1: Double Integrals over Rectangles
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CHAPTER 16
VECTOR CALCULUS
Similarly, if we keep v constant by putting v v0 , we get a grid curve C2 given by
r͑u, v0 ͒ that lies on S, and its tangent vector at P0 is
ru
5
Ѩx
Ѩy
Ѩz
͑u0 , v0 ͒ i ϩ
͑u0 , v0 ͒ j ϩ
͑u0 , v0 ͒ k
Ѩu
Ѩu
Ѩu
If ru ϫ rv is not 0, then the surface S is called smooth (it has no “corners”). For a smooth
surface, the tangent plane is the plane that contains the tangent vectors ru and rv , and the
vector ru ϫ rv is a normal vector to the tangent plane.
Figure 13 shows the self-intersecting
surface in Example 9 and its tangent plane
at ͑1, 1, 3͒.
EXAMPLE 9 Find the tangent plane to the surface with parametric equations x u 2,
y v 2, z u ϩ 2v at the point ͑1, 1, 3͒.
v
SOLUTION We first compute the tangent vectors:
z
(1, 1, 3)
ru
Ѩy
Ѩz
Ѩx
iϩ
jϩ
k 2u i ϩ k
Ѩu
Ѩu
Ѩu
rv
Ѩx
Ѩy
Ѩz
iϩ
jϩ
k 2v j ϩ 2 k
Ѩv
Ѩv
Ѩv
y
x
Thus a normal vector to the tangent plane is
Խ Խ
i
ru ϫ rv 2u
0
FIGURE 13
j k
0 1 Ϫ2v i Ϫ 4u j ϩ 4uv k
2v 2
Notice that the point ͑1, 1, 3͒ corresponds to the parameter values u 1 and v 1, so
the normal vector there is
Ϫ2 i Ϫ 4 j ϩ 4 k
Therefore an equation of the tangent plane at ͑1, 1, 3͒ is
Ϫ2͑x Ϫ 1͒ Ϫ 4͑y Ϫ 1͒ ϩ 4͑z Ϫ 3͒ 0
x ϩ 2y Ϫ 2z ϩ 3 0
or
Surface Area
Now we define the surface area of a general parametric surface given by Equation 1. For
simplicity we start by considering a surface whose parameter domain D is a rectangle, and
we divide it into subrectangles Rij . Let’s choose ͑u i*, vj*͒ to be the lower left corner of Rij.
(See Figure 14.)
√
z
R ij
r
Ỵ√
Pij
Sij
Ỵu
(u *i , √ *j )
FIGURE 14
The image of the
subrectangle Rij is the patch Sij .
0
0
u
x
y
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PARAMETRIC SURFACES AND THEIR AREAS
SECTION 16.6
1129
The part Sij of the surface S that corresponds to Rij is called a patch and has the point Pij
with position vector r͑u i*, vj*͒ as one of its corners. Let
ru* ru͑u i*, vj*͒
Sij
Pij
(a)
rv* rv͑u i*, vj*͒
and
be the tangent vectors at Pij as given by Equations 5 and 4.
Figure 15(a) shows how the two edges of the patch that meet at Pij can be approximated
by vectors. These vectors, in turn, can be approximated by the vectors ⌬u r*u and ⌬v r*v
because partial derivatives can be approximated by difference quotients. So we approximate Sij by the parallelogram determined by the vectors ⌬u r*u and ⌬v r*v . This parallelogram
is shown in Figure 15(b) and lies in the tangent plane to S at Pij . The area of this parallelogram is
͑⌬u ru*͒ ϫ ͑⌬v r*͒
ru* ϫ r*v ⌬u ⌬v
v
Խ
Խ Խ
Խ
and so an approximation to the area of S is
m
n
͚ ͚ Խ r* ϫ r* Խ ⌬u ⌬v
u
v
i1 j1
Ỵ√ r √*
Ỵu r u*
(b)
Our intuition tells us that this approximation gets better as we increase the number of subrectangles, and we recognize the double sum as a Riemann sum for the double integral
xxD ru ϫ rv du dv. This motivates the following definition.
