Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

248 0 CHAPTER 15 MULTIPLE INTEGRALS
(b) The subrectangles are shown in the figure.
In each subrectangle, the sample point closest to the origin
is the lower left comer, and the area of each subrectangle is .6.A = ~.
Thus we estimate
4 4
ffnf(x,y) dA ~ L:: L:: f(xii,Yii) .6.A
i= l j = l .
= f (O, 2) .6.A + f(O , 2.5) .6.A + f(O , 3) . .6.A + f(O, 3.5) .6.A
+ /(1, 2) .6.A + f(1, 2.5) .6.A + /(1, 3) .6.A + /(1, 3.5) .6.A
+ / (2, 2) .6.A + /(2, 2.5) .6.A + /(2, 3) .6.A + /(2, 3.5) .6.A
y
4
3
2
0 I 2
3 4 X
+ / (3, 2) .6.A + /(3, 2.5) .6.A + /(3, 3) .6.A + !(3, 3.5) .6.A
= - 3(~) + < -5) G) + < - 6) (~) + < -4)(~) + < - 1) (~) + < - 2){ ~) + < -3)(~) + < -1)G)
= - 8
+ 1G) + o(~) + ( -1 ){~) + 1 (~) + 2(~) + 2G) ·;~- 1 (~) + 3G)
7. The val~es off ( x, y) = .j52 - x 2 - y 2 get smaller as we move farther from the origin, so on any of the subrectangles in the
problem, the function will have its largest value at the lower left comer of the subrectangle and its smallest value at the upper
right comer, and any otl).er value will lie between these two. So using these subrectangles we have U < V < L . (Note that this
is true no matter how R is divided into subrectangles.)
9. (a) With m = n = 2, we h~ve .6.A = 4. Using the contolir map to estimate the value off at the center of each subrectangle,
we have
2 2
ffn f(x, y) dA ~ L:; L:; J(xi, Yi) .6.A = .6.A[f(1, 1) + /(1, 3) -t /(3, 1) + /(3, 3)] i'::j 4(27 + 4 + 14 + 17) = 248
i = lj=l
(b) /ave= Aln) ffn f(x, y) dA i'::j ft(248) = 15.5
11. z = 3 > 0; so we can interpret the integral as the volume of the solid S that lies below the plane z = 3 and above the
rectangle [- 2, 2] x [1, 6]. Sis a rectangular solid, thus ffn 3 dA = 4 · 5 · 3 = 60.
13. z = f(x, y) = 4- 2y :2: 0 for 0 :::; y :5 1. Thus the integral represents the volume of that
part ofthe rectangular.solid [0, 1] x [0, 1] x (0, 4] which lies below the plane z = 4 - 2y.
So
ffn(4 - 2y) dA = (1)(1)(2) + H 1)(1)(2) = 3
X
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