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

CHAPTER 15 REVIEW D 293
33. Using the wedge above the plane z = 0 and below the plane z = mx and noting that we have the same volume for .m < 0 as
for m > 0 (so use m > 0), we have
V = 2 foa/3 J/a2-9y2 mx.dxdy = 2 foal~ tm(a2 - 9y2) dy = m[a2y - 3y3)~/3 = m(ia3 - ia3) = ~ma3 .
35. (a) m = fo1 fo1 - v2 ydxdy = J~1 (y - y3) dy = ~- ~ = ~
rl rl- v 2 d d ri 1 ( 2)2 d 1 ( 1 2)3) 1 1
(b) Mv = Jo Jo xy x Y = .Jo 2Y 1 - Y Y = -u - 'y o = 12'
M, = Jo r l Jo r1 - v 2 y 2 d x d y = r1 c 2 4) d. 2 H c- -) ( 1 s )
10 y - y y = 15 . ence x, y = _ 3
, 15
.
r1 r1-v 2 3 J·1 ( 3 s) 1
(c)lx=JoJo y dxdy= 0 y - y dy=u,
r1 rl- v 2 2 d d _ r1 1 ( 2)3 d _ 1 ( 1 2)4] 1 _ 1
ly = Jo Jo yx x Y - Jo 3Y 1 - Y Y- -24 - Y o - 24'
.l _ I I _ 1 =2 _ .!.L_g _ 1 = _ 1 d = 2 _ 1/24 _ 1 = 1
o - "'+ v- 8• Y - 1/ 4 - 3 =? Y - 73• an x - 1/ 4 - 6 =? x = ?s·
37. (a) The equation of the cone with the suggested orientation is ( h - z) = ~ .j x 2 + y 2 , 0 $ z $ h. Then V = ~ 1ra 2 h is the
• !! 11•-(11/a),jx2+y2 12,.1a1(hfa)(a-r) 1a. h2 ;
M;r;y = z dz dA = rzdzdrd/J = 1r r 2 (a- r) dr
o ooo o· a
;r;2+y2 ~ 112 ·
volume of one frustum of a cone; by symmetry M 11 :: = !vfx:: = 0; and
1rh 2 1a 2 2 3 1rh 2 (a 4 2a 4 a 4 ) 1rh 2 a 2
= - (a r- 2ar + r ) dr = - - - - + - = --
a2 0 a 2 2 3 4 12
Hence the centroid is (x,y,z) = (0,0, ~h).
2"1"1(/•/.a)(a-r) 3 1" h 3 4 27rh (a5 a5) 7ra4 h
10 o o 0 a a 4 5 10
(b) I , = r. dz dr d() = 27r - (ar - r ) dr = - - - - = - -
39. Let D represent the given triangle; then D can be described as the area enclosed by the x- and y-axes and the line y = 2 - 2x,
or equivalently D = { (x, y) I 0 $ x $ 1; 0 $ y $ 2 - 2x }. We wantto find the surface area of the part of the graph of
z = x 2 + y that lies over D, so using Equation 15.6.3 we have
A(S)= Jfv 1+ (~;y + (~~Y dA = Jfv v 1+(2x)2+(1)2dA=
112
-
2
;r; )2 + 4x2dydx
= .r; .../2 + 4x 2 [y J~:~- 2 ;r; dx = J;(2 - 2x) .../2 + 4x 2 dx = · J 0
1
2 J2 + 4x2 dx- JC: 2x J2 + 4x2 dx
Using Formula 21 in the Table oflntegrals with a = J2. u. = 2x, and du = 2 dx, we have
J 2 J2 + 4x2 dx = x J2 + 4x 2 + ln(2x + J2 + 4x 2 ). lfwe substitute u = 2 + 4x 2 in the second integral, then
du = Bx dx and J 2x ,J2 + 4x 2 dx = ~ J .JU du = i · tu 3 1 2 = i (2 + 4x 2 ? 1 2 . Thus
. 1
A(S) = [xJ2 + 4x 2 + 1n(2x + J2 + 4x2) - t{2 + 4x 2 ) 3 1 2 Jo
= J6 + ln (2 + JB) - ~(6) 3 1 2 - lnJ2 + 4- = ln 2 t# + {1
= ln( J2 + v'3) + 4- ~ 1.6176
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