Խ
Խ
6
Definition If a smooth parametric surface S is given by the equation
r͑u, v͒ x͑u, v͒ i ϩ y͑u, v͒ j ϩ z͑u, v͒ k
FIGURE 15
Approximating a patch
by a parallelogram
͑u, v͒ ʦ D
and S is covered just once as ͑u, v͒ ranges throughout the parameter domain D,
then the surface area of S is
Խ
Խ
A͑S͒ yy ru ϫ rv dA
D
where
ru
Ѩx
Ѩy
Ѩz
iϩ
jϩ
k
Ѩu
Ѩu
Ѩu
rv
Ѩx
Ѩy
Ѩz
iϩ
jϩ
k
Ѩv
Ѩv
Ѩv
EXAMPLE 10 Find the surface area of a sphere of radius a.
SOLUTION In Example 4 we found the parametric representation
x a sin cos
y a sin sin
z a cos
where the parameter domain is
D ͕͑, ͒
Խ 0 ഛ ഛ , 0 ഛ ഛ 2 ͖
We first compute the cross product of the tangent vectors:
Խ
i
Ѩx
r ϫ r Ѩ
Ѩx
Ѩ
j
Ѩy
Ѩ
Ѩy
Ѩ
ԽԽ
k
Ѩz
i
Ѩ Ϫa cos cos
Ѩz
Ϫa sin sin
Ѩ
j
a cos sin
a sin cos
k
Ϫa sin
0
Խ
a 2 sin 2 cos i ϩ a 2 sin2 sin j ϩ a 2 sin cos k
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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1130
CHAPTER 16
VECTOR CALCULUS
Thus
Խr
Խ
ϫ r sa 4 sin 4 cos 2 ϩ a 4 sin 4 sin 2 ϩ a 4 sin 2 cos 2
sa 4 sin 4 ϩ a 4 sin 2 cos 2 a 2 ssin 2 a 2 sin
since sin ജ 0 for 0 ഛ ഛ . Therefore, by Definition 6, the area of the sphere is
Խ
Խ
A yy r ϫ r dA y
D
a2 y
2
0
2
0
y
0
a 2 sin d d
d y sin d a 2͑2͒2 4a 2
0
Surface Area of the Graph of a Function
For the special case of a surface S with equation z f ͑x, y͒, where ͑x, y͒ lies in D and f has
continuous partial derivatives, we take x and y as parameters. The parametric equations are
xx
rx i ϩ
so
and
yy
ͩ ͪ
Ѩf
Ѩx
z f ͑x, y͒
ry j ϩ
k
ͩ ͪ
Ѩf
Ѩy
k
Խ Խ
7
i
j
rx ϫ ry 1
0
0
1
k
Ѩf
Ѩf
Ѩf
Ϫ
iϪ
jϩk
Ѩx
Ѩx
Ѩy
Ѩf
Ѩy
Thus we have
8
Խr
x
Խ
ϫ ry
ͱͩ ͪ ͩ ͪ
Ѩf
Ѩx
2
ϩ
Ѩf
Ѩy
2
ϩ1
ͱ ͩ ͪ ͩ ͪ
Ѩz
Ѩx
1ϩ
2
ϩ
Ѩz
Ѩy
2
and the surface area formula in Definition 6 becomes
Notice the similarity between the surface area
formula in Equation 9 and the arc length formula
L
y
b
a
from Section 8.1.
ͱ ͩ ͪ
1ϩ
dy
dx
2
9
dx
A͑S͒
yy
D
v
ͱ ͩ ͪ ͩ ͪ
1ϩ
Ѩz
Ѩx
2
ϩ
Ѩz
Ѩy
2
dA
EXAMPLE 11 Find the area of the part of the paraboloid z x 2 ϩ y 2 that lies under
the plane z 9.
SOLUTION The plane intersects the paraboloid in the circle x 2 ϩ y 2 9, z 9. There-
fore the given surface lies above the disk D with center the origin and radius 3. (See
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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PARAMETRIC SURFACES AND THEIR AREAS
SECTION 16.6
z
1131
Figure 16.) Using Formula 9, we have
9
A
yy
D
ͱ ͩ ͪ ͩ ͪ
Ѩz
Ѩx
1ϩ
Ѩz
Ѩy
2
ϩ
2
dA
yy s1 ϩ ͑2x͒ 2 ϩ ͑2y͒ 2 dA
D
yy s1 ϩ 4͑x 2 ϩ y 2 ͒ dA
D
x
3
D
y
Converting to polar coordinates, we obtain
FIGURE 16
Ay
2
0
y
3
0
s1 ϩ 4r 2 r dr d y
2
0
2 ( 18 ) 23 ͑1 ϩ 4r 2 ͒3͞2
]
3
0
3
d y rs1 ϩ 4r 2 dr
0
(37s37 Ϫ 1)
6
The question remains whether our definition of surface area 6 is consistent with the
surface area formula from single-variable calculus (8.2.4).
