 12
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I = ± ² ³ ´ µ 1 6 e u du u = 0 u = 12 ² ³ · ´ µ ¸ = 6 e u du 0 12 = 6 e u ² ³ ´ µ 0 12 = 6 e 12 ¹ e 0 () = 6 e 12 ¹ 1 º 85,218 3) Find the surface area of the portion of the graph of z = 7 x + 1 2 y 2 that lies over R , where R
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This note was uploaded on 09/08/2011 for the course MATH 252 taught by Professor Staff during the Spring '11 term at Mesa CC.

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Math252Sols4F08 - QUIZ 4 (CHAPTER 17) SOLUTIONS MATH 252...

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