Chapter 14 - SECTION 14.1 (PAGE 759) R. A. ADAMS: CALCULUS...

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SECTION 14.1 (PAGE 759) R. A. ADAMS: CALCULUS CHAPTER 14. MULTIPLE INTEGRATION Section 14.1 Double Integrals (page 759) 1. f ( x , y ) = 5 x y R = 1 × ± f ( 0 , 1 ) + f ( 0 , 2 ) + f ( 1 , 1 ) + f ( 1 , 2 ) + f ( 2 , 1 ) + f ( 2 , 2 ) ² = 4 + 3 + 3 + 2 + 2 + 1 = 15 2. R = 1 × ± f ( 1 , 1 ) + f ( 1 , 2 ) + f ( 2 , 1 ) + f ( 2 , 2 ) + f ( 3 , 1 ) + f ( 3 , 2 ) ² = 3 + 2 + 2 + 1 + 1 + 0 = 9 3. R = 1 × ± f ( 0 , 0 ) + f ( 0 , 1 ) + f ( 1 , 0 ) + f ( 1 , 1 ) + f ( 2 , 0 ) + f ( 2 , 1 ) ² = 5 + 4 + 4 + 3 + 3 + 2 = 21 4. R = 1 × ± f ( 1 , 0 ) + f ( 1 , 1 ) + f ( 2 , 0 ) + f ( 2 , 1 ) + f ( 3 , 0 ) + f ( 3 , 1 ) ² = 4 + 3 + 3 + 2 + 2 + 1 = 15 5. R = 1 × ± f ( 1 2 , 1 2 ) + f ( 1 2 , 3 2 ) + f ( 3 2 , 1 2 ) + f ( 3 2 , 3 2 ) + f ( 5 2 , 1 2 ) + f ( 5 2 , 3 2 ) ² = 4 + 3 + 3 + 2 + 2 + 1 = 15 6. I = ³³ D ( 5 x y ) dA is the volume of the solid in the figure. x y z 3 5 3 2 2 z = 5 x y Fig. 14.1.6 The solid is split by the vertical plane through the z - axis and the point ( 3 , 2 , 0 ) into two pyramids, each with a trapezoidal base; one pyramid’s base is in the plane y = 0 and the other’s is in the plane z = 0. I is the sum of the volumes of these pyramids: I = 1 3 ´ 5 + 2 2 ( 3 )( 2 ) µ + 1 3 ´ 5 + 3 2 ( 2 )( 3 ) µ = 15 . 7. J = ³³ D 1 R = 4 × 1 × ± 5 + 5 + 5 + 5 + 4] = 96 8. R = 4 × 1 × ± 4 + 4 + 4 + 3 + 0] = 60 9. R = 4 × 1 × ± 5 + 5 + 4 + 4 + 2] = 80 10. J = area of disk = π( 5 2 ) 78 . 54 11. R = 1 × ( e 1 / 2 + e 1 / 2 + e 3 / 2 + e 3 / 2 + e 5 / 2 + e 5 / 2 ) 32 . 63 12. f ( x , y ) = x 2 + y 2 R = 4 × 1 × ± f ( 1 2 , 1 2 ) + f ( 3 2 , 1 2 ) + f ( 5 2 , 1 2 ) + f ( 7 2 , 1 2 ) + f ( 9 2 , 1 2 ) + f ( 1 2 , 3 2 ) + f ( 3 2 , 3 2 ) + f ( 5 2 , 3 2 ) + f ( 7 2 , 3 2 ) + f ( 9 2 , 3 2 ) + f ( 1 2 , 5 2 ) + f ( 3 2 , 5 2 ) + f ( 5 2 , 5 2 ) + f ( 7 2 , 5 2 ) + f ( 1 2 , 7 2 ) + f ( 3 2 , 7 2 ) + f ( 5 2 , 7 2 ) + f ( 1 2 , 9 2 ) + f ( 3 2 , 9 2 ) ² = 918 13. ¶¶ R = area of R = 4 × 5 = 20 . y x 3 1 1 4 R Fig. 14.1.13 14. ³³ D ( x + 3 ) = ³³ D xdA + 3 ³³ D = 0 + 3 ( area of D ) = 3 × π 2 2 2 = 6 π. The integral of x over D is zero because D is symmetri- cal about x = 0. 528
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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 14.1 (PAGE 759) y x 2 D y = 4 x 2 2 2 Fig. 14.1.14 15. T is symmetric about the line x + y = 0. Therefore, ±± T ( x + y ) dA = 0. y x T ( 2 , 2 ) ( 1 , 1 ) ( 2 , 2 ) ( 1 , 1 ) Fig. 14.1.15 16. ±± | x |+| y |≤ 1 ² x 3 cos ( y 2 ) + 3 sin y π ³ = 0 + 0 π ² area bounded by | x |+| y |= 1 =− π × 4 × 1 2 ( 1 )( 1 ) 2 π. (Each of the first two terms in the integrand is an odd function of one of the variables, and the square is sym- metrical about each coordinate axis.) y x 1 1 1 1 Fig. 14.1.16 17. x 2 + y 2 1 ( 4 x 2 y 3 x + 5 ) = 0 0 + 5(area of disk) (by symmetry) = 5 y x x 2 + y 2 = 1 1 Fig. 14.1.17 18. x 2 + y 2 a 2 ´ a 2 x 2 y 2 = volume of hemisphere shown in the figure = 1 2 µ 4 3 π a 3 = 2 3 π a 3 . x y z a z = a 2 x 2 y 2 x 2 + y 2 = a 2 a Fig. 14.1.18 19. x 2 + y 2 a 2 ² a ´ x 2 + y 2 ³ = volume of cone shown in the figure = 1 3 π a 3 . x y z a y = a x 2 + y 2 x 2 + y 2 = a 2 a Fig. 14.1.19 20. By the symmetry of S with respect to x and y we have ±± S ( x + y ) = 2 ±± S xdA = 2 × ( volume of wedge shown in the figure ) = 2 × 1 2 ( a 2 ) a = a 3 . 529
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SECTION 14.1 (PAGE 759) R. A. ADAMS: CALCULUS x y z z = x S ( a , a , 0 ) Fig. 14.1.20 21. ±± T ( 1 x y ) dA = volume of the tetrahedron shown in the figure = 1 3 ² 1 2 ( 1 )( 1 ) ³ ( 1 ) = 1 6 . x y z z = 1 x y ( 0 , 0 , 1 ) ( 0 , 1 , 0 ) ( 1 , 0 , 0 ) T Fig. 14.1.21 22. R ´ b 2 y 2 = volume of the quarter cylinder shown in the figure = 1 4 b 2 ) a = 1 4 π ab 2 .
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This note was uploaded on 12/16/2009 for the course FEW, FEWEB 400567 taught by Professor Moerdersen during the Fall '09 term at Vrije Universiteit Amsterdam.

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Chapter 14 - SECTION 14.1 (PAGE 759) R. A. ADAMS: CALCULUS...

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