SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS
¤
321
i j k
(b) From (a), v = w × r = 0 0 = (0 · − ) i + ( − 0 · ) j + (0 · − · 0) k = − i + j
j
k
i
(c) curl v = ∇ × v =
−
0
=
(0) −
() i +
(−) −
(0) j +
() −
(−) k
= [ − (−)] k = 2 k = 2w
39. For any continuous function on R3 , define a vector field G( ) = h( ) 0 0i where ( ) =
Then div G =
0
( ) .
(( )) +
(0) +
(0) =
( ) = ( ) by the Fundamental Theorem of
0
Calculus. Thus every continuous function on R3 is the divergence of some vector field.
16.6 Parametric Surfaces and Their Areas
1. (7 10 4) lies on the parametric surface r( ) = h2 + 3 1 + 5 − 2 + + i if and only if there are values for
and where 2 + 3 = 7, 1 + 5 − = 10, and 2 + + = 4. But solving the first two equations simultaneously gives
= 2, = 1 and these values do not satisfy the third equation, so does not lie on the surface.
(5 22 5) lies on the surface if 2 + 3 = 5, 1 + 5 − = 22, and 2 + + = 5 for some values of and . Solving the
first two equations simultaneously gives = 4, = −1 and these values satisfy the third equation, so lies on the surface.
3. r( ) = ( + ) i + (3 − ) j + (1 + 4 + 5) k = h0 3 1i + h1 0 4i + h1 −1 5i. From Example 3, we recognize
this as a vector equation of a plane through the point (0 3 1) and containing vectors a = h1 0 4i and b = h1 −1 5i. If we
i j k
wish to find a more conventional equation for the plane, a normal vector to the plane is a × b = 1 0 4 = 4 i − j − k
1 −1 5
and an equation of the plane is 4( − 0) − ( − 3) − ( − 1) = 0 or 4 − − = −4.
5. r( ) = 2 − 2 , so the corresponding parametric equations for the surface are = , = , = 2 − 2 . For any
point ( ) on the surface, we have = 2 − 2 . With no restrictions on the parameters, the surface is = 2 − 2 , which
we recognize as a hyperbolic paraboloid.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
322
¤
CHAPTER 16
VECTOR CALCULUS
7. r( ) = 2 2 + , −1 ≤ ≤ 1, −1 ≤ ≤ 1.
The surface has parametric equations = 2 , = 2 , = + , −1 ≤ ≤ 1, −1 ≤ ≤ 1.
In Maple, the surface can be graphed by entering
plot3d([uˆ2,vˆ2,u+v],u=-1..1,v=-1..1);.
In Mathematica we use the ParametricPlot3D command.
If we keep constant at 0 , = 20 , a constant, so the
corresponding grid curves must be the curves parallel to the
-plane. If is constant, we have = 02 , a constant, so these
grid curves are the curves parallel to the -plane.
9. r( ) = cos sin 5 .
The surface has parametric equations = cos , = sin ,
= 5 , −1 ≤ ≤ 1, 0 ≤ ≤ 2. Note that if = 0 is constant
then = 50 is constant and = 0 cos , = 0 sin describe a
circle in , of radius |0 |, so the corresponding grid curves are
circles parallel to the -plane. If = 0 , a constant, the parametric
equations become = cos 0 , = sin 0 , = 5 . Then
= (tan 0 ), so these are the grid curves we see that lie in vertical
planes = through the -axis.
11. = sin , = cos sin 4, = sin 2 sin 4, 0 ≤ ≤ 2, − 2 ≤ ≤
.
2
Note that if = 0 is constant, then = sin 0 is constant, so the
corresponding grid curves must be parallel to the -plane. These
are the vertically oriented grid curves we see, each shaped like a
“figure-eight.” When = 0 is held constant, the parametric
equations become = sin , = cos 0 sin 4,
= sin 20 sin 4. Since is a constant multiple of , the
corresponding grid curves are the curves contained in planes
= that pass through the -axis.
13. r( ) = cos i + sin j + k. The parametric equations for the surface are = cos , = sin , = . We look at
the grid curves first; if we fix , then and parametrize a straight line in the plane = which intersects the -axis. If is
held constant, the projection onto the -plane is circular; with = , each grid curve is a helix. The surface is a spiraling
ramp, graph IV.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS
¤
323
15. r( ) = sin i + cos sin 2 j + sin sin 2 k. Parametric equations for the surface are = sin , = cos sin 2,
= sin sin 2. If = 0 is fixed, then = sin 0 is constant, and = (sin 20 ) cos and = (sin 20 ) sin describe a
circle of radius |sin 20 |, so each corresponding grid curve is a circle contained in the vertical plane = sin 0 parallel to the
-plane. The only possible surface is graph II. The grid curves we see running lengthwise along the surface correspond to
holding constant, in which case = (cos 0 ) sin 2, = (sin 0 ) sin 2
⇒ = (tan 0 ), so each grid curve lies in a
plane = that includes the -axis.