We consider the surface S obtained by rotating the curve y f ͑x͒, a ഛ x ഛ b, about
the x-axis, where f ͑x͒ ജ 0 and f Ј is continuous. From Equations 3 we know that parametric equations of S are
y f ͑x͒ cos
xx
z f ͑x͒ sin
aഛxഛb
0 ഛ ഛ 2
To compute the surface area of S we need the tangent vectors
rx i ϩ f Ј͑x͒ cos j ϩ f Ј͑x͒ sin k
r Ϫf ͑x͒ sin j ϩ f ͑x͒ cos k
Խ
Thus
i
j
rx ϫ r 1 f Ј͑x͒ cos
0 Ϫf ͑x͒ sin
k
f Ј͑x͒ sin
f ͑x͒ cos
Խ
f ͑x͒ f Ј͑x͒ i Ϫ f ͑x͒ cos j Ϫ f ͑x͒ sin k
and so
Խr
x
Խ
ϫ r s͓ f ͑x͔͒ 2 ͓ f Ј͑x͔͒ 2 ϩ ͓ f ͑x͔͒ 2 cos 2 ϩ ͓ f ͑x͔͒ 2 sin 2
s͓ f ͑x͔͒ 2 ͓1 ϩ ͓ f Ј͑x͔͒ 2 ͔ f ͑x͒s1 ϩ ͓ f Ј͑x͔͒ 2
because f ͑x͒ ജ 0. Therefore the area of S is
Խ
Խ
A yy rx ϫ r dA
D
y
2
0
y
b
a
f ͑x͒s1 ϩ ͓ f Ј͑x͔͒ 2 dx d
b
2 y f ͑x͒s1 ϩ ͓ f Ј͑x͔͒ 2 dx
a
This is precisely the formula that was used to define the area of a surface of revolution in
single-variable calculus (8.2.4).
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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1132
VECTOR CALCULUS
CHAPTER 16
16.6
Exercises
1–2 Determine whether the points P and Q lie on the given
surface.
z
I
II
z
1. r͑u, v͒ ͗2u ϩ 3v, 1 ϩ 5u Ϫ v, 2 ϩ u ϩ v ͘
P͑7, 10, 4͒, Q͑5, 22, 5͒
2. r͑u, v͒ ͗u ϩ v, u 2 Ϫ v, u ϩ v 2 ͘
x
P͑3, Ϫ1, 5͒, Q͑Ϫ1, 3, 4͒
x
y
y
3–6 Identify the surface with the given vector equation.
3. r͑u, v͒ ͑u ϩ v͒ i ϩ ͑3 Ϫ v͒ j ϩ ͑1 ϩ 4u ϩ 5v͒ k
4. r͑u, v͒ 2 sin u i ϩ 3 cos u j ϩ v k,
z
III
IV
z
0ഛvഛ2
5. r͑s, t͒ ͗s, t, t Ϫ s ͘
2
2
6. r͑s, t͒ ͗ s sin 2t, s 2, s cos 2t ͘
x
; 7–12 Use a computer to graph the parametric surface. Get a
printout and indicate on it which grid curves have u constant and
which have v constant.
z
V
7. r͑u, v͒ ͗u 2, v 2, u ϩ v ͘,
Ϫ1 ഛ u ഛ 1, Ϫ1 ഛ v ഛ 1
y
x
y
VI
z
8. r͑u, v͒ ͗u, v 3, Ϫv ͘ ,
Ϫ2 ഛ u ഛ 2, Ϫ2 ഛ v ഛ 2
9. r͑u, v͒ ͗u cos v, u sin v, u 5 ͘ ,
Ϫ1 ഛ u ഛ 1, 0 ഛ v ഛ 2
y
x
x
10. r͑u, v͒ ͗u, sin͑u ϩ v͒, sin v ͘ ,
Ϫ ഛ u ഛ , Ϫ ഛ v ഛ
y
11. x sin v,
y cos u sin 4 v, z sin 2u sin 4 v,
0 ഛ u ഛ 2, Ϫ͞2 ഛ v ഛ ͞2
y cos u sin v, z sin v,
0 ഛ u ഛ 2, 0 ഛ v ഛ 2
12. x sin u,
19–26 Find a parametric representation for the surface.
