Solução_Calculo_Stewart_6e

F.
260 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
1. Asshowninthecontourmap,wedivideR into 9 equally sized subsquares, each with area ∆A =1. Then we approximate
f(x, y) dA by a Riemann sum with m = n =3and the sample points the upper right corners of each square, so
3.
5.
R
R f(x, y) dA ≈ 3
i =1j =1
3
f(x i,y j) ∆A
= ∆A [f(1, 1) + f(1, 2) + f(1, 3) + f(2, 1) + f(2, 2) + f(2, 3) + f(3, 1) + f(3, 2) + f(3, 3)]
Using the contour lines to estimate the function values, we have
f(x, y) dA ≈ 1[2.7+4.7+8.0+4.7+6.7+10.0+6.7+8.6+11.9] ≈ 64.0
2
1
1
0
R
2 (y 0 +2xey ) dx dy = 2
1 xy + x 2 e y x=2
dy = 2
(2y x=0 1 +4ey ) dy = y 2 +4e y 2
1
=4+4e 2 − 1 − 4e =4e 2 − 4e +3
x
0 cos(x2 ) dy dx = 1
0 cos(x 2 )y y=x
dx = 1
x y=0 0 cos(x2 ) dx = 1 2 sin(x2 ) 1
= 1 sin 1
0 2
7.
π
0
1
0
√ 1−y 2
0
y sin xdzdydx = π
0
= π
0
1
z=
√1−y
0 (y sin x)z 2
dy dx = π
1 y 1 − y
z=0
0 0 2 sin xdydx
y=1
− 1 (1 − 3 y2 ) 3/2 sin x dx = π 1
sin xdx= − 1 cos x π
= 2
0 3 3 0 3
y=0
9. The region R is more easily described by polar coordinates: R = {(r, θ) | 2 ≤ r ≤ 4, 0 ≤ θ ≤ π}. Thus
R f(x, y) dA = π 4
f(r cos θ, r sin θ) rdrdθ.
0 2
11. The region whose area is given by π/2
sin 2θ
rdrdθis
0 0
(r, θ) | 0 ≤ θ ≤
π
, 0 ≤ r ≤ sin 2θ , which is the region contained in the
2
loop in the first quadrant of the four-leaved rose r =sin2θ.
13.
1
0
1
x cos(y2 ) dy dx = 1
y
0 0 cos(y2 ) dx dy
= 1
0 cos(y2 ) x x=y
x=0 dy = 1
0 y cos(y2 ) dy
= 1
2 sin(y2 ) 1
0 = 1 2 sin 1
15.
R yexy dA = 3
0
2
0 yexy dx dy = 3
0 e
xy x=2
dy = 3
x=0 0 (e2y − 1) dy = 1
2 e2y − y 3
= 1 0 2 e6 − 3 − 1 = 1 2 2 e6 − 7 2