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Solução_Calculo_Stewart_6e

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F.<br />

TX.10<br />

SECTION 16.6 TRIPLE INTEGRALS ET SECTION 15.6 ¤ 247<br />

47. (a) m = 1<br />

0<br />

√ 1−x 2<br />

0<br />

y<br />

0<br />

<br />

(b) (x, y, z) = m −1 1 √ 1−x 2<br />

0 0<br />

(c) I z =<br />

1<br />

0<br />

3π<br />

(1 + x + y + z) dz dy dx = + 11<br />

32 24<br />

m −1 1<br />

0<br />

y<br />

0<br />

√ 1−x 2<br />

0<br />

x(1 + x + y + z) dz dy dx,<br />

y<br />

0<br />

y(1 + x + y + z) dz dy dx,<br />

m −1 1 √ <br />

1−x 2 y z(1 + x + y + z) dz dy dx 0 0 0<br />

<br />

28 30π +128 45π +208<br />

= , ,<br />

9π +44 45π +220 135π + 660<br />

√ 1−x 2<br />

0<br />

y<br />

0<br />

(x 2 + y 2 )(1 + x + y + z) dz dy dx =<br />

68 + 15π<br />

240<br />

49. (a) f(x, y, z) is a joint density function, so we know R 3 f(x, y, z) dV =1.Herewehave<br />

R 3 f(x, y, z) dV = ∞<br />

−∞<br />

∞<br />

−∞<br />

Then we must have 8C =1 ⇒ C = 1 8 .<br />

(b) P (X ≤ 1,Y ≤ 1,Z ≤ 1) = 1<br />

−∞<br />

1<br />

−∞<br />

= 1 8<br />

∞<br />

−∞ f(x, y, z) dz dy dx = 2<br />

= C 2<br />

0 xdx 2<br />

0 ydy 2<br />

0 zdz= C 1<br />

2 x2 2<br />

0<br />

1 f(x, y, z) dz dy dx = 1<br />

1<br />

1<br />

−∞ 0 0 0<br />

1<br />

0 xdx 1<br />

0 ydy 1<br />

0 zdz= 1 8<br />

1<br />

2 x2 1<br />

0<br />

1<br />

2 y2 1<br />

0<br />

2<br />

2<br />

0 0 0<br />

1<br />

2 y2 2<br />

0<br />

1<br />

xyz dz dy dx<br />

8<br />

Cxyz dz dy dx<br />

1 z2 2<br />

=8C<br />

2 0<br />

1<br />

2 z2 1<br />

0 = 1 8<br />

1<br />

2<br />

3<br />

= 1 64<br />

(c) P (X + Y + Z ≤ 1) = P ((X, Y, Z) ∈ E) where E is the solid region in the first octant bounded by the coordinate planes<br />

and the plane x + y + z =1. The plane x + y + z =1meets the xy-plane in the line x + y =1,sowehave<br />

1−x<br />

1−x−y<br />

P (X + Y + Z ≤ 1) = f(x, y, z) dV = 1<br />

1<br />

E 0 0 0<br />

<br />

= 1 1<br />

1−x<br />

8<br />

xy 1<br />

z2 z=1−x−y<br />

0 0 2<br />

dy dx = 1 z=0 16<br />

51. V (E) =L 3 ⇒ f ave = 1 L 3 L<br />

0<br />

<br />

= 1 1<br />

1−x<br />

16 0<br />

<br />

= 1 1<br />

16 0<br />

8 xyzdzdydx<br />

1<br />

0<br />

1−x<br />

0<br />

xy(1 − x − y) 2 dy dx<br />

[(x 3 − 2x 2 + x)y +(2x 2 − 2x)y 2 + xy 3 ] dy dx<br />

0<br />

(x 3 − 2x 2 + x) 1 2 y2 +(2x 2 − 2x) 1 3 y3 + x 1<br />

y4 y=1−x<br />

dx<br />

4 y=0<br />

<br />

= 1 1 (x − <br />

192 0 4x2 +6x 3 − 4x 4 + x 5 ) dx = 1 1<br />

<br />

192 30 =<br />

1<br />

5760<br />

L<br />

L<br />

0<br />

0<br />

xyz dx dy dz = 1 L<br />

xdx<br />

L 3<br />

0<br />

L<br />

0<br />

ydy<br />

L<br />

0<br />

zdz<br />

= 1 L 3 x<br />

2<br />

2<br />

L<br />

0<br />

y<br />

2<br />

2<br />

L<br />

0<br />

z<br />

2<br />

2<br />

L<br />

0<br />

= 1 L 3 L 2<br />

2<br />

L 2<br />

2<br />

L 2<br />

2 = L3<br />

8<br />

53. The triple integral will attain its maximum when the integrand 1 − x 2 − 2y 2 − 3z 2 is positive in the region E and negative<br />

everywhere else. For if E contains some region F where the integrand is negative, the integral could be increased by excluding<br />

F from E,andifE fails to contain some part G of the region where the integrand is positive, the integral could be increased<br />

by including G in E. Sowerequirethatx 2 +2y 2 +3z 2 ≤ 1. This describes the region bounded by the ellipsoid<br />

x 2 +2y 2 +3z 2 =1.

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