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

1028 Chapter 15 Multiple Integrals
z
Using Formula 2 with f sx, yd − x 2 1 2y, we get
A − y ss2xd 2 1 s2d 2 1 1 dA − y 1
y x
s4x 2 1 5 dy dx
0 0
T
− y 1
0
xs4x 2 1 5 dx − 1 8 ? 2 3 s4x 2 1 5d 3y2 g 0
1
−
1
12 (27 2 5s5 )
x
T
y
Figure 4 shows the portion of the surface whose area we have just computed.
■
FIGURE 4
ExamplE 2 Find the area of the part of the paraboloid z − x 2 1 y 2 that lies under the
plane z − 9.
z
9
SOLUTION The plane intersects the paraboloid in the circle x 2 1 y 2 − 9, z − 9. Therefore
the given surface lies above the disk D with center the origin and radius 3. (See
Figure 5.) Using Formula 3, we have
A − y Î1 1S
Dy −xD
−z
2
1S
−yD
−z
2
dA − yys1 1 s2xd 2 1 s2yd 2
D
dA
D
3
x
FIGURE 5
y
− y s1 1 4sx 2 1 y 2 d dA
D
Converting to polar coordinates, we obtain
A − y 2
0
y 3
s1 1 4r 2 r dr d − y 2
0 0
d y 3
0
1
8 s1 1 4r 2 s8rd dr
− 2( 1 8) 2 3 s1 1 4r 2 d 3y2 g 0
3
−
6 (37s37 2 1) ■
1–12 Find the area of the surface.
1. The part of the plane 5x 1 3y 2 z 1 6 − 0 that lies above the
rectangle f1, 4g 3 f2, 6g
2. The part of the plane 6x 1 4y 1 2z − 1 that lies inside the
cylinder x 2 1 y 2 − 25
3. The part of the plane 3x 1 2y 1 z − 6 that lies in the
first octant
4. The part of the surface 2y 1 4z 2 x 2 − 5 that lies above the
triangle with vertices s0, 0d, s2, 0d, and s2, 4d
5. The part of the paraboloid z − 1 2 x 2 2 y 2 that lies above the
plane z − 22
6. The part of the cylinder x 2 1 z 2 − 4 that lies above the square
with vertices s0, 0d, s1, 0d, s0, 1d, and s1, 1d
7. The part of the hyperbolic paraboloid z − y 2 2 x 2 that lies
between the cylinders x 2 1 y 2 − 1 and x 2 1 y 2 − 4
8. The surface z − 2 3 sx 3y2 1 y 3y2 d, 0 < x < 1, 0 < y < 1
9. The part of the surface z − xy that lies within the cylinder
x 2 1 y 2 − 1
10. The part of the sphere x 2 1 y 2 1 z 2 − 4 that lies above the
plane z − 1
11. The part of the sphere x 2 1 y 2 1 z 2 − a 2 that lies within the
cylinder x 2 1 y 2 − ax and above the xy-plane
12. The part of the sphere x 2 1 y 2 1 z 2 − 4z that lies inside the
paraboloid z − x 2 1 y 2
13–14 Find the area of the surface correct to four decimal places
by expressing the area in terms of a single integral and using your
calculator to estimate the integral.
13. The part of the surface z − 1ys1 1 x 2 1 y 2 d that lies above the
disk x 2 1 y 2 < 1
14. The part of the surface z − cossx 2 1 y 2 d that lies inside the
cylinder x 2 1 y 2 − 1
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