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

61 . Since m :S j(x, y) :S M, ffD mdA ~ ffD j(x, y) dA ~ ffD M dA by (8) =>
SECTION 15.3 DOUBLE INTEGRALS OVER GENERAL REGIONS 0 257
mjf 0
ldA ~ fj~ j(x,y)dA ~ lVI .ffD ldA by(?) => mA(D) ~ ffDJ(x,y)dA ~ MA(D) by (10).
63.
y
- 3 X
First we' can write JJ 0 (x + 2) dA = JJD xdA + JJ 0
2dA. But f(x, y) =xis
an odd function with respect to x [that is, f( -x, y) = - f(x, y)] and Dis
symmetric with respect to x. Consequently, the volume above D and below the
graph off is the same as the voiume below D and above the graph off, so
JJD x dA = 0. Also, .f.fD 2 dA = 2 ·A( D) = 2 · ~7r(3) 2
= 97r5ince Dis a half
disk of radius 3. Thus JJD(x + 2) dA = 0 + 97r = 97r.
65. We can write J J 0
(2x + 3y) dA = J J 0
2x dA + J J D 3y dA. J J 0 2x dA represents the volume of the solid lying under the
plane z = 2x and above the rectangle D . This solid region is a triangular cylinder with length band whose cross-section is a
triangle with width a and height 2a. (See the first figure.)
(a,0,2a) (O,b, 3b)
z= 3y
y
X
Thus its volume is~· a· 2a · b = a 2 b. Similarly, JJ 0
3ydA represents the volume of a triangular cylinder with length a,
triangular cross-section with width band height 3b, and volume t · b · 3b · a ~ ~ab 2 • (Sec the second figure.) T~us
JJ 0
(2x + 3y) dA = a 2 b + ~ab 2
67. ff 0
(ax 3 + In/+ va 2 - x 2 ) dA = ff 0
ax 3 dA + JJ 0
by 3 dA + JJ'a Ja 2 - x 2 dA. Now ax 3 is odd with respect
to x and by 3 is odd witb respect to y , and the region of integration is symmetric with respect to both x and y,
so ffD ax 3 dA = ffD'by 3 dA = 0.
JJ'v J a 2 -
x2 dA represents the volume of the solid region under the
graph of z = Ja 2 -
x 2 and above the rectangle p, namely a half circular
cylinder with radius a and length 2b (see the figure) whose volume is
t · 1rr 2 h = ~7ra 2 (2b) = 1ra 2 b. Th~s
JJD (ax 3 + by 3 + Ja 2 - x 2 ) dA = 0 + 0 + 1ra 2 b = 1ra 2 b.
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