Question

(1 point) Consider the solid that lies above the rectangle (in the xy-plane) R = [-2, 2] x [0, 2], and below the surface z = x2 - 7y + 14. (A) Estimate the volume by dividing Rinto 4 rectangles of equal size, each twice as wide as high, and choosing the sample points to result in the largest possible Riemann sum. Riemann sum = (B) Estimate the volume by dividing Rinto 4 rectangles of equal size, each twice as wide as high, and choosing the sample points to result in the smallest possible Riemann sum. Riemann sum = (C) Using iterated integrals, compute the exact value of the volume. Volume =

Answer 1

Sluding twice as Consider the solid lies above, the rectangle Cin the xy-plane) R = [-2.2] * [0, 2], and below the surface = x²-74 +14 Ginen R = [-2,2] [0, 2] here -2<xse osysa and sur urface 25 = x²-74 + 14 recteingles of wide Chigh. AY (012) , (2,2) (0,1) {(2,1) 157 god (-2,0) Coro) (2,0) X area at each point of 2= The each rectangle is 2. for the dogest possibed Riemann sum we find the nature of a the 1st rectangle fex,y) = x²_gy +34 at (ooo f(0,0) = 07-7(0) +14 = 14 at co,1) fo, 1) 02-7(1) +14 = 7 at c- 2,1) +(-2,1) - 4 - 7+ 14 at (-2,0) +1-2,0) 40+24 Now, nolume a { flo,o) + flo, 2) + f(-2,0) + f(-2, 1)) (1 4+ 7 + 10 + 1 1)2 =100 11 18

as high США find the value of (B) estimate noleme by buiding R Pato e rectangle equal size each twice as wide. have to find the smallest possible Réemann sum, so we have rectangle, now, at each point of the 4th rectangle 2 = fex,y) 1x ²-74 414 at (0,1) je 0:2) 0²_7(1) + 14 = 7 at ( 2,1) |(2,1). 4 7(1) + 14 = 11 at (2, 2) $12,2) 7(2) + 24 at (0,2) fro, 2) = o - 7(2) + 14 = 0 4 xow, volume a {f(0,1)+ f(2,3)+(2, 2) + f(0.2)} DA (7 + 11 + 4 to 1 * (2) 44 the (c) Exact valume of region VE SI fex,y) dA ff (x²-7y+19) dx dy R 2 2 2 0 2 2 142 २ erig_1444.28]dy : 69 [92 y. 144² 2 ya N

nolume - 2. 192(2) - 14 [92622–1923 184 Re nolume क 3 nolume a 2 (100) nolume = 66.66