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hw-15-sec-16.1-solns

# hw-15-sec-16.1-solns - S6 C/6 16 El MULTIPLE INTEGRALS D ET...

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Unformatted text preview: S6 C /6. ( 16 El MULTIPLE INTEGRALS D ET 15 16.1 Double Integrals over Rectangles ET 15.1 1. (a) The subrectangles are shown in the ﬁgure. The surface is the graph of 1"(177 y) = my and AA : 4, so we estimate 3 2 V” Z Z f(\$iayj)AA i:1j:1 = f(2, 2) AA + f(2, 4) AA + f(4, 2) AA + f(4, 4) AA + f(6, 2) AA + f(6, 4) AA : 4(4) + 8(4) + 8(4) + 16(4) + 12(4) + 24(4) 2 288 (b) V m i f Hajj) AA : f(1, 1) AA+ f(1,3) AA + f(3, 1) AA + f(3,3) AA + f(5, 1) AA + f(5,3) AA 1:1 ]:1 = 1(4) + 3(4) + 3(4) + 9(4) + 5(4) + 15(4) : 144 The subrectangles are shown in the ﬁgure. Since AA : 1, we estimate 4 2 11,. (y? — 242) 44 w z 21%,”) 4A : f(—1,1)AA + f(~1,2) AA + f(0, 1) AA + f(0,2) AA +f(1,1)AA+f(1,2)AA+f(2.1)AA+f(2,2)AA : w1(1) + 2(1) + 1(1) + 4(1) — 1(1) + 2(1) — 7(1) 4 4(1) : 44 ® (a) The subrectangles are shown in the ﬁgure. Since AA : 7r2 / 4, we estimate 2 2 ffRsin(m + y) dA z z 2 (\$23.31;?) AA 4:1 1:1 463 464 CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15 (E) (a) The subrectangles are shown in the ﬁgure. The surface is the graph of f (x, y) = a: + 2y2 and AA = 2, so we estimate V: ffR(93+2y2)dA~ Z 2 f(x xfj,y:j)AA i=1j= =ﬂLmAA+ﬂLmAA+ﬂzmAA+ﬂzmAA = 1(2) + 9(2) + 2(2) + 10(2) 2 44 (b) V = ffR(x+2y2) (1A R: :21 f(§.,yj)AA = f(§,1)AA+f(%,3)AA+ f(%71) AA+f(%»3) AA 3(2) + 3—272” %(2> + em) = 88 H p 5. (a) Each subrectangle and its midpoint are shown in the ﬁgure. The area of each subrectangle is AA = 2, so we evaluate f at each midpoint and estimate ffRf<m,y>dA m i i f(i?‘i,yj)AA i=1j=1 : f(1.5, 1) AA + f(1.5, 3) AA +f(2.52.1)AA+f(5 ,3)AA =1M2)+(—8X2)+5(2)<+(~1X2)::~6 (b) The subrectangles are shown in the ﬁgure. In each subrectangle, the sample point farthest from the origin is the upper right comer, and the area of each subrectangle is AA = %. Thus we estimate ffRf (my) dA~ Z Zf(cci,yj) AA i=1j=1 : f(1.5, 1) AA + f(1.5, 2) AA + f(1.5, 3) AA + f(1.5, 4) AA + f(2, 1) AA + f(2, 2) AA + f(2, 3) AA + f(2, 4) AA + f(2.5, 1) AA + f(2.5, 2) AA + f(2.5, 3) AA + f(2.5, 4) AA +f(3 1)A +f(3, 2) AA+f(3,3) AA+f(3,4) AA =1%—()+( 4>(%+) -8)-(%)+( 6>(% )+ 3%()+0(%)+(-5>(%)+(—8)(%) +5(%)+ 3(%)+( 1)%+() (-196 )+8(% )+6(% )+3(%)+0(%) = —3.5 SECTION 16.1 DOUBLE INTEGRALS OVER RECTANGLES ET SECTION 15.1 7 467 @z : 3 > 0, so we can interpret the integral as the volume of the solid S that lies below the plane 2 = 3 and above the rectangle [—2, 2] x [1, 6]. S is a rectangular solid, thus ij 3 dA : 4 - 5 ~ 3 = 60. ® 2 = 5 — :c 2 0 for 0 3 cc 3 5, so we can interpret the integral as the volume of the solid 5' that lies below the plane 2 2 5 — a: and above the rectangle [0, 5] X [0, 3]. S is a triangular cylinder whose volume is 3(area of triangle) : 3% - 5 - 5) : 37.5. Thus ffR(5 — at) (M : 37.5 13. z 2 f(x, y) : 4 — 2y 2 0 forO g y S 1. Thus the integral represents the volume ofthat part of the rectangular solid [0, 1] X [0, 1] X [0,4] which lies below the plane 2 = 4 — 2y. So ffR(4 - 23/) M = (1)(1)(2) + %(1)(1)(2) = 3 Here 2 = «9 - 3/2, so 22 + 11/2 = 9, z 2 0. Thus the integral represents the volume of the top half of the part of the circular cylinder 22 + 3,12 : 9 that lies above the rectangle [0,4] x [0,2]. 15. To calculate the estimates using a programmable calculator, we can use an algorithm similar to that of Exercise 5.1.7 [ET 5.1.7]. In Maple, we can deﬁne the function f(:c, y) = \/ 1 + ate‘y (calling it f), load the student package, and then use the 1141606 command 1.143191 middlesum(middlesum(f,x=0. .l,m) , 1.143535 y=0. .l,m); 1.143617 1.143637 1.143642 to get the estimate with n = m2 squares of equal size. Mathematica has no special Riemann sum command, but we can deﬁne f and then use nested Sum commands to calculate the estimates. 0.934591 64 0.860490 16. 4 0881991 256 0.858745 16 0.865750 1024 0.858157 ...
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