Section 16.6 Parametric Surfaces and Their Areas
670 ¤ CHAPTER 16 VECTOR CALCULUS
(c) Using Exercise 29, we have that curl curl E = grad div E − ∇2E ⇒
∇2E= grad div E − curl curl E = grad 0 + 1
2
2E
2 [from part (a)] = 1
2
2E
2. (d) As in part (c), ∇2H= grad div H − curl curl H = grad 0 + 1
2
2H
2 [using part (b)] = 1
2
2H
2 . 39. For any continuous function on R3, define a vector field G( ) = h( ) 0 0i where ( ) =
0 ( ) .
Then div G =
(( )) +
(0) +
(0) =
0 ( ) = ( )by the Fundamental Theorem of Calculus. Thus every continuous function on R3is the divergence of some vector field.
16.6 Parametric Surfaces and Their Areas
1. (4 −5 1) lies on the parametric surface r( ) = h + − 2 3 + − i if and only if there are values for and where + = 4, − 2 = −5, and 3 + − = 1. From the first equation we have = 4 − and substituting into the second equation gives 4 − − 2 = −5 ⇔ = 3. Then = 1, and these values satisfy the third equation, so does lie on the surface.
(0 4 6)lies on r( ) if and only if + = 0, − 2 = 4, and 3 + − = 6, but solving the first two equations simultaneoulsy gives =43, = −43 and these values do not satisfy the third equation, so does not lie on the surface.
2. (1 2 1)lies on the parametric surface r( ) =
1 + − + 2 2− 2if and only if there are values for and where 1 + − = 1, + 2 = 2, and 2− 2= 1. From the first equation we have = and substituting into the third equation gives 0 = 1, an impossibility, so does not lie on the surface.
(2 3 3)lies on r( ) if and only if 1 + − = 2, + 2= 3, and 2− 2= 3. From the first equation we have
= + 1and substituting into the second equation gives + 1 + 2= 3 ⇔ 2+ − 2 = 0 ⇔ ( + 2)( − 1) = 0, so = −2 ⇒ = −1 or = 1 ⇒ = 2. The third equation is satisfied by = 2, = 1 so does lie 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
wish to find a more conventional equation for the plane, a normal vector to the plane is a × b =
i j k
1 0 4
1 −1 5
= 4 i − j − k
and an equation of the plane is 4( − 0) − ( − 3) − ( − 1) = 0 or 4 − − = −4.
4. r( ) = 2i+ cos j + sin k, so the corresponding parametric equations for the surface are = 2, = cos ,
= sin . For any point ( ) on the surface, we have 2+ 2 = 2cos2 + 2sin2 = 2= . Since no restrictions are placed on the parameters, the surface is = 2+ 2, which we recognize as a circular paraboloid whose axis is the -axis.
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SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS ¤ 673
= (20)20= (130)2, and = 2− 20= (10) − 02. Each grid curve lies in the plane = (10) − 02and its projection onto the -plane is a parabola = 2with axis the -axis. The surface is graph VI.
15.r( ) = (3− ) i + 2j+ 2k. The parametric equations for the surface are = 3− , = 2, = 2. If we fix then and are constant so each corresponding grid curve is contained in a line parallel to the -axis. (Since = 2≥ 0, the grid curves are half-lines.) If is held constant, then = 2= constant, so each grid curve is contained in a plane parallel to the -plane. Since and are functions of only, the grid curves all have the same shape. The surface is the cylinder shown in graph I.
16. = (1 − ||) cos , = (1 − ||) sin , = . Then 2+ 2 = (1 − ||)2cos2 + (1 − ||)2sin2 = (1 − ||)2, so if is held constant, each grid curve is a circle of radius (1 − ||) in the horizontal plane = . The graph then must be graph VI.
If is held constant, so = 0, we have = (1 − ||) cos 0and = (1 − ||) sin 0. Then = (tan 0), so the grid curves we see running vertically along the surface in the planes = correspond to keeping constant.
17. = cos3 cos3, = sin3 cos3, = sin3. If = 0is held constant then = sin30is 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 ellipses nor straight lines, so graph III is the only possibility. (In fact, the horizontal grid curves here are members of the family = cos3, = sin3and are called astroids.) The vertical grid curves we see on the surface correspond to = 0held constant, as then we have = cos30 cos3, = sin30 cos3so the corresponding grid curve lies in the vertical plane = (tan30)through the -axis.
18. = sin , = cos sin , = sin . If = 0is fixed, then = sin 0is constant, and = sin , = (sin 0) cos describe an ellipse that is contained in the horizontal plane = sin 0. If = 0is fixed, then = sin 0is constant, and
= (cos 0) sin , = sin ⇒ = (cos 0), so the grid curves are portions of lines through the -axis contained in the plane = sin 0(parallel to the -plane). The surface is graph II.
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) = −.