17. = cos3 cos3 , = sin3 cos3 , = sin3 . If = 0 is held constant then = sin3 0 is constant, so the
corresponding grid curve lies in a horizontal plane. Several of the graphs exhibit horizontal grid curves, but the curves for this
surface are neither circles nor straight lines, so graph III is the only possibility. (In fact, the horizontal grid curves here are
members of the family = cos3 , = sin3 and are called astroids.) The vertical grid curves we see on the surface
correspond to = 0 held constant, as then we have = cos3 0 cos3 , = sin3 0 cos3 so the corresponding grid curve
lies in the vertical plane = (tan3 0 ) through the -axis.
19. From Example 3, parametric equations for the plane through the point (0 0 0) that contains the vectors a = h1 −1 0i and
b = h0 1 −1i are = 0 + (1) + (0) = , = 0 + (−1) + (1) = − , = 0 + (0) + (−1) = −.
21. Solving the equation for gives 2 = 1 + 2 + 14 2
⇒ =
1 + 2 + 14 2 . (We choose the positive root since we want
the part of the hyperboloid that corresponds to ≥ 0.) If we let and be the parameters, parametric equations are = ,
= , = 1 + 2 + 14 2 .
23. Since the cone intersects the sphere in the circle 2 + 2 = 2, =
can parametrize the surface as = , = , =
√
2 and we want the portion of the sphere above this, we
4 − 2 − 2 where 2 + 2 ≤ 2.
Alternate solution: Using spherical coordinates, = 2 sin cos , = 2 sin sin , = 2 cos where 0 ≤ ≤
4
and
0 ≤ ≤ 2.
25. Parametric equations are = , = 4 cos , = 4 sin , 0 ≤ ≤ 5, 0 ≤ ≤ 2.
27. The surface appears to be a portion of a circular cylinder of radius 3 with axis the -axis. An equation of the cylinder is
2 + 2 = 9, and we can impose the restrictions 0 ≤ ≤ 5, ≤ 0 to obtain the portion shown. To graph the surface on a
CAS, we can use parametric equations = , = 3 cos , = 3 sin with the parameter domain 0 ≤ ≤ 5, 2 ≤ ≤ 3
.
2
√
Alternatively, we can regard and as parameters. Then parametric equations are = , = , = − 9 − 2 , where
0 ≤ ≤ 5 and −3 ≤ ≤ 3.
29. Using Equations 3, we have the parametrization = , = − cos ,
= − sin , 0 ≤ ≤ 3, 0 ≤ ≤ 2.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
324
¤
CHAPTER 16
VECTOR CALCULUS
31. (a) Replacing cos by sin and sin by cos gives parametric equations
= (2 + sin ) sin , = (2 + sin ) cos , = + cos . From the graph, it
appears that the direction of the spiral is reversed. We can verify this observation by
noting that the projection of the spiral grid curves onto the -plane, given by
= (2 + sin ) sin , = (2 + sin ) cos , = 0, draws a circle in the clockwise
direction for each value of . The original equations, on the other hand, give circular
projections drawn in the counterclockwise direction. The equation for is identical in
both surfaces, so as increases, these grid curves spiral up in opposite directions for
the two surfaces.
(b) Replacing cos by cos 2 and sin by sin 2 gives parametric equations
= (2 + sin ) cos 2, = (2 + sin ) sin 2, = + cos . From the graph, it
appears that the number of coils in the surface doubles within the same parametric
domain. We can verify this observation by noting that the projection of the spiral grid
curves onto the -plane, given by = (2 + sin ) cos 2, = (2 + sin ) sin 2,
= 0 (where is constant), complete circular revolutions for 0 ≤ ≤ while the
original surface requires 0 ≤ ≤ 2 for a complete revolution. Thus, the new
surface winds around twice as fast as the original surface, and since the equation for
is identical in both surfaces, we observe twice as many circular coils in the same
-interval.
33. r( ) = ( + ) i + 32 j + ( − ) k.
r = i + 6 j + k and r = i − k, so r × r = −6 i + 2 j − 6 k. Since the point (2 3 0) corresponds to = 1, = 1, a
normal vector to the surface at (2 3 0) is −6 i + 2 j − 6 k, and an equation of the tangent plane is −6 + 2 − 6 = −6 or
3 − + 3 = 3.
35. r( ) = cos i + sin j + k
√
⇒ r 1 3 = 12 23 3 .
