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

D PROBLEMS PLUS
1.
y
5
4
3
2
x + y = 5
Let R, = u; = 1 Ri, where
R , = { (x, y) I x+ y ~ i + 2, x +y < i+3, 1 ~ x ~ 3, 2 ~ y ~ 5}.
5 5
ffn[x+y) dA = I: f fn .[x+y)dA= L: [x+ylffn . dA, since
i =1 1. i=l '
[x + y] = cons tant = ~+ 2 for (x, y) E R;. Therefore
0
2 3 X
I rb I t [ t ?
]
3. f,, .• = b _ala f (x)dx = I- 0
Jo J. cos(t-)dt dx
Jfn[x + y) dA = E?= 1 (i + 2) [A(R;)]
= 3A{RI) + 4A(R2) + 5A(R3) + 6A(&) + 7 A(Rs)
= 3(4) + 4(~) + 5{2) + 6(~) + 7 ( ~) = 30
= J; J: cos(t 2 ) dt dx = ,{ 0
1 J; cos(t 2 ) dx dt [changing the ordeqlf intcsrntion]
x = t
0
5. Since lxvl < 1, except at (1, 1), the formula for the sum of a geometric series gives -I- 1 - = f (xy)", so
. - ~ n~
00 00
= "'
L.., n+1
1
' n+1
1 = "'
L.., (n+
1
1)2 = 1!! 1 + 21
1
+ 32"
1
+ · · · =
"'""
L..,n = 1 ~ 1
n=O
n= O
•.
7. (a) Since ixyz i < 1 except at (I , I , I), the formula for the sum of a geometric se;ies gives - - 1 - = E (xyz )n, so
1- X 1JZ n=O
1
11 11 1 11lll100 00 111·1
- _--d:r;dydz= I: (xyztdxdy dz = I: (xyztdxdydz
o o o 1 xyz o . o o n =O n = O 0 o 0
00
00
= I:
[.f1 ] [ 1 ] [ f 1 ] 1 1 1
0
x" dx ,[ 0
y"' dy Jo z" dz = I: - - · - - · - - 1 1 1
n = O . n =O n + n + n +
00
1 1 1 1 .
00
1
= ... ~o (n + 1)3 = 13 + 23 + 33 + ... = n~1 n 3
(b) Since 1-xyzl < 1, except at (1, 1, I ), the formula for the sum of a geometric series gives
1 +xyz n = O
1
111 1 11·1·1 00 00 11.1111
1
1
= f= (- x yz)" , so
dx dy dz = I: ( -xyz)" dx dy dz = 2: ( -xyz)" dx dy dz ·
o o o + xyz o o o n=O n=O o o o
= f (-1)". (1~ 1 x" dx] [J; y"dy] [J; z" dz] = f (-1)"- 1 -- -
1 - . -
1 -
n=o n=O n + 1 n + 1 n + 1
00
(-1)" 1 1 1
00
(-1)" - 1
= n~o (n + 1)3 = 13 - 23 + :P - . . . = n~o n 3
[continued]
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