James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 15.2 Double Integrals over General Regions 1005
Then (5) gives
y
D
y xy dA − y 4 22 yy11 1
2 y 2 23 xy dx dy − y 4 22F x 2
x−y11
2 y Gx− 1 2 y 2 23
dy
− 1 2 y 4 22
yfsy 1 1d 2 2 ( 1 2 y2 2 3) 2 g dy
− 1 2 y 4 22S2 y 5
4 1 4y 3 1 2y 2 2 8yD dy
− 1 2F2 y 6
24 1 y 4 1 2 y 3
3
4
2 4y − 36
2G22
If we had expressed D as a type I region using Figure 12(a), then we would have
obtained
y xy dA − y 21
23 ys2x16 2s2x16
D
xy dy dx 1 y 5 21 ys2x16 x21
but this would have involved more work than the other method.
xy dy dx
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x=2y
z
0
FIGURE 13
y
1
0
(0, 0, 2)
T
FIGURE 14
D
x
x+2y+z=2
(0, 1, 0)
1
”1, 2, 0’
x+2y=2
(or y=1-x/2)
y=x/2
1
1
”1, ’
2
y
x
ExamplE 4 Find the volume of the tetrahedron bounded by the planes
x 1 2y 1 z − 2, x − 2y, x − 0, and z − 0.
SOLUTION In a question such as this, it’s wise to draw two diagrams: one of the threedimensional
solid and another of the plane region D over which it lies. Figure 13 shows
the tetrahedron T bounded by the coordinate planes x − 0, z − 0, the vertical plane
x − 2y, and the plane x 1 2y 1 z − 2. Since the plane x 1 2y 1 z − 2 intersects the
xy-plane (whose equation is z − 0) in the line x 1 2y − 2, we see that T lies above
the triangular region D in the xy-plane bounded by the lines x − 2y, x 1 2y − 2, and
x − 0. (See Figure 14.)
The plane x 1 2y 1 z − 2 can be written as z − 2 2 x 2 2y, so the required
volume lies under the graph of the function z − 2 2 x 2 2y and above
Therefore
D − hsx, yd |
V − y s2 2 x 2 2yd dA
D
− y 1
− y 1
− y
1
0 y12xy2 xy2
s2 2 x 2 2yd dy dx
0 f2y 2 xy 2 y 2 g y−xy2
0 < x < 1, xy2 < y < 1 2 xy2j
y−12xy2
dx
F2 2 x 2 xS1 2 x 2S1 2
0 2D 2D
x 2
2 x 1 x 2
− y 1
sx 2 2 2x 1 1d dx − x 3
0
3 2 x 2 1 − 1 xG0
3
1
2 4G 1 x 2
dx
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