19. The plane through the origin that contains the vectors i Ϫ j
and j Ϫ k
13–18 Match the equations with the graphs labeled I–VI and
give reasons for your answers. Determine which families of grid
curves have u constant and which have v constant.
13. r͑u, v͒ u cos v i ϩ u sin v j ϩ v k
Ϫ ഛ u ഛ
15. r͑u, v͒ sin v i ϩ cos u sin 2v j ϩ sin u sin 2v k
16. x ͑1 Ϫ u͒͑3 ϩ cos v͒ cos 4 u,
y ͑1 Ϫ u͒͑3 ϩ cos v͒ sin 4 u,
z 3u ϩ ͑1 Ϫ u͒ sin v
Խ Խ
y sin 3 u cos 3 v,
18. x (1 Ϫ u ) cos v,
;
Խ Խ
21. The part of the hyperboloid 4x 2 Ϫ 4y 2 Ϫ z 2 4 that lies in
22. The part of the ellipsoid x 2 ϩ 2y 2 ϩ 3z 2 1 that lies to the
left of the xz-plane
23. The part of the sphere x 2 ϩ y 2 ϩ z 2 4 that lies above the
cone z sx 2 ϩ y 2
24. The part of the sphere x 2 ϩ y 2 ϩ z 2 16 that lies between
z sin 3 v
y (1 Ϫ u ) sin v,
Graphing calculator or computer required
contains the vectors ͗2, 1, 4 ͘ and ͗Ϫ3, 2, 5͘
front of the yz-plane
14. r͑u, v͒ u cos v i ϩ u sin v j ϩ sin u k,
17. x cos 3 u cos 3 v,
20. The plane that passes through the point ͑0, Ϫ1, 5͒ and
zu
the planes z Ϫ2 and z 2
25. The part of the cylinder y 2 ϩ z 2 16 that lies between the
planes x 0 and x 5
CAS Computer algebra system required
1. Homework Hints available at stewartcalculus.com
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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SECTION 16.6
26. The part of the plane z x ϩ 3 that lies inside the cylinder
38. r͑u, v͒ ͑1 Ϫ u 2 Ϫ v 2 ͒ i Ϫ v j Ϫ u k;
x ϩy 1
2
PARAMETRIC SURFACES AND THEIR AREAS
2
1133
͑Ϫ1, Ϫ1, Ϫ1͒
39–50 Find the area of the surface.
CAS
27–28 Use a computer algebra system to produce a graph that
39. The part of the plane 3x ϩ 2y ϩ z 6 that lies in the
looks like the given one.
first octant
27.
28.
40. The part of the plane with vector equation
r͑u, v͒ ͗u ϩ v, 2 Ϫ 3u, 1 ϩ u Ϫ v ͘ that is given by
0 ഛ u ഛ 2, Ϫ1 ഛ v ഛ 1
3
z
z 0
_3
_3
y
0 5
x
41. The part of the plane x ϩ 2y ϩ 3z 1 that lies inside the
0
cylinder x 2 ϩ y 2 3
42. The part of the cone z sx 2 ϩ y 2 that lies between the
_1
_1
0
y
0
1 1
0
plane y x and the cylinder y x 2
_1
x
43. The surface z 3 ͑x 3͞2 ϩ y 3͞2 ͒, 0 ഛ x ഛ 1, 0 ഛ y ഛ 1
2
44. The part of the surface z 1 ϩ 3x ϩ 2y 2 that lies above the
triangle with vertices ͑0, 0͒, ͑0, 1͒, and ͑2, 1͒
; 29. Find parametric equations for the surface obtained by
Ϫx
rotating the curve y e , 0 ഛ x ഛ 3, about the x-axis and
use them to graph the surface.