√
r = cos i + sin j and r = − sin i + cos j + k, so a normal vector to the surface at the point 12 23 3 is
√
√
√
r 1 3 × r 1 3 = 12 i + 23 j × − 23 i + 12 j + k = 23 i − 12 j + k. Thus an equation of the tangent plane at
√
√
√
√
1
3
is 23 − 12 − 12 − 23 + 1 − 3 = 0 or 23 − 12 + = 3 .
2 2 3
37. r( ) = 2 i + 2 sin j + cos k
⇒ r(1 0) = (1 0 1).
r = 2 i + 2 sin j + cos k and r = 2 cos j − sin k,
so a normal vector to the surface at the point (1 0 1) is
r (1 0) × r (1 0) = (2 i + k) × (2 j) = −2 i + 4 k.
Thus an equation of the tangent plane at (1 0 1) is
−2( − 1) + 0( − 0) + 4( − 1) = 0 or − + 2 = 1.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS
¤
39. The surface is given by = ( ) = 6 − 3 − 2 which intersects the -plane in the line 3 + 2 = 6, so is the
triangular region given by ( ) 0 ≤ ≤ 2 0 ≤ ≤ 3 − 32 . By Formula 9, the surface area of is
2 2
() =
1+
+
√
√
√
√
= 1 + (−3)2 + (−2)2 = 14 = 14 () = 14 12 · 2 · 3 = 3 14
41. Here we can write = ( ) =
() =
1+
√
14
3
=
1
3
− 13 − 23 and is the disk 2 + 2 ≤ 3, so by Formula 9 the area of the surface is
() =
2
√
14
3
2
2 2
=
1 + − 13 + − 23 =
√ 2 √
· 3 = 14
+
√
14
3
+ 32 ) and = {( ) | 0 ≤ ≤ 1 0 ≤ ≤ 1 }. Then = 12 , = 12 and
11√
√ 2 √ 2
() = 1 + ( ) +
= 0 0 1 + +
=1
1
1
= 23 0 ( + 2)32 − ( + 1)32
= 0 23 ( + + 1)32
43. = ( ) =
32
2
3 (
2
3
=
=0
2
(
5
52
+ 2)
−
2
(
5
+ 1)52
1
=
0
45. = ( ) = with 2 + 2 ≤ 1, so = , =
() =
4
(352
15
− 252 − 252 + 1) =
4
(352
15
− 272 + 1)
⇒
=1
√
2 + 2 = 2 1
2 + 1 = 2 1 ( 2 + 1)32
1
+
3
0
0
0
=0
2 √
= 0 13 2 2 − 1 =
√
2
2 2−1
3
47. A parametric representation of the surface is = , = 4 + 2 , = with 0 ≤ ≤ 1, 0 ≤ ≤ 1.
Hence r × r = (i + 4 j) × (2 j + k) = 4 i − j + 2 k.
2 2
Note: In general, if = ( ) then r × r =
i−j+
k and () =
1+
+
. Then
1√
1 1 √
() = 0 0 17 + 4 2 = 0 17 + 4 2
√
1
√
√
√
√
ln 2 + 21 − ln 17
ln2 + 4 2 + 17 0 = 221 + 17
= 12 17 + 4 2 + 17
2
4
49. r = h2 0i, r = h0 i, and r × r = 2 −2 22 . Then
1 2 √
1 2
4 + 42 2 + 44 = 0 0
(2 + 22 )2
0 0
1
=2
1 2
1
1
= 0 0 ( 2 + 22 ) = 0 13 3 + 22 =0 = 0 83 + 42 = 83 + 43 3 0 = 4
() =
|r × r | =
1 + ( )2 + ( )2 . But if | | ≤ 1 and | | ≤ 1 then 0 ≤ ( )2 ≤ 1,
√
0 ≤ ( )2 ≤ 1 ⇒ 1 ≤ 1 + ( )2 + ( )2 ≤ 3 ⇒ 1 ≤ 1 + ( )2 + ( )2 ≤ 3. By Property 15.3.11,
√
√
1 ≤ 1 + ( )2 + ( )2 ≤ 3 ⇒ () ≤ () ≤ 3 () ⇒
√
2 ≤ () ≤ 32 .
51. From Equation 9 we have () =
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
325
326
¤
CHAPTER 16
VECTOR CALCULUS
53. = ( ) = −
2 − 2
with 2 + 2 ≤ 4.
2
2
1 + −2−2 −2 + −2−2 −2 = 1 + 4(2 + 2 )−2(2 +2 )
2 2
2
2
2
= 0 0 1 + 42 −22 = 0 0 1 + 42 −22 = 2 0 1 + 42 −22 ≈ 139783
() =
2
=
Using the Midpoint Rule with ( ) =
1+
55. (a) () =
1+
() ≈
2
+
6
0
4
1+
0
42 + 4 2
.
