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

SECTION 15.8 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES 0 277
3. (a) From Equations 2 we have r 2 = ( - 1) 2 + 1 2 = 2 so r = v'2; tan{;/= _: 1 = -1 and the point ( -1, 1) is in the second
quadrant of the xy-plane, so 8 = s; + 2mr; z = 1. Thus, one set of cylindrical coordinates is ( v'2, 3 4 '~~' , 1).
(b) r 2 = ( -2) 2 + (2J3) 2 = i6 so r = 4; tanB = ~ = - v'3 and the point ( - 2, 2v'3) is in the second quadrant of the
xy-plane, so (} = 2 ; + 2mr; z = 3. Thus, one set of cylindrical coordinates is ( 4, 2 ; , 3).
5. Since (;I = f but r and z may vary, the surface is a vertical half-plane i~clud ing the z -axis and intersecting the xy-plane in the
half-line y = x, x ~ 0.
1. z = 4- r 2 = 4 - (x 2 + y 2 ) or 4 - x 2 - y 2 , SQ the surface is a circular paraboloid with vertex (0, 0, 4), axis the z-axis, and
opening downward.
9. (a) Substituting x 2 + y 2 = r 2 and x = r cos 8, the equation x 2 - x + y 2 + z 2 = 1 becomes r 2 - r cos 8 + z 2 = 1 or
z 2 = 1 + r cos 8 - r 2 .
(b) Substituting x = r cosO andy = r sinO, the equation z = x 2 -
y 2 becomes
z = (rcos8) 2 - (rsinB? = r 2 (cos 2 8- sin 2 B) or z = r 2 cos 28. 1
11.
0 ~ r ~ 2 and 0 ~ :z ~ 1 describe a solid circular cylinder with
radius 2, axis the z -axis, and height 1, but - 11' /2 ~ 8 ~ 11' /2 restricts
the solid to the first and fourth quadrants of the xy-plane, so we have
a half-cylinder.
13. We can position the cylindrical shell vertically so that its axis coincides with the z-axis and its base lies in the xy-plane. If we
use centimeters as the unit of measurement, then cylindrical coordinates conveniently describe the shell as 6 ~ r ~ 7,
o ~ o ~ 211', o .~ z ~ 20.
15.
The region of integration is given in cylindrical coordinates by
E = {(r,B,z) 1- 11'/2 ~ 8 ~ 11'/2, 0 ~ r ~ 2, 0 ~ z ~ r 2 } . This
represents the solid region above quadrants I and IV of the xy-plane enclosed
by the circular cylinder r = 2, bounded above by the circular paraboloid
z = r 2 (z = x 2 + y 2 ), and bounded below by the xy-plane (z = 0).
I
T<
-71'/ 2 0 Jo rdz r = - -rr/2 Jo rz z=O r = -71'/ 2 Jo r drd
/ 2 j '2 rr
2
d d(} I" / 2 f 2 [ ] z = r
2 d dB I 'll' / 2 f 2 3 (}
= I 1f 12 dB f 2 r 3 dr = toJr. 12 (lr 4 ] 2
c-'lf/2 Jo - 71'/2 4 o
= 11'(4- 0) = 411'
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