20.From Example 3, parametric equations for the plane through the point (0 −1 5) that contains the vectors a = h2 1 4i and b= h−3 2 5i are = 0 + (2) + (−3) = 2 − 3, = −1 + (1) + (2) = −1 + + 2,
= 5 + (4) + (5) = 5 + 4 + 5.
21.Parametric equations are = , = 4 cos , = 4 sin , 0 ≤ ≤ 5, 0 ≤ ≤ 2.
22.In spherical coordinates, parametric equations are = 4 sin cos , = 4 sin sin , = 4 cos . The intersection of the sphere with the plane = 2 corresponds to = 4 cos = 2 ⇒ cos =12 ⇒ = 3. By symmetry, the intersection of the sphere with the plane = −2 corresponds to = −3 =23 . Thus the surface is described by 0 ≤ ≤ 2,
3 ≤ ≤ 23 .
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674 ¤ CHAPTER 16 VECTOR CALCULUS
22. Solving the equation for gives 2= 12(1 − 2− 32) ⇒ = −
1
2(1 − 2− 32)(since we want the part of the ellipsoid that corresponds to ≤ 0). If we let and be the parameters, parametric equations are = , = ,
= −
1
2(1 − 2− 32).
Alternate solution: The equation can be rewritten as 2+ 2
1√
22 + 2 (1√
3)2 = 1, and if we let = cos and
= 1
√3 sin , then = −
1
2(1 − 2− 32) = −
1
2(1 − 2cos2 − 2sin2) = −
1
2(1 − 2), where 0 ≤ ≤ 1 and 0 ≤ ≤ 2.
Second alternate solution: We can adapt the formulas for converting from spherical to rectangular coordinates as follows.
We let = sin cos , = 1
√2sin sin , = 1
√3cos ; the surface is generated for 0 ≤ ≤ , ≤ ≤ 2.
23. Since the cone intersects the sphere in the circle 2+ 2= 2, =√
2and we want the portion of the sphere above this, we can parametrize the surface as = , = , =
4 − 2− 2where 2+ 2≤ 2.
Alternate solution: Using spherical coordinates, = 2 sin cos , = 2 sin sin , = 2 cos where 0 ≤ ≤ 4 and 0 ≤ ≤ 2.
24. We can parametrize the cylinder as = 3 cos , = , = 3 sin . To restrict the surface to that portion above the -plane and between the planes = −4 and = 4 we require 0 ≤ ≤ , −4 ≤ ≤ 4.
25. In spherical coordinates, parametric equations are = 6 sin cos , = 6 sin sin , = 6 cos . The intersection of the sphere with the plane = 3√
3corresponds to = 6 cos = 3√
3 ⇒ cos = √23 ⇒ = 6, and the plane = 0 (the -plane) corresponds to =2. Thus the surface is described by6 ≤ ≤ 2, 0 ≤ ≤ 2.
26. Using and as the parameters, = , = , = + 3 where 0 ≤ 2+ 2≤ 1. Also, since the plane intersects the cylinder in an ellipse, the surface is a planar ellipse in the plane = + 3. Thus, parametrizing with respect to and , we have = cos , = sin , = 3 + cos where 0 ≤ ≤ 1 and 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 ≤ ≤32. Alternatively, we can regard and as parameters. Then parametric equations are = , = , = −√
9 − 2, where 0 ≤ ≤ 5 and −3 ≤ ≤ 3.
28. The surface appears to be a portion of a sphere of radius 1 centered at the origin. In spherical coordinates, the sphere has equation = 1, and imposing the restrictions2 ≤ ≤ 2,4 ≤ ≤ will give only the portion of the sphere shown. Thus, to graph the surface on a CAS we can either use spherical coordinates with the stated restrictions, or we can use parametric equations: = sin cos , = sin sin , = cos ,2 ≤ ≤ 2,4 ≤ ≤ .
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SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS ¤ 675
29. Using Equations 3, we have the parametrization = ,
= 1
1 + 2cos , = 1
1 + 2sin , −2 ≤ ≤ 2, 0 ≤ ≤ 2.
30. Letting be the angle of rotation about the -axis (adapting Equations 3), we have the parametrization = (1) cos , = ,
= (1) sin , ≥ 1, 0 ≤ ≤ 2.
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.
32. First we graph the surface as viewed from the front, then from two additional viewpoints.
The surface appears as a twisted sheet, and is unusual because it has only one side. (The Möbius strip is discussed in more detail in Section 16.7.)
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676 ¤ CHAPTER 16 VECTOR CALCULUS
33. r( ) = ( + ) i + 32j+ ( − ) k.
r= i + 6 j + kand 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.