45. The part of the surface z xy that lies within the
cylinder x 2 ϩ y 2 1
; 30. Find parametric equations for the surface obtained by
46. The part of the paraboloid x y 2 ϩ z 2 that lies inside the
rotating the curve x 4y 2 Ϫ y 4, Ϫ2 ഛ y ഛ 2, about the
y-axis and use them to graph the surface.
cylinder y 2 ϩ z 2 9
47. The part of the surface y 4x ϩ z 2 that lies between the
; 31. (a) What happens to the spiral tube in Example 2 (see Fig-
planes x 0, x 1, z 0, and z 1
ure 5) if we replace cos u by sin u and sin u by cos u ?
(b) What happens if we replace cos u by cos 2u and sin u
by sin 2u?
48. The helicoid (or spiral ramp) with vector equation
r͑u, v͒ u cos v i ϩ u sin v j ϩ v k, 0 ഛ u ഛ 1, 0 ഛ v ഛ
49. The surface with parametric equations x u 2 , y u v,
z 12 v 2, 0 ഛ u ഛ 1, 0 ഛ v ഛ 2
; 32. The surface with parametric equations
x 2 cos ϩ r cos͑͞2͒
50. The part of the sphere x 2 ϩ y 2 ϩ z 2 b 2 that lies inside the
y 2 sin ϩ r cos͑͞2͒
cylinder x 2 ϩ y 2 a 2, where 0 Ͻ a Ͻ b
z r sin͑͞2͒
51. If the equation of a surface S is z f ͑x, y͒, where
where Ϫ 12 ഛ r ഛ 12 and 0 ഛ ഛ 2, is called a Möbius
strip. Graph this surface with several viewpoints. What is
unusual about it?
Խ Խ
52–53 Find the area of the surface correct to four decimal places
33–36 Find an equation of the tangent plane to the given
by expressing the area in terms of a single integral and using
your calculator to estimate the integral.
parametric surface at the specified point.
33. x u ϩ v,
34. x u 2 ϩ 1,
y 3u 2,
z u Ϫ v ; ͑2, 3, 0͒
y v 3 ϩ 1,
cylinder x 2 ϩ y 2 1
37–38 Find an equation of the tangent plane to the given
parametric surface at the specified point. Graph the surface and
the tangent plane.
37. r͑u, v͒ u 2 i ϩ 2u sin v j ϩ u cos v k;
2
u 1, v 0
2
53. The part of the surface z eϪx Ϫy that lies above the
u 1, v ͞3
36. r͑u, v͒ sin u i ϩ cos u sin v j ϩ sin v k;
u ͞6 , v ͞6
CAS
52. The part of the surface z cos͑x 2 ϩ y 2 ͒ that lies inside the
z u ϩ v ; ͑5, 2, 3͒
35. r͑u, v͒ u cos v i ϩ u sin v j ϩ v k;
Խ Խ
x 2 ϩ y 2 ഛ R 2, and you know that fx ഛ 1 and fy ഛ 1,
what can you say about A͑S͒?
disk x ϩ y ഛ 4
2
CAS
2
54. Find, to four decimal places, the area of the part of the sur-
face z ͑1 ϩ x 2 ͒͑͞1 ϩ y 2 ͒ that lies above the square
x ϩ y ഛ 1. Illustrate by graphing this part of the
surface.
Խ Խ Խ Խ
55. (a) Use the Midpoint Rule for double integrals (see Sec-
tion 15.1) with six squares to estimate the area of the
surface z 1͑͞1 ϩ x 2 ϩ y 2 ͒, 0 ഛ x ഛ 6, 0 ഛ y ഛ 4.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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1134
CAS
CAS
CHAPTER 16
VECTOR CALCULUS
61. Find the area of the part of the sphere x 2 ϩ y 2 ϩ z 2 4z
(b) Use a computer algebra system to approximate the surface area in part (a) to four decimal places. Compare
with the answer to part (a).
that lies inside the paraboloid z x 2 ϩ y 2.
62. The figure shows the surface created when the cylinder
y 2 ϩ z 2 1 intersects the cylinder x 2 ϩ z 2 1. Find the
area of this surface.
56. Find the area of the surface with vector equation
r͑u, v͒ ͗ cos 3u cos 3v, sin 3u cos 3v, sin 3v ͘ , 0 ഛ u ഛ ,
0 ഛ v ഛ 2. State your answer correct to four decimal
z
places.
