Solução_Calculo_Stewart_6e

F.
232 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15 TX.10
59.
D (x2 tan x + y 3 +4)dA = D x2 tan xdA+ D y3 dA + 4 dA. D Butx2 tan x is an odd function of x and D is
symmetric with respect to the y-axis, so D x2 tan xdA=0. Similarly, y 3 isanoddfunctionofy and D is symmetric with
respect to the x-axis, so D y3 dA =0.Thus
D (x2 tan x + y 3 +4)dA =4 dA =4(areaofD) =4· π√ 2 2
D =8π
61. Since 1 − x 2 − y 2 ≥ 0, we can interpret
D 1 − x2 − y 2 dA as the volume of the solid that lies below the graph of
z = 1 − x 2 − y 2 and above the region D in the xy-plane. z = 1 − x 2 − y 2 is equivalent to x 2 + y 2 + z 2 =1, z ≥ 0
which meets the xy-plane in the circle x 2 + y 2 =1, the boundary of D. Thus, the solid is an upper hemisphere of radius 1
which has volume 1 4 π (1)3 = 2 π.
2 3 3
16.4 Double Integrals in Polar Coordinates ET 15.4
1. The region R is more easily described by polar coordinates: R =
(r, θ) | 0 ≤ r ≤ 4, 0 ≤ θ ≤ 3π 2 .
Thus f(x, y) dA = 3π/2 4
f(r cos θ, r sin θ) rdrdθ.
R 0 0
3. The region R is more easily described by rectangular coordinates: R = (x, y) | −1 ≤ x ≤ 1, 0 ≤ y ≤ 1 2 x + 1 2
.
Thus f(x, y) dA = 1
(x+1)/2
f(x, y) dy dx.
R −1 0
5. The integral 2π
7
π 4
rdrdθrepresents the area of the region
R = {(r, θ) | 4 ≤ r ≤ 7, π ≤ θ ≤ 2π},thelowerhalfofaring.
2π
7
rdrdθ= 2π
7
dθ rdr π 4
π
4
= θ 2π 1 r2 7
= π · 1
33π
(49 − 16) =
π 2 4 2 2
7. The disk D can be described in polar coordinates as D = {(r, θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}. Then
D xy dA = 2π
3
(r cos θ)(r sin θ) rdrdθ= 2π
3
sin θ cos θdθ
0 0 0 0 r3 dr = 1
2 sin2 θ 2π 1 r4 3
=0.
0 4 0
9.
R cos(x2 + y 2 ) dA = π
3
0 0 cos(r2 ) rdrdθ= π
dθ
3 r 0 0 cos(r2 ) dr
= θ π 1
0 2 sin(r2 ) 3
= π · 1 (sin 9 − sin 0) = π sin 9
0 2 2
11.
D e−x2 −y 2 dA = π/2
2
π/2
−π/2 0 e−r2 rdrdθ= dθ 2
−π/2 0 re−r2 dr
= θ π/2
−π/2
− 1 2 e−r2 2
= π
− 1 2 (e −4 − e 0 )= π (1 − 2 e−4 )
0
13. R is the region shown in the figure, and can be described
by R = {(r, θ) | 0 ≤ θ ≤ π/4, 1 ≤ r ≤ 2}. Thus
R arctan(y/x) dA = π/4
2
arctan(tan θ) rdrdθsince y/x =tanθ.
0 1
Also, arctan(tan θ) =θ for 0 ≤ θ ≤ π/4, so the integral becomes
π/4
2 θrdrdθ= π/4
θdθ 2
rdr= 1
θ2 π/4 1 r2 2
= π2 · 3 = 3
0 1 0 1 2 0 2 1 32 2 64 π2 .