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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 - 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 find 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 find 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 first find an over- and underestimate of the integral, using four subrectangles. On the first 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 . A better estimate of the integral is the average of the overestimate and the underestimate: Better estimate = 4248 + 3156 2 = 3702 . (b) The average value of f ( x, y ) on this region is the value of the integral divided by the area of the region. Since the area of R is (6)(8) = 48 , we approximate Average value = 1 Area Z R f ( x, y ) dA 1 48 · 3702 = 77 . 125 . We see in the table that the values of f ( x, y ) on this region vary between 55 and 100, so an average value of 77 . 125 is reasonable. 5. Partition R into subrectangles with the lines x = 0 , x = 0 . 5 , x = 1 , x = 1 . 5 , and x = 2 and the lines y = 0 , y = 1 , y = 2 , y = 3 , and y = 4 . Then we have 16 subrectangles, each of which we denote R ( a,b ) , where ( a, b ) is the location of the lower-left corner of the subrectangle.
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16.1 SOLUTIONS 1107 We want to find a lower bound and an upper bound for the volume above each subrectangle. The lower bound for the
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