CAS
57. Find the exact area of the surface z 1 ϩ 2x ϩ 3y ϩ 4y 2,
1 ഛ x ഛ 4, 0 ഛ y ഛ 1.
x
58. (a) Set up, but do not evaluate, a double integral for the area
of the surface with parametric equations x au cos v,
y bu sin v, z u 2, 0 ഛ u ഛ 2, 0 ഛ v ഛ 2.
(b) Eliminate the parameters to show that the surface is an
elliptic paraboloid and set up another double integral for
the surface area.
(c) Use the parametric equations in part (a) with a 2 and
b 3 to graph the surface.
(d) For the case a 2, b 3, use a computer algebra system
to find the surface area correct to four decimal places.
;
CAS
59. (a) Show that the parametric equations x a sin u cos v,
y b sin u sin v, z c cos u, 0 ഛ u ഛ , 0 ഛ v ഛ 2,
63. Find the area of the part of the sphere x 2 ϩ y 2 ϩ z 2 a 2
that lies inside the cylinder x 2 ϩ y 2 ax.
64. (a) Find a parametric representation for the torus obtained
;
represent an ellipsoid.
(b) Use the parametric equations in part (a) to graph the
ellipsoid for the case a 1, b 2, c 3.
(c) Set up, but do not evaluate, a double integral for the surface area of the ellipsoid in part (b).
;
y
by rotating about the z-axis the circle in the xz-plane
with center ͑b, 0, 0͒ and radius a Ͻ b. [Hint: Take as
parameters the angles and ␣ shown in the figure.]
(b) Use the parametric equations found in part (a) to graph
the torus for several values of a and b.
(c) Use the parametric representation from part (a) to find
the surface area of the torus.
z
(x, y, z)
60. (a) Show that the parametric equations x a cosh u cos v,
y b cosh u sin v, z c sinh u, represent a hyperboloid
;
16.7
of one sheet.
(b) Use the parametric equations in part (a) to graph the
hyperboloid for the case a 1, b 2, c 3.
(c) Set up, but do not evaluate, a double integral for the surface area of the part of the hyperboloid in part (b) that
lies between the planes z 3 and z 3.
0
ồ
ă
x
y
(b, 0, 0)
Surface Integrals
The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length. Suppose f is a function of three variables
whose domain includes a surface S. We will define the surface integral of f over S in such
a way that, in the case where f ͑x, y, z͒ 1, the value of the surface integral is equal to the
surface area of S. We start with parametric surfaces and then deal with the special case
where S is the graph of a function of two variables.
Parametric Surfaces
Suppose that a surface S has a vector equation
r͑u, v͒ x͑u, v͒ i ϩ y͑u, v͒ j ϩ z͑u, v͒ k
͑u, v͒ ʦ D
We first assume that the parameter domain D is a rectangle and we divide it into subrect-
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97817_16_ch16_p1130-1137.qk_97817_16_ch16_p1130-1137 11/9/10 9:23 AM Page 1135
SECTION 16.7
√
1135
angles Rij with dimensions ⌬u and ⌬v. Then the surface S is divided into corresponding
patches Sij as in Figure 1. We evaluate f at a point Pij* in each patch, multiply by the area
⌬Sij of the patch, and form the Riemann sum
R ij
Ỵ√
D
SURFACE INTEGRALS
Ỵu
m
n
͚ ͚ f ͑P*͒ ⌬S
ij
0
ij
i1 j1
u
Then we take the limit as the number of patches increases and define the surface integral
of f over the surface S as
r
m
z
S
P *ij
S
Sij
0
x
yy f ͑x, y, z͒ dS
1
y
lim
n
͚ ͚ f ͑P*͒ ⌬S
ij
m, n l ϱ i1 j1
ij
Notice the analogy with the definition of a line integral (16.2.2) and also the analogy with
the definition of a double integral (15.1.5).
To evaluate the surface integral in Equation 1 we approximate the patch area ⌬Sij by the
area of an approximating parallelogram in the tangent plane. In our discussion of surface
area in Section 16.6 we made the approximation
Խ
Խ
⌬Sij Ϸ ru ϫ rv ⌬u ⌬v
FIGURE 1
where
ru
Ѩx
Ѩy
Ѩz
iϩ
jϩ
k
Ѩu
Ѩu
Ѩu
rv
Ѩx
Ѩy
Ѩz
iϩ
jϩ
k
Ѩv
Ѩv
Ѩv
are the tangent vectors at a corner of Sij . If the components are continuous and ru and rv
are nonzero and nonparallel in the interior of D, it can be shown from Definition 1, even
when D is not a rectangle, that
We assume that the surface is covered only
once as ͑u, v͒ ranges throughout D. The value
of the surface integral does not depend on the
parametrization that is used.
