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

1118 Chapter 16 Vector Calculus
Thus
| r 3 r | − sa 4 sin 4 cos 2 1 a 4 sin 4 sin 2 1 a 4 sin 2 cos 2
− sa 4 sin 4 1 a 4 sin 2 cos 2 − a 2 ssin 2 − a 2 sin
since sin > 0 for 0 < < . Therefore, by Definition 6, the area of the sphere is
A − yy | r 3 r | dA − y 2
y
a 2 sin d d
0 0
D
− a 2 y 2
d y
0
0 sin d − a 2 s2d2 − 4a 2 ■
Surface Area of the Graph of a Function
For the special case of a surface S with equation z − f sx, yd, where sx, yd lies in D and f
has continuous partial derivatives, we take x and y as parameters. The parametric equations
are
so
x − x y − y z − f sx, yd
r x − i 1S
−xD −f k
and
Z
Z
i j k
−f
1 0
7 r x 3 r y − −x −
−f
0 1
−y
Thus we have
8 | r x 3 r y | ÎS
−xD
− −f 2
1S
and the surface area formula in Definition 6 becomes
r y − j 1S
−yD −f k
2 −f
−x i 2 −f
−y j 1 k
−yD
−f 2
1 1 −Î1 1S
−xD
−z 2
1S
−yD
−z 2
Notice the similarity between the
surface area formula in Equation 9 and
the arc length formula
b
L − y Î1 1S dy
2
dx
a dxD
from Section 8.1.
9 AsSd − y Î1 1S
Dy −xD
−z 2
1S
−yD
−z 2
dA
ExamplE 11 Find the area of the part of the paraboloid z − x 2 1 y 2 that lies under the
plane 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
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