Solução_Calculo_Stewart_6e

F.
TX.10
16 MULTIPLE INTEGRALS ET 15
16.1 Double Integrals over Rectangles ET 15.1
1. (a) The subrectangles are shown in the figure.
The surface is the graph of f(x, y) =xy and ∆A =4, so we estimate
V ≈
3 2
f(x i ,y j ) ∆A
(b) V ≈
i =1 j =1
= f(2, 2) ∆A + f(2, 4) ∆A + f(4, 2) ∆A + f(4, 4) ∆A + f(6, 2) ∆A + f(6, 4) ∆A
= 4(4) + 8(4) + 8(4) + 16(4) + 12(4) + 24(4) = 288
3
i =1 j =1
2
f
x i, y j ∆A = f(1, 1) ∆A + f(1, 3) ∆A + f(3, 1) ∆A + f(3, 3) ∆A + f(5, 1) ∆A + f(5, 3) ∆A
=1(4)+3(4)+3(4)+9(4)+5(4)+15(4)=144
3. (a) The subrectangles are shown in the figure. Since ∆A = π 2 /4, we estimate
R sin(x + y) dA ≈ 2 2
f x ∗ ij,yij ∗ ∆A
i=1 j=1
(b) R sin(x + y) dA ≈ 2
= f(0, 0) ∆A + f
0, π 2 ∆A + f π , 0 2
∆A + f π
, π
2 2 ∆A
π
=0
2 π
+1
2 π
+1
2 π
+0
2
= π2 ≈ 4.935
4
4
4
4 2
i=1 j=1
2
f(x i , y j ) ∆A
= f π
, π
4 4 ∆A + f π , 3π 4 4
=1
π 2
4
+0
π 2
4
+0
∆A + f
3π
4 , π 4
π 2
4
+(−1)
π 2
=0
4
∆A + f
3π
4 , 3π 4
∆A
5. (a) Each subrectangle and its midpoint are shown in the figure. The area of each
subrectangle is ∆A =2,soweevaluatef at each midpoint and estimate
R f(x, y) dA ≈ 2 2
f
x i , y j ∆A
i =1j =1
= f(1.5, 1) ∆A + f(1.5, 3) ∆A
+ f(2.5, 1) ∆A + f(2.5, 3) ∆A
=1(2)+(−8)(2) + 5(2) + (−1)(2) = −6
223