Estimate the volume of the solid that lies below the surface z = xy and above th

Question

Estimate the volume of the solid that lies below the surface z = xy and above the following rectangle. R = {(x, y) | 2 x 8, 8 y 12} Use a Riemann sum with m = 3, n = 2, and take the sample point to be the upper right corner of each square. Use the Midpoint Rule to estimate the volume of the solid. Estimate the volume of the solid that lies below the surface z = 7x + 2y^2 and above the rectangle R = [0, 2] Times [0, 4]. Use a Riemann sum with m = n = 2 and choose t (b) sample points to be lower right corners. Use the Midpoint Rule to estimate the volume in part (a). If R = [-3, 1] Times [-1, 1], use a Riemann sum with m = 4, n = 2 to estimate the value of double Integral _R. (y^2 - 2x^2) dA. Take the sample points to be the upper left corners of the squares.

Explanation / Answer

Sol. (1)

x = (8 - 2)/3 = 2 and y = (12 - 8)/2 = 2.

So, our region R is partitioned as follows:
(2, 12) (4, 12) (6, 12) (8, 12) ,

(2, 10) (4, 10) (6, 10) (8, 10) ,

(2, 8) (4, 8) (6, 8) (8, 8)
---------------
(a) Using upper right endpoints for each of the little 'grids' yields (with f(x,y) = xy)
R xy dA
2 * 2 [f(2, 10) + f(4, 10) + f(6, 10) +f(8, 10) + f(2, 12) + f(4, 12) + f(6, 12) + f(8,12)]
= 4 [20 + 40 + 60 + 80 + 24 + 48 + 72 + 96]
= 1760.

(b) Using the midpoint for each of the little 'grids' yields
R xy dA
2 * 2 [f(3, 9) + f(5, 9) + f(7, 9) + f(3, 11) + f(5, 11) + f(7, 11)]
= 4 [27 + 45 + 63 + 33 + 55 + 77]
= 1200

Sol.(2)

With m = n = 2, we have x = (2 - 0)/2 = 1 and y = (4 - 0)/2 = 2.

we partition R = [0, 2] x [0, 4] into 4 squares with side lengths 2 with vertices

(0, 4).....(1, 4)....(2, 4)
...............................
(0, 2).....(1, 2)....(2, 2)
................................
(0, 0).....(1, 0)....(2, 0)

So, using lower right endpoints yields (with f(x,y) = 7x + 2y^2
f(x*, y*) x y
= 1 * 2 [f(1, 0) + f(2, 0) + f(1, 2) + f(2, 2)]
= 2 [7 + 14 + 15 + 22]
= 116
------------------
B) Using the midpoints of each square instead yields
1 * 2 [f(1/2, 1) + f(1/2, 3) + f(3/2, 1) + f(3/2, 3)]
= 2 [(7/2 + 2) + (7/2 + 18) + (21/2 + 2) + (21/2 + 18)]
= 136

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