Calculus Multivariable McCallum, Hughes-Hallett, Gleason 4th Ed ch16 solutions

Calculus: Multivariable

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16.1 SOLUTIONS 1105 CHAPTER SIXTEEN Solutions for Section 16.1 Exercises 1. Mark the values of the function on the plane, as shown in Figure 16.1, so that you can guess respectively at the smallest and largest values the function takes on each small rectangle. Lower sum = X f ( x i , y i x Δ y = 4Δ x Δ y + 6Δ x Δ y + 3Δ x Δ y + 4Δ x Δ y = 17Δ x Δ y = 17(0 . 1)(0 . 2) = 0 . 34 . Upper sum = X f ( x i , y i x Δ y = 7Δ x Δ y + 10Δ x Δ y + 6Δ x Δ y + 8Δ x Δ y = 31Δ x Δ y = 31(0 . 1)(0 . 2) = 0 . 62 . 1 . 0 1 . 1 1 . 2 2 . 0 2 . 2 2 . 4 x y 5 4 3 7 6 5 10 8 4 - ¾ Δ x = 0 . 1 6 ? Δ y = 0 . 2 Figure 16.1 2. In the subrectangle in the top left in Figure 16.4, it appears that f ( x, y ) has a maximum value of about 9. In the subrect- angle in the top middle, f ( x, y ) has a maximum value of 10. Continuing in this way, and multiplying by Δ x and Δ y , we have Overestimate = (9 + 10 + 12 + 7 + 8 + 10 + 5 + 7 + 8)(10)(5) = 3800 . Similarly, we ±nd Underestimate = (7 + 7 + 8 + 4 + 5 + 7 + 1 + 3 + 6)(10)(5) = 2400 . Thus, we expect that 2400 Z R f ( x, y ) dA 3800 . 3. (a) If we take the partition of R consisting of just R itself, we get Lower bound for integral = min R f · A R = 0 · (4 - 0)(4 - 0) = 0 . Similarly, we get Upper bound for integral = max R f · A R = 4 · (4 - 0)(4 - 0) = 64 .
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1106 Chapter Sixteen /SOLUTIONS (b) The estimates asked for are just the upper and lower sums. We partition R into subrectangles R ( a,b ) of width 2 and height 2, where ( a, b ) is the lower-left corner of R ( a,b ) . The subrectangles are then R (0 , 0) , R (2 , 0) , R (0 , 2) , and R (2 , 2) . (0 , 0) (0 , 4) (4 , 0) (4 , 4) Figure 16.2 Then we Fnd the lower sum Lower sum = X ( a,b ) A R ( a,b ) · min R ( a,b ) f = X ( a,b ) 4 · ( Min of f on R ( a,b ) ) = 4 X ( a,b ) ( Min of f on R ( a,b ) ) = 4( f (0 , 0) + f (2 , 0) + f (0 , 2) + f (2 , 2)) = 4( 0 · 0 + 2 · 0 + 0 · 2 + 2 · 2) = 8 . Similarly, the upper sum is Upper sum = 4 X ( a,b ) ( Max of f on R ( a,b ) ) = 4( f (2 , 2) + f (4 , 2) + f (2 , 4) + f (4 , 4)) = 4( 2 · 2 + 4 · 2 + 2 · 4 + 4 · 4) = 24 + 16 2 46 . 63 . The upper sum is an overestimate and the lower sum is an underestimate, so we can get a better estimate by averaging them to get 16 + 8 2 27 . 3 . 4. (a) We Frst Fnd an over- and underestimate of the integral, using four subrectangles. On the Frst subrectangle ( 0 x 3 , 0 y 4) , the function f ( x, y ) appears to have a maximum of 100 and a minimum of 79. Continuing in this way, and using the fact that Δ x = 3 and Δ y = 4 , we have Overestimate = (100 + 90 + 85 + 79)(3)(4) = 4248 , and Underestimate = (79 + 68 + 61 + 55)(3)(4) = 3156 .
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Calculus Multivariable McCallum, Hughes-Hallett, Gleason 4th Ed ch16 solutions

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