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

252 D CHAPTER 15 ' MULTIPLE INTEGRALS
7. IIv y2 dA = I~t I~~ -2 Y2 dx dy = I~t (xy2J ::~v - 2 dy = I~l y2 [y - ( - y- 2)] dy
= J 1 (2y3 + 2y2)dy = [ly4 + 1y3J 1 = 1 + 1 - l + 1 = :!.
- 1 2 3 - 1 2 3 2 3 3
9. ffv x dA = f " ro•ln"' x dy dx = ~ " [xy]Y:'"oin "' dx = f" x sin x dx [ . integrate by parts ]
· .lo J t Jo Y- Jo wtth v.=x, rlv = sm x dx
= [ - x cos x + sin x J ~ = -1r cos 1r + sin 1r + 0 - sin 0 = 1r
11. (a) At the right we sketch an example of a region D that can be described as lying
y
betw~en the graphs of two continuous functions of X (a type I region) but not as
lying between graphs of two continuous functions ofy (a type IJ region). The
regions shown in Figures 6 and 8 in the text are additional examples.
(b) Now we sketch an example of a region D that can be described as lying between
the graphs of two continuous functions ofy but not as lying between graphs of two
continuous functions of x. The first region shown in Figure 7 is another example.
0
y
X
0
X
13.
y
(1, 1)
As a type I region, D lies between the lower boundary y = 0 and the upper
boundary y = x for 0 ::; x ::; 1, soD= {(x, y) I 0 $ x $ 1, 0 $ y $ x}. lfwe
x =l
describeD as a type ll region, D lies between the left boundary x = y and the
right boundary x = 1 for 0 ::; y ::; 1, soD = {(x, y) I 0 ::; y::; 1, y ::; x::; 1}.
X
1 1 2
Thus IIv xdA =I; IC: xdydx = I 0 [xy] ~ =~ dx = I 0 x dx = ! x 3 ]~ = !(1 - 0) = ~or
1 1
.rf'v x dA = f 0
I: xdxdy = I 0
(tx 2 J : : ~ dy = t J;(1 - y 2 ) dy = t[y - h 3 ] ~ = t((1-!) - 0] = ~ -
15.
y
The curves y = x - 2 or x = y + 2 and x = y 2 intersect when y + 2 = y 2 ~
y 2 - y- 2 =0 ~ (y-2)(y+ 1)=0 ~ y = -1, y =2,so thepoints.of
intersection arc (1, - 1) and (4, 2). Tf we de~c ribe D as a type I region, the upper
boundary curve is y = ..fii but the lower boundary curve consists of two parts,
y = -vx for 0 ::; x ::; 1 andy = x - 2 for 1 ::; x ::; 4.
Thus D = {(x, y) I 0::; X::; 1, - vx::; y::; vx} u {(x, y) 11 ::; X::; 4, X- 2::; y::; vx} and
I I 0
y dA = Io 1 I!5z y dy dx + f 1
4
f.,~ y dy dx. If we describe D as a type II region, D is enclosed by the left boundary
x = y 2 and the right boundary x = y + 2 for -1 ::; y ::; 2, so D = {( x, y) I - 1 ::; y ::; 2, y 2 ::; x ::; y + 2} and
. . .