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

CHAPTER 15 REVIEW 0 291
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9. The volume enclosed by the cone z = ...jx2 + y2 and the plane z = 2 is, in cylindrical coordinates,
V = J~" J~ J: r dz dr dO i= f 0
2
" J~ J: dz dr dO, so the assertion is false.
EXERCISES
1. As shown in the_contour map, we divideR into 9 equally sized subsquares, each with area b.A = 1. Then we approximate
ffn f(x, y) dA by a Riemann sum with m = n = 3 and the sample points the upper right comers of each square, so
3 3
ffnf(x,y)dA>:::J L: L: f(xi, yj)b.A
i =lj=l
= b.A [!(1, 1) + /(1, 2) + /(1, 3) + !(2, 1) + !(2, 2) + f(2, 3) + !(3, 1) + !(3, 2) + !(3, 3)]
Using the contour lines to estimate the function values, we have
ffnf(x,y) dA >:::! 1(2.7 + 4.7 + 8.0 + 4.7 + 6.7 + 10.0 + 6.7 + 8.6 + 11 .9] ~ 64.0
3. f 1
2
J~ (y + 2xeY) dx dy = J? [xy + x 2 e 11 ] ::~ dy = f 1
2 (2y + 4eY) dy = [y 2 + 4eY] ~
= 4 + 4e 2 - 1 - 4e = 4e 2 - 4e + 3
7. J; J; J0~ ysinx.dzdy dx = f 0
" J; [(ysinx)zJ::F dydx·= f 0
" f 0
1
y ...jl- y 2 sinxdydx
= f 0
" [- ~ (1- y 2 ) 3 1 2 sinx)·u:l dx = f 0
" ~ sin x dx = -~ cosx]~ = ~
. . y~
9. The regi?n R is more easily d~sc ribed by polar coordinates: R = { ( r, 0) I 2 s; r s; 4, 0 s; 8. s; 71"}. Thus
ffn f(x, y) dA = fo" f 2
4
f (rcosO, r sinO) rdrdO.
11.
Th . h . . b fTr / 2 r•in 20 d dO .
e reg1on w ose area IS g.tven y Jo Jo r r IS
· { ( r, 0) I 0 s; 0 s; %, 0 s; r s; sill 20}, which is th~ region contained in the
loop in the first quadrant of the four-leaved rose r =sin 28 .
13.
./~ J: 1 cos(y 2 ) dy dx = J; g cos(y 2 ) dx dy
= f 0
1
cos(y 2 ) [ x] ::~ dy = J 0
1
y cos(y 2 ) dy
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