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

250 0 CHAPTER 15 MULTIPLE INTEGRALS
13. J~ f 0
1f r sin 2 8 d8 dr = f~ r dr J; sin 2 8 d8 [as in Example 5] = J 0
2
r dr J 0
1f t (1 - cos 28) d8
= [ tr 2 )~· ~ [0- ~ sin28)~ = (2 - 0) · ~ [(1r- ~ sin 27r) - (0- ~sinO)]
= 2. ~[(7r- 0)- (0 - 0)] = 7r
. 15. ffn sin(x - y) dA = Io1f 12 Iorr~ 2 sin(x- y) dyd:r: = I; 12 [cos(x - y)]~~~l 2 dx = Iorrl 2 [cos(x - ~)- cosx] dx
17.ffn X:~ 1 dA = 11 /_33 X::
19. Io1riG j~1rl 3 x sin(x + y) dy dx
1f 12 .
= [ sin(x-~)-si?x ] 0
= sinO-sin~-,- (sin(-~) - sinO]
=0-1-(-1-0) = 0
1 dydx = 11 :r;2:
= ~ (In 2 -In 1) . t(27 + 27) = 91n 2
1 dx ;_: y2 dy = [t ln(x2 + 1)]: [h3] ~3
= IorriG [-xcos(x + y)]~=~ 13 dx = J; 16 [x cosx- xcos(x +f)] dx
= x (sin x - sin ( x + f) )~ 16 - J 0
1f
16 [sin x - sin ( x + f)] dx [by integrating by parts separately for each term]
= i [~- 1)- (-cosx + cos(x + f)]~ 16 = - -&- [ -~ + 0 - (-1 + ~)] = -q- 1 - f2
21. IInye-:z:il dA =I: I: ye-"' 11 dxdy = J 0
3
(-e -"' 11 J ::~ dy = I:(-e- 2 il + 1) dy = [~e- 2 Y + y]~
= ~ e- 6 + 3 - ( ~ + 0) = t~-6 + ~
23. z = f(x, y) = 4- x - 2y ~ 0 for 0 ~ x ~ 1 and 0 ~ y ~ 1. So the solid
is the region in the first octant which lies below the plane z = 4 - x -
and above [0, 1] x {0, 1].
2y
X
25. The solid lies under the plane 4x + 6y - 2z + 15 = 0 or z = 2x + 3y + 1/ so
V = fin(2x + 3y + 1 ;)'dA = I~ 1 I~ 1 (2x + 3y + Jf) dxdy = j~ 1 [x 2 + 3xy + ¥xJ::~ 1 dy
= j~ 1 ((19 + 6y) - ( - 1 23 - 3y)) dy = I~ 1 (¥' + 9y) dy = [Yf y + ¥Y 2 )~ 1 = 30- ( -21) = 51
27. V = I~ 2 f~ 1 (1 - tx 2 - h 2 ) dx dy = 4 J; I 0
1
(1 - ix 2 - ~y 2 ) dx dy
4 f2 [ 1 3 1 2 ] :z; = 1 d 4 f2 ( 11 1 2) d [ 11 l 3] 2 83 166
= Jo X- fiX - iiY X :z:=O y = Jo i2- iiY y = 4 i2Y - vY 0 = 4. 54= 27
29. Here we need the volume of the solid lying under the surface z = x sec 2 y and above the rectangle R = (0, 2] ~ [0, 1r / 4] in
the xy-plane.
V = I: Iorr 14 x sec 2 y dy dx = I: x dx Io" 14 sec 2 y dy = [ ~ x 2 ] ~ [tan y] ~ 14
= (2- O)(tan f - tan 0) = 2(1 - 0) = 2
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