2
yy f ͑x, y, z͒ dS yy f ͑r͑u, v͒͒ Խ r
u
S
Խ
ϫ rv dA
D
This should be compared with the formula for a line integral:
y
C
Խ
b
Խ
f ͑x, y, z͒ ds y f ͑r͑t͒͒ rЈ͑t͒ dt
a
Observe also that
yy 1 dS yy Խ r
u
S
D
Խ
ϫ rv dA A͑S͒
Formula 2 allows us to compute a surface integral by converting it into a double integral over the parameter domain D. When using this formula, remember that f ͑r͑u, v͒͒ is
evaluated by writing x x͑u, v͒, y y͑u, v͒, and z z͑u, v͒ in the formula for f ͑x, y, z͒.
EXAMPLE 1 Compute the surface integral xxS x 2 dS, where S is the unit sphere
x 2 ϩ y 2 ϩ z 2 1.
SOLUTION As in Example 4 in Section 16.6, we use the parametric representation
x sin cos
y sin sin
z cos
0ഛ ഛ
0 ഛ ഛ 2
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97817_16_ch16_p1130-1137.qk_97817_16_ch16_p1130-1137 11/9/10 9:23 AM Page 1136
1136
CHAPTER 16
VECTOR CALCULUS
r͑, ͒ sin cos i ϩ sin sin j ϩ cos k
that is,
As in Example 10 in Section 16.6, we can compute that
Խr
Խ
ϫ r sin
Therefore, by Formula 2,
yy x
2
S
D
y
Here we use the identities
2
0
cos2 12 ͑1 ϩ cos 2 ͒
y
sin 1 Ϫ cos
2
Խ
Խ
dS yy ͑sin cos ͒2 r ϫ r dA
Instead, we could use Formulas 64 and 67 in
the Table of Integrals.
sin 2 cos 2 sin d d y
0
2 1
2
0
2
y
2
cos 2 d
0
͑1 ϩ cos 2 ͒ d
[
y
0
0
sin 3 d
͑sin Ϫ sin cos 2͒ d
2
0
] [Ϫcos ϩ
12 ϩ 12 sin 2
y
1
3
]
cos 3 0
4
3
Surface integrals have applications similar to those for the integrals we have previously
considered. For example, if a thin sheet (say, of aluminum foil) has the shape of a surface
S and the density (mass per unit area) at the point ͑x, y, z͒ is ͑x, y, z͒, then the total mass
of the sheet is
m yy ͑x, y, z͒ dS
S
and the center of mass is ͑x, y, z͒, where
x
1
m
yy x ͑x, y, z͒ dS
y
S
1
m
yy y ͑x, y, z͒ dS
z
S
1
m
yy z ͑x, y, z͒ dS
S
Moments of inertia can also be defined as before (see Exercise 41).
Graphs
Any surface S with equation z t͑x, y͒ can be regarded as a parametric surface with parametric equations
xx
and so we have
rx i ϩ
yy
ͩ ͪ
Ѩt
Ѩx
z t͑x, y͒
ry j ϩ
k
ͩ ͪ
Ѩt
Ѩy
k
Thus
rx ϫ ry Ϫ
3
and
Խr
x
Խ
ϫ ry
Ѩt
Ѩt
iϪ
jϩk
Ѩx
Ѩy
ͱͩ ͪ ͩ ͪ
Ѩz
Ѩx
2
ϩ
Ѩz
Ѩy
2
ϩ1
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_16_ch16_p1130-1137.qk_97817_16_ch16_p1130-1137 11/9/10 9:23 AM Page 1137
SECTION 16.7
SURFACE INTEGRALS
1137
Therefore, in this case, Formula 2 becomes
4
ͱͩ ͪ ͩ ͪ
Ѩz
Ѩx
yy f ͑x, y, z͒ dS yy f ( x, y, t͑x, y͒)
S
D
Ѩz
Ѩy
2
ϩ
2
ϩ 1 dA
Similar formulas apply when it is more convenient to project S onto the yz-plane or
xz-plane. For instance, if S is a surface with equation y h͑x, z͒ and D is its projection
onto the xz-plane, then
ͱͩ ͪ ͩ ͪ
Ѩy
Ѩx
yy f ͑x, y, z͒ dS yy f ( x, h͑x, z͒, z)
S
z
D
Ѩy
Ѩz
2
ϩ
2
ϩ 1 dA
EXAMPLE 2 Evaluate xxS y dS, where S is the surface z x ϩ y 2, 0 ഛ x ഛ 1, 0 ഛ y ഛ 2.
