Solução_Calculo_Stewart_6e

F.
258 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
16 Review ET 15
1. (a) A double Riemann sum of f is m
i =1j =1
n
f
x ∗ ij,yij ∗ ∆A,where∆A is the area of each subrectangle and x
∗
ij ,yij ∗ is a
sample point in each subrectangle. If f(x, y) ≥ 0, this sum represents an approximation to the volume of the solid that lies
above the rectangle R and below the graph of f.
(b) f(x, y) dA = lim
R
m
m,n→∞ i =1j =1
n
f x ∗ ij,yij ∗ ∆A
(c) If f(x, y) ≥ 0, f(x, y) dA represents the volume of the solid that lies above the rectangle R and below the surface
R
z = f(x, y). Iff takes on both positive and negative values, f(x, y) dA is the difference of the volume above R but
R
below the surface z = f(x, y) and the volume below R but above the surface z = f(x, y).
(d) We usually evaluate f(x, y) dA as an iterated integral according to Fubini’s Theorem (see Theorem 16.2.4
R
[ET 15.2.4]).
(e) The Midpoint Rule for Double Integrals says that we approximate the double integral f(x, y) dA by the double
R
Riemann sum m
i =1j =1
n
f
x i , y j ∆A where the sample points xi , y j are the centers of the subrectangles.
(f ) f ave = 1
f(x, y) dA where A (R) is the area of R.
A (R) R
2. (a) See (1) and (2) and the accompanying discussion in Section 16.3 [ET 15.3].
(b) See (3) and the accompanying discussion in Section 16.3 [ET 15.3].
(c) See (5) and the preceding discussion in Section 16.3 [ET 15.3].
(d) See (6)–(11) in Section 16.3 [ET 15.3].
3. We may want to change from rectangular to polar coordinates in a double integral if the region R of integration is more easily
described in polar coordinates. To accomplish this, we use R f(x, y) dA = β
α
given by 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β.
4. (a) m = ρ(x, y) dA
D
(b) M x = yρ(x, y) dA, M D y = xρ(x, y) dA
D
b
a
f(r cos θ, r sin θ) rdrdθwhere R is
(c) The center of mass is (x, y) where x = M y
m and y = M x
m .
(d) I x = D y2 ρ(x, y) dA, I y = D x2 ρ(x, y) dA, I 0 = D (x2 + y 2 )ρ(x, y) dA
5. (a) P (a ≤ X ≤ b, c ≤ Y ≤ d) = b
a
(b) f(x, y) ≥ 0 and R 2 f(x, y) dA =1.
d
f(x, y) dy dx
c
(c) The expected value of X is μ 1 = R 2 xf(x, y) dA; the expected value of Y is μ 2 = R 2 yf(x, y) dA.