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

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(b) The w<strong>ed</strong>ge in question is the shad<strong>ed</strong> area rotat<strong>ed</strong> from B = 81 to B = 82.<br />

Letting<br />

V.i = <strong>vol</strong>ume of the region bound<strong>ed</strong> by the sphere of radius P;<br />

and the cone with angle ¢; (B = 81 to 82)<br />

SECTION 15.1 0 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS 0 285<br />

and letting V be the <strong>vol</strong>ume of the w<strong>ed</strong>ge, we have<br />

y<br />

V = (1122- 1l21)- (V12 - Vn)<br />

· = 1(82 - B1)[pg(1- cos¢ 2 ) - p~(l - cos¢ 1 )- p~(1- cos¢ 2 ) + p~(1- cos¢ 1<br />

) ]<br />

= 1(82- 81) [(P~- p~)(1- cos ¢ 2 ) - (P~- pi) (1- cos ¢1)] = i(B2 - (Jt)[(p~- pi) (cos ¢ 1 - cos¢ 2 )]<br />

Or: Show that V =<br />

1<br />

021 P2 s in 21 r cot ¢ 1<br />

01 p 1 s in 4> 1 r cot 4> 2<br />

r dz dr dB.<br />

(c) By the Mean Value Theorem with f(p) = p 3 there exists some p with p 1 ~ p ~ p 2 such that<br />

f(p 2 ) - f(p 1 ) = !'(p)(p 2 - p 1 ) or p~ - p~ = 3Ji D.p. Similarly there exists¢ with ¢1 ~ ~ ~ ¢ 2<br />

such that cos ¢ 2 - cos ¢ 1 ·= (-sin¢) D.¢. Substituting into the result from (b) gives<br />

D. V = (p 2 D.p)(B2- B1)(sin¢) D.¢ = p 2 sin¢> D.pD.¢ D.B.<br />

15.10 Change of Variables in Multiple Integrals<br />

1. x = 5u- v, y = u + 3v.<br />

TheJacobianis<br />

a(x<br />

~( ,y)<br />

1)<br />

=<br />

l8xf8u 8xf8v I<br />

= =5(3) -(-1)(1) =16.<br />

v u,v 8yf8u oyfav 1 3<br />

1<br />

3. x = e-r sinB, y :;= er cos B.<br />

5 -ll<br />

8(x,y) 18xf8r 8xj8fJ I ~ -e - ''sinB e-rcosBI · ·<br />

-- - - r . = e-•·er sin 2 B- e-re'· cos 2 B = sin 2 fJ- cos 2 ()or- cos 2()<br />

8(r, B) - ayj8r ayjaB - er cos() -e smfJ . .<br />

5. x = ufv, y = vfw, z = wfu.<br />

a(x,y,z) -<br />

a(u,v,w)-<br />

oxfou 8xfav 8x/8w 1/ v - u/v 2 0<br />

ayfau ayf8v 8yf8w 0 1/w -vjw 2<br />

azjou 8z/8v 8z/8w - wfu2 0 1/u<br />

= .!_11/w - vjw 2 ! ( u) I 0 -vjw 2 ! I 0 1/w I<br />

v 0 1/u - - V 2 .:..wfu 2 1/ u + 0 -wfu 2 0<br />

=.!. (...!_ - o) + ~ (o- ~) + o = - 1 - - - 1 - = o<br />

v uw v 2 u 2 w uvw uvw<br />

1. The transformation maps the boundary of S to the boundary of the imageR, so we first look at side 8 1 in the uv-plane. S 1 is<br />

describ<strong>ed</strong> by v = 0, 0.::; u.::; 3, sox = 2u + 3v = 2u andy = u- v = u. Eliminating u, we have.x = 2y, 0 _::; x ~ 6. 8 2 is<br />

@ 20 12 Ccng:~ge Learning. All Right5 Resen-cd. MBy not be sccum<strong>ed</strong>. copi<strong>ed</strong>. or duplicat<strong>ed</strong>. or post<strong>ed</strong> 10 a publicly acccssiblo wcb~ itc, in whole or in prut.

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