(See Figure 2.)
SOLUTION Since
Ѩz
1
Ѩx
y
Formula 4 gives
ͱ ͩ ͪ ͩ ͪ
x
yy y dS yy y
FIGURE 2
S
D
y
Ѩz
2y
Ѩy
and
1
0
y
2
0
1
1ϩ
Ѩz
Ѩy
2
ϩ
2
dA
ys1 ϩ 1 ϩ 4y 2 dy dx
y dx s2
0
Ѩz
Ѩx
y
2
0
ys1 ϩ 2y 2 dy
]
2
s2 ( 14 ) 23 ͑1 ϩ 2y 2 ͒3͞2 0
13s2
3
If S is a piecewise-smooth surface, that is, a finite union of smooth surfaces S1 , S2, . . . ,
Sn that intersect only along their boundaries, then the surface integral of f over S is defined
by
yy f ͑x, y, z͒ dS yy f ͑x, y, z͒ dS ϩ и и и ϩ yy f ͑x, y, z͒ dS
S
z
S£ (z=1+x )
S¡ (≈+¥=1)
x
SOLUTION The surface S is shown in Figure 3. (We have changed the usual position of
the axes to get a better look at S.) For S1 we use and z as parameters (see Example 5
in Section 16.6) and write its parametric equations as
0
FIGURE 3
Sn
v EXAMPLE 3 Evaluate xxS z dS, where S is the surface whose sides S1 are given by the
cylinder x 2 ϩ y 2 1, whose bottom S2 is the disk x 2 ϩ y 2 ഛ 1 in the plane z 0, and
whose top S3 is the part of the plane z 1 ϩ x that lies above S2 .
y
S™
S1
x cos
y sin
zz
where
0 ഛ ഛ 2
and
0 ഛ z ഛ 1 ϩ x 1 ϩ cos
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_16_ch16_p1138-1147.qk_97817_16_ch16_p1138-1147 11/9/10 9:25 AM Page 1138
1138
CHAPTER 16
VECTOR CALCULUS
Therefore
Խ
Խ
i
r ϫ rz Ϫsin
0
Խr
and
j
k
cos 0 cos i ϩ sin j
0
1
Խ
ϫ rz scos 2 ϩ sin 2 1
Thus the surface integral over S1 is
yy z dS yy z Խ r ϫ r Խ dA
z
S1
D
y
2
y
0
1ϩcos
0
12 y
2
0
z dz d y
2 1
2
͑1 ϩ cos ͒2 d
0
͓1 ϩ 2 cos ϩ 12 ͑1 ϩ cos 2 ͔͒ d
[
2
0
]
12 32 ϩ 2 sin ϩ 14 sin 2
3
2
Since S2 lies in the plane z 0, we have
yy z dS yy 0 dS 0
S2
S2
The top surface S3 lies above the unit disk D and is part of the plane z 1 ϩ x. So,
taking t͑x, y͒ 1 ϩ x in Formula 4 and converting to polar coordinates, we have
yy
ͱ ͩ ͪ ͩ ͪ
z dS yy ͑1 ϩ x͒
S3
D
y
2
0
y
1
0
2
y y
s2
y (
s2
0
0
ͫ
Ѩz
Ѩx
2
ϩ
Ѩz
Ѩy
2
dA
͑1 ϩ r cos ͒s1 ϩ 1 ϩ 0 r dr d
s2
1
0
2
1ϩ
1
2
͑r ϩ r 2 cos ͒ dr d
ϩ 13 cos ) d
sin
ϩ
2
3
ͬ
2
s2
0
Therefore
yy z dS yy z dS ϩ yy z dS ϩ yy z dS
S
S1
S2
S3
3
ϩ 0 ϩ s2 ( 32 ϩ s2 )
2
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.