Solução_Calculo_Stewart_6e

F.
248 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
16.7 Triple Integrals in Cylindrical Coordinates ET 15.7
1. (a)
(b)
x =2cos π 4 = √ 2, y =2sin π 4 = √ 2, z =1,
so the point is √ 2, √ 2, 1 in rectangular coordinates.
x =4cos − π 3
=2, y =4sin
−
π
3
= −2
√
3,
and z =5,sothepointis 2, −2 √ 3, 5 in rectangular
coordinates.
3. (a) r 2 = x 2 + y 2 =1 2 +(−1) 2 =2so r = √ 2; tan θ = y x = −1
1
= −1 and the point (1, −1) is in the fourth quadrant of
the xy-plane, so θ = 7π 4 +2nπ; z =4. Thus, one set of cylindrical coordinates is √ 2, 7π 4 , 4 .
(b) r 2 =(−1) 2 + − √ 3 2
=4so r =2; tan θ =
− √ 3
−1
= √ 3 and the point −1, − √ 3 is in the third quadrant of the
xy-plane, so θ = 4π 3 +2nπ; z =2. Thus, one set of cylindrical coordinates is 2, 4π 3 , 2 .
5. Since θ = π but r and z may vary, the surface is a vertical half-plane including the z-axis and intersecting the xy-plane in the
4
half-line y = x, x ≥ 0.
7. z =4− r 2 =4− (x 2 + y 2 ) or 4 − x 2 − y 2 , so the surface is a circular paraboloid with vertex (0, 0, 4), axis the z-axis, and
opening downward.
9. (a) x 2 + y 2 = r 2 , so the equation becomes z = r 2 .
(b) Substituting x 2 + y 2 = r 2 and y = r sin θ, the equation x 2 + y 2 =2y becomes r 2 =2r sin θ or r =2sinθ.
11. 0 ≤ r ≤ 2 and 0 ≤ z ≤ 1 describe a solid circular cylinder with
radius 2,axisthez-axis, and height 1,but−π/2 ≤ θ ≤ π/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-axisanditsbaseliesinthexy-plane. If we
use centimeters as the unit of measurement, then cylindrical coordinates conveniently describe the shell as 6 ≤ r ≤ 7,
0 ≤ θ ≤ 2π, 0 ≤ z ≤ 20.