Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

CHAPTER 15 REVIEW 0 289
15 Review
CONCEPT CHECK
m
n
1. (a) A double Riemann sum off is ,L ,L f (xi;, Yi;) b.A, where b. A is the area of each subrectangle and (xi;, Yi;) is a
i=lj = l
S3J!lple point in each subrectangle. If j(x, y) ;::: 0, this sum represents an approximation to the volume of the solid that lies
above the. rectangle R and below the graph off.
(b) ffn. f(x, y) dA = lim
m,n-+oo i = 1 j = 1
f f= f(xi;. Yi;) b.A
(c) If f(x, y) ;::: 0, ffn f(x, y) dA represents the volume of the solid that lies above the rectangle Rand below the surface
z = f(x, y). Iff takes on both positive and negative values, ffn j(x, y) dA is the difference of the volume above R but
below the surface z = f(x, y) and the volume below R but above the surface z = f(x, y).
(d) We usually evaluate ffn. j(x, y) dA as an iterated integral according to Fubini's Theorem (see Theorem 15.2.4).
(e) The Midpoint Rule for Double Integrals says that we approximate the double integral JJR f(x, y) dA by the double
Riemann sum f f: f(xi, fi;) b.A where the sample points (:x,, fi;) are the centers of the subrectangles.
i=lj= l
(f) fnvc = A ~R) I In f(x, y) dA where A (R) is the area of R.
2. (a) See ( I) and (2) and the accompanying discussion in Section 15.3.
(b) See (3) and the accompanying discussion in Section 15.3.
(c) See (5) and the preceding discussion in Section 15.3.
(d) See (6)- (1 I) in Section 15.3 .
3. We may want to change from re~tangular to polar coord inates in a double integral if the region R of integration is more easily
described in polar coordinates. To accomplish this, we use ffn f(x, y) dA = J: J: f (rcosB, rsin 9) r dr d(} where R is
given by 0 ::::; a ::::; r ::::; b, a ::::; B ::::; {3.
4. (a) m = JJD p(x, y) dA
(b) M , = ffv yp(x, y) dA, M 11 = ffv xp(x, y) dA
(c) The center of mass is (x, y) where x == M" and y= lvfx .
m m
(d) I, = ffv y 2 p(x, y) dA I v = ffv x 2 p(x, y) dA, Io = ffv (x 2 + y 2 )p(x, y) dA
5. (a) P(a ::::; X::::; b, c:=; Y::::; d)= 1: J: j(x, y)dydx
(b) f(x , y) ;::: 0 and JJ~~ f(x, y) dA = 1.
· (c) The expected value of X is JJ. 1 = ffR2 xf(x, y) dA; the expected value ofY is p..z = JJT:/.2 yf(x, y) dA.
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