(1 + 2 + 2 )4
42 + 4 2
, = 3, = 2 we have
(1 + 2 + 2 )4
∆ = 4 [ (1 1) + (1 3) + (3 1) + (3 3) + (5 1) + (5 3)] ≈ 242055
3
2
=1 =1
(b) Using a CAS we have () =
6
0
4
1+
0
42 + 4 2
≈ 242476. This agrees with the estimate in part (a)
(1 + 2 + 2 )4
to the first decimal place.
57. = 1 + 2 + 3 + 4 2 , so
() =
1+
2
+
Using a CAS, we have
4 1
14 + 48 + 64 2 =
1
0
or
45
8
√
14 +
15
16
√
2
45
8
=
4
1
√
14 +
15
16
√
3 70
√
ln 113 √55 +
.
+ 70
1
0
1 + 4 + (3 + 8)2 =
√ √
√
ln 11 5 + 3 14 5 −
59. (a) = sin cos , = sin sin , = cos
⇒
1
15
16
4
0
1
14 + 48 + 64 2 .
√ √
√
ln 3 5 + 14 5
(b)
2
2
2
+
+
= (sin cos )2 + (sin sin )2 + (cos )2
2
2
2
= sin2 + cos2 = 1
and since the ranges of and are sufficient to generate the entire graph,
the parametric equations represent an ellipsoid.
(c) From the parametric equations (with = 1, = 2, and = 3),
we calculate r = cos cos i + 2 cos sin j − 3 sin k and
r = − sin sin i + 2 sin cos j. So r × r = 6 sin2 cos i + 3 sin2 sin j + 2 sin cos k, and the surface
2
2
36 sin4 cos2 + 9 sin4 sin2 + 4 cos2 sin2
area is given by () = 0 0 |r × r | = 0 0
61. To find the region : = 2 + 2 implies + 2 = 4 or 2 − 3 = 0. Thus = 0 or = 3 are the planes where the
surfaces intersect. But 2 + 2 + 2 = 4 implies 2 + 2 + ( − 2)2 = 4, so = 3 intersects the upper hemisphere.
Thus ( − 2)2 = 4 − 2 − 2 or = 2 +
4 − 2 − 2 . Therefore is the region inside the circle 2 + 2 + (3 − 2)2 = 4,
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS
¤
that is, = ( ) | 2 + 2 ≤ 3 .
() =
1 + [(−)(4 − 2 − 2 )−12 ]2 + [(−)(4 − 2 − 2 )−12 ]2
=
2
0
=
√
3
0
2
0
2
1+
=
4 − 2
(−2 + 4) = 2
2
0
2
0
√
3
0
2
√
=
4 − 2
2
=√3
−2(4 − 2 )12
=0
0
= 4
63. Let (1 ) be the surface area of that portion of the surface which lies above the plane = 0. Then () = 2(1 ).
Following Example 10, a parametric representation of 1 is = sin cos , = sin sin ,
= cos and |r × r | = 2 sin . For , 0 ≤ ≤
2
2
2
and for each fixed , − 12 + 2 ≤ 12 or
2
sin cos − 12 + 2 sin2 sin2 ≤ (2)2 implies 2 sin2 − 2 sin cos ≤ 0 or
sin (sin − cos ) ≤ 0. But 0 ≤ ≤
Hence = ( ) | 0 ≤ ≤
,
2
2
−
(1 ) =
,
2
so cos ≥ sin or sin 2 + ≥ sin or −
≤≤
− . Then
2
2 (2) −
0
− (2)
2 sin = 2
2
0
2
≤≤
2
− .
( − 2) sin
= 2 [(− cos ) − 2(− cos + sin )]2
= 2 ( − 2)
0
Thus () = 22 ( − 2).
Alternate solution: Working on 1 we could parametrize the portion of the sphere by = , = , =
Then |r × r | =
1+
2
2
+
=
and
2
2 − 2 − 2
2 − 2 − 2
− 2 − 2
(1 ) =
0 ≤ ( − (2))2 + 2 ≤ (2)2
=
=
Thus () = 42
Notes:
2
2
−2
2
−2
−(2 − 2 )12
=
2 − 2 − 2
= cos
2 (1 − |sin |) = 22
− 1 = 22 ( − 2).
=
=0
2
0
2
−2
2
−2
cos
0
2 − 2 − 2 .
√
2 − 2
2 [1 − (1 − cos2 )12 ]
(1 − sin ) = 22
2
−1
(1) Perhaps working in spherical coordinates is the most obvious approach here. However, you must be careful
in setting up .
(2) In the alternate solution, you can avoid having to use |sin | by working in the first octant and then
multiplying by 4. However, if you set up 1 as above and arrived at (1 ) = 2 , you now see your error.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
327
© Copyright 2024 Paperzz