Section 15.1
Double Integrals over Rectangles
4. (a). The surface is the graph of f (x, y) = x + 2y2 and ∆A = 2, so we estimate V =∫ ∫
R(x + 2y2)dA≈∑2 i=1
∑2
j=1f (x∗ij, yij∗)∆A
= f (1, 0)∆A + f (1, 2)∆A + f (2, 0)∆A + f (2, 2)∆A
= 1(2) + 9(2) + 2(2) + 10(2) = 44.
(b). V =∫ ∫
R(x + 2y2)dA≈∑2
i=1
∑2
j=1f (xi, yj)∆A
= f (12, 1)∆A + f (12, 3)∆A + f (32, 1)∆A + f (32, 3)∆A
= 52(2) + 372(2) + 72(2) + 392 (2) = 88.
8. Divide R into 4 equal rectangles (squares) and identify the midpoint of each sub- rectangle as shown in the figure.
The area of each subrectangle is ∆A = 1, so using the contour map to estimate the function values at each midpoint, we have
∫ ∫
Rf (x, y)dA≈∑2 i=1
∑2
j=1f (xi, yj)∆A = f (12,12)∆A + f (12,32)∆A + f (32,12)∆A + f (32,32)∆A≈ (1.3)(1) + (3.3)(1) + (3.2)(1) + (5.2)(1) = 13.0.
You could improve the estimate by increasing m and n to use a larger number of smaller subrectangles.
10. As in Example 4, we place the origin at the southwest corner of the state. Then R = [0, 388]× [0, 276](in miles) is the rectangle corresponding to Colorado and we
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define f (x, y) to the temperature at the location (x, y). The average temperature is given by
fave = A(R)1 ∫ ∫
Rf (x, y)dA = 3881·276∫ ∫
Rf (x, y)dA.
To use the Midpoint Rule with m = n = 4, we divide R into 16 regions of equal size, as shown in the figure, with the center of each subrectangle indicated.
The area of each subrectangle is ∆A = 3884 · 2764 = 6693, so using the contour map to estimate the function values at each midpoint, we have
∫ ∫
Rf (x, y)dA≈∑4 i=1
∑4
j=1f (xi, yj)∆A
≈ ∆A[31+28+52+43+43+25+57+46+36+20+42+45+30+23+43+41] = 6639·605 Therefore, fave ≈ 6639388·276·605 ≈ 37.8, so the average temperature in Colorado at 4:00 PM on February 26, 2007, was approximately 37.8◦F .
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