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

49. Since u = x- y and v = x + y, x = t(u + v) andy= ~(v- u).
dv
8(u,v) -1/2 ·1/2 2 Rx+y 2 _ 2 v 2 , 2 v ·
8(x y) I 1/2 1/21 1 fl x- y 14
/ .o u (1) £4
Thus --'- = = - and --dA = - - du dv = - - = -In 2
CHAPTER 15 REVIEW 0 295
51 . Let U = y - X and V = y +X SO X = y :... U = ( V - X) - U =:> X = ~ ( V - U) and y = V - ~ ( V - U) = ~ ( V + u).
I~~::~~ I = I ~:~ -. ~~~~I = 1-~ ( ~) - ~(~)I = J -'~I = ~. · R is the image under this transformation .of the square
with vertices (u, v) = (0, 0), ( -2, 0), (0, 2), and ( -2, 2). So
This result could have been anticipated by symmetry, since the integrand is an odd function of y and R is symmetric about
the x-axis.
53. For each r such that Dr lies within the domain, A(Dr) = 71'1' 2 , and by the Mean Value Theorem for Double Integrals there
exists (xr, Yr) in D,. such that f (xr, Yr) = ~ j' { f(x, y) dA. But lim (xr, Yr ) = (a, b),
7TT lv,. r-o+
so lim ~ j' { f (x, y) dA = lim f(xr, y,.) = f(a, b) by the continuity of f.
r-o+ 1fT } Dr r--+O+
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