Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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248 0 CHAPTER 15 MULTIPLE INTEGRALS (b) The subrectangles are shown in the figure. In each subrectangle, the sample point closest to the origin is the lower left comer, and the area of each subrectangle is .6.A = ~. Thus we estimate 4 4 ffnf(x,y) dA ~ L:: L:: f(xii,Yii) .6.A i= l j = l . = f (O, 2) .6.A + f(O , 2.5) .6.A + f(O , 3) . .6.A + f(O, 3.5) .6.A + /(1, 2) .6.A + f(1, 2.5) .6.A + /(1, 3) .6.A + /(1, 3.5) .6.A + / (2, 2) .6.A + /(2, 2.5) .6.A + /(2, 3) .6.A + /(2, 3.5) .6.A y 4 3 2 0 I 2 3 4 X + / (3, 2) .6.A + /(3, 2.5) .6.A + /(3, 3) .6.A + !(3, 3.5) .6.A = - 3(~) + < -5) G) + < - 6) (~) + < -4)(~) + < - 1) (~) + < - 2){ ~) + < -3)(~) + < -1)G) = - 8 + 1G) + o(~) + ( -1 ){~) + 1 (~) + 2(~) + 2G) ·;~- 1 (~) + 3G) 7. The val~es off ( x, y) = .j52 - x 2 - y 2 get smaller as we move farther from the origin, so on any of the subrectangles in the problem, the function will have its largest value at the lower left comer of the subrectangle and its smallest value at the upper right comer, and any otl).er value will lie between these two. So using these subrectangles we have U < V < L . (Note that this is true no matter how R is divided into subrectangles.) 9. (a) With m = n = 2, we h~ve .6.A = 4. Using the contolir map to estimate the value off at the center of each subrectangle, we have 2 2 ffn f(x, y) dA ~ L:; L:; J(xi, Yi) .6.A = .6.A[f(1, 1) + /(1, 3) -t /(3, 1) + /(3, 3)] i'::j 4(27 + 4 + 14 + 17) = 248 i = lj=l (b) /ave= Aln) ffn f(x, y) dA i'::j ft(248) = 15.5 11. z = 3 > 0; so we can interpret the integral as the volume of the solid S that lies below the plane z = 3 and above the rectangle [- 2, 2] x [1, 6]. Sis a rectangular solid, thus ffn 3 dA = 4 · 5 · 3 = 60. 13. z = f(x, y) = 4- 2y :2: 0 for 0 :::; y :5 1. Thus the integral represents the volume of that part ofthe rectangular.solid [0, 1] x [0, 1] x (0, 4] which lies below the plane z = 4 - 2y. So ffn(4 - 2y) dA = (1)(1)(2) + H 1)(1)(2) = 3 X © 2012 C.ngnge L1cd too publicly accessible website, in whole or in port.
SECTION 15.2 ITERATED INTEGRALS D 249 15. To calculate the estimates using a programmable calculator; we can use an algorithm similar to that of Exercise 4.1.9 [ET 5.1.9). In Maple, we cari define the function f(x, y) = v'1 + xe 11 (calling it f), load the student package, and then use the command middlesum(middlesum(f, x=0 . . 1,m), y=O .. 1 , m); to get the estimate with n = m 2 squares of equal size. Mathematica has no special Riemann sum command, but we can define f and then use nested Sum commands to calculate the estimates. ,. n estimate ' 1 1.141606 4 1.143191 16 1.143535 64 1.143617 256 1.143637 1024 1.143642 17. If we divideR into mn subrectangles, ffn k dA:::::: .f _t f (xi; . Yi;) ~A for any choice of sample points (xi;, Yi;).
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- STUDENT SOLUTIONS MANUAL for STEW
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.. BROOKS/COLE ~ I ~~r CENGAGE Lear
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D ABBREVIATIONS AND SYMBOLS CD cu D
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viii o CONTENTS 12.4 The Cross Prod
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10 D PARAMETRIC EQUATIONS AND POLAR
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SECTION 10.1 CURVES DEFINED BY PARA
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SECTION 10.1 CURVES DEFINED BY PARA
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SECTION 10.5 CONIC SECTIONS 0 27 5.
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x2 y2 y2 a:2 _ a2 b 61. ;_2 - - = 1
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CHAPTER 10 REVIEW 0 35 the length o
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CHAPTER 10 REVIEW 0 37 EXERCISES 1.
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CHAPTER 10 REVIEW 0 39 25. x = t +
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0 PROBLEMS PLUS l lt sin u dx cost
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11 . D INFINITE SEQUENCES AND SERIE
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SECTION 11.1 SEQUENCES 0 47 35. a,.
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61 _ x_ x · sin x - x- tx 3 + 1~
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CHAPTER 11 REVIEW 0 97 J'(xn)(xn -
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49./- 1 - dx = -ln{4- x) + C and 4-
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112 0 CHAPTER 12 VECTORS AND THE GE
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SECTION 13.1 VECTOR FUNCTIONS AND S
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0 APPENDIX Appendix H Complex Numbe
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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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