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

15 D MULTIPLE INTEGRALS
15.1 Double Integrals over Rectangles
1. (a) The subrectangles are shown in the figure.
The surface is the graph of f(x, y) = xy and 6.A = 4, so we estimate
3 2
v~ I: I: f(x;,y 1 )6.A
i =l j=l
= j(2, 2) 6.A + !(2, 4) 6.A + f(4, 2) 6.A + /(4,4) 6.A + !(6, 2) LU + !(6, 4) 6.A
= 4(4) + 8(4) + 8(4) + 16(4) + 12(4) + 24(4) = 288
y
4
2
0 2 4 6 X
3 2
(b) V ~ ,L: ,L: j (x.:, v 1 ) 6.A = /(1, 1) 6.A + f(1, 3) 6.A + !(3, 1) 6.A + !(3, 3) 6.A + !(5, 1) LU + !(5, 3) 6.A
i=lj=l
= 1(4) + 3(4) + 3(4) + 9(4) + 5(4) + 15(4) = 144
3. (a) The subrectangles are shown in the figure. Since 6.A = 1 · ~ = ~ .we estimate
2 2
J fn. xe-"' 11 dA ~ I: I: f ( x;1, Yi;) 6.A
i=l j = l
= /(1, ~) 6.A + /(1, 1) 6.A + /(2, ~) 6.A + /(2, 1) 6.A
= e - 1 1 2 ( ~ ) +e - 1 (~) +2e - 1 (~) + 2e - 2 ( ~) ~ 0 .990
. 2 2
(b) ffn xe-"' 11 dA ~ I: I: f(x;, v 1 ) 6.A
. i=l j =l
= !(~, t) 6.A + !(~ , ~) 6.A + f(~, t) 6.A + ! (~. ~) 6.A
= ~e- l /8a) + ~e - 318(~) + ~e-3 18(~) + ~e - 9/ sa) ~ u 51
y
I
I
2
0 1 2 X
y
1
. .
I
2
. .
0 1 2 X
5. (a) Each subrectangle and its midpoint are shown in the figure.
The area of each subrectangle is 6.A = 2, so we evaluate f
at each midpoint and estimate
2 2
.ffn.f(x,y)dA ~ I: I: f(x.:,y 1
) 6.A
i = lj = l
= /(1, 2.5) 6.A + /(1, 3.5) AA
+ /(3, 2.5) AA + /(3, 3.5) AA
= - 2(2) + ( -1)(2) + 2(2) + 3(2) = 4
y
4
. 3
. .
2
0 2 4 X
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