Solução_Calculo_Stewart_6e

F.
166 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
(c) Because the range of g(x, y) =3xy is R, and the range of e x is (0, ∞), the range of e g(x,y) = e 3xy is (0, ∞).
The range of x 2 is [0, ∞), so the range of the product x 2 e 3xy is [0, ∞).
√
6−2
9. (a) f(2, −1, 6) = e
2 −(−1) 2 = e √1 = e.
√
z−x
(b) e
2 −y 2 is defined when z − x 2 − y 2 ≥ 0 ⇒ z ≥ x 2 + y 2 . Thus the domain of f is (x, y, z) | z ≥ x 2 + y 2 .
(c) Since √
z − x 2 − y 2 z−x
≥ 0,wehavee
2 −y 2 ≥ 1. Thus the range of f is [1, ∞).
√
11. x + y is defined only when x + y ≥ 0,ory ≥−x. So
the domain of f is {(x, y) | y ≥−x}.
13. ln(9 − x 2 − 9y 2 ) is definedonlywhen
9 − x 2 − 9y 2 > 0,or 1 9 x2 + y 2 < 1. So the domain of f
is (x, y) 1
9 x2 + y 2 < 1 , the interior of an ellipse.
√
15. 1 − x2 is defined only when 1 − x 2 ≥ 0,orx 2 ≤ 1
17. y − x 2 is defined only when y − x 2 ≥ 0,ory ≥ x 2 .
⇔ −1 ≤ x ≤ 1,and 1 − y 2 is defined only when
1 − y 2 ≥ 0,ory 2 ≤ 1 ⇔ −1 ≤ y ≤ 1. Thusthe
domain of f is {(x, y) | −1 ≤ x ≤ 1, − 1 ≤ y ≤ 1}.
In addition, f is not defined if 1 − x 2 =0
x = ±1. Thus the domain of f is
(x, y) | y ≥ x 2 ,x6=±1 .
⇒
19. We need 1 − x 2 − y 2 − z 2 ≥ 0 or x 2 + y 2 + z 2 ≤ 1,
21. z =3, a horizontal plane through the point (0, 0, 3).
so D = (x, y, z) | x 2 + y 2 + z 2 ≤ 1 (the points inside
or on the sphere of radius 1, center the origin).