34. r( ) = (2+ 1) i + (3+ 1) j + ( + ) k.
r= 2 i + kand r= 32j+ k, so r× r= −32i− 2 j + 62k. Since the point (5 2 3) corresponds to = 2,
= 1, a normal vector to the surface at (5 2 3) is −3 i − 4 j + 12 k, and an equation of the tangent plane is
−3( − 5) − 4( − 2) + 12( − 3) = 0 or 3 + 4 − 12 = −13.
35. r( ) = cos i + sin j + k ⇒ r 13
=
1
2√233 .
r= cos i + sin jand r= − sin i + cos j + k, so a normal vector to the surface at the point
1
2√233 is
r 13
× r 13
=
1
2i+√23j
×
−√23i+12j+ k
=√23i−12j+ k. Thus an equation of the tangent plane at
1
2√233 is√23
−12
−12
−√23
+ 1
−3
= 0 or √23 −12 + =3.
36. r( ) = sin i + cos sin j + sin k ⇒ r
66
=
1 2√4312
.
r= cos i − sin sin j and r= cos cos j + cos k, so a normal vector to the surface at the point
1 2√4312
is
r
66
× r
66
=√ 3 2 i−14j
×
3
4j+√23k
= −√83i−34j+3√83k.
Thus an equation of the tangent plane at
1 2√4312
is −√83
−12
−34
−√43
+3√83
−12
= 0 or
√3 + 6 − 3√
3 =√23 or 2 + 4√
3 − 6 = 1.
37. r( ) = 2i+ 2 sin j + cos k ⇒ r(1 0) = (1 0 1).
r= 2 i + 2 sin j + cos kand 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.
38. r( ) = (1 − 2− 2) i − j − k.
r= −2 i − k and r= −2 i − j. Since the point (−1 −1 −1) corresponds to = 1, = 1, a normal vector to the surface at (−1 −1 −1) is
r(1 1) × r(1 1) = (−2 i − k) × (−2 i − j) = −i + 2 j + 2 k.
Thus an equation of the tangent plane is −1( + 1) + 2( + 1) + 2( + 1) = 0 or − + 2 + 2 = −3.
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SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS ¤ 677
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
() =
1 +
2
+
2
=
1 + (−3)2+ (−2)2 =√ 14
=√
14 () =√ 141
2· 2 · 3
= 3√ 14
40. r( ) = h + 2 − 3 1 + − i ⇒ r= h1 −3 1i, r= h1 0 −1i, and r× r= h3 2 3i. Then by Definition 6,
() =
| r× r| =2 0
1
−1| h3 2 3i | =√ 222
0 1
−1 =√
22 (2)(2) = 4√ 22 41. Here we can write = ( ) =13−13 −23and is the disk 2+ 2≤ 3, so by Formula 9 the area of the surface is
() =
1 +
2
+
2
=
1 +
−13
2
+
−23
2
= √314
= √314() =√314· √
32
=√ 14
42. = ( ) =
2+ 2 ⇒
=1 2
2+ 2−12
· 2 =
2+ 2,
=
2+ 2, and
1 +
2
+
2
=
1 + 2
2+ 2 + 2
2+ 2 =
1 +2+ 2
2+ 2 =√ 2
Here is given by ( )
0 ≤ ≤ 1 2≤ ≤ , so by Formula 9, the surface area of is
() =
√2 =1 0
2
√2 =√ 21
0
− 2
=√ 21
22−1331
0=√
21
2−13
=√62
43. = ( ) =23(32+ 32)and = {( ) | 0 ≤ ≤ 1 0 ≤ ≤ 1 }. Then = 12, = 12and
() =
1 + (√ )2+√2
=1 0
1 0
√1 + +
=1 0
2
3( + + 1)32=1
=0 =231 0
( + 2)32− ( + 1)32
=23
2
5( + 2)52−25( + 1)521
0= 154 (352− 252− 252+ 1) =154(352− 272+ 1) 44. = ( ) = 4 − 22+ and = {( ) | 0 ≤ ≤ 1 0 ≤ ≤ }. Thus, by Formula 9,
() =
1 + (−4)2+ (1)2 =1 0
0
√162+ 2 =1 0 √
162+ 2
=321 ·23(162+ 2)321
0=481(1832− 232) =481(54√ 2 − 2√
2 ) = 1312√ 2
45. = ( ) = with 2+ 2≤ 1, so = , = ⇒
() =
1 + 2+ 2 =2
0
1 0
√2+ 1 =2
0
1
3(2+ 1)32=1
=0
=2
0 1 3
2√ 2 − 1
=23 2√
2 − 1
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1
678 ¤ CHAPTER 16 VECTOR CALCULUS
46. A parametric representation of the surface is = 2+ , = , = with 0 ≤ ≤ 2, 0 ≤ ≤ 2.
Hence r× r = (i + j) × (2 i + k) = i − j − 2 k.
Then
() =
| r× r| =2 0
2 0
√1 + 1 + 42 =2 0 2√
2 + 42
= 2 ·12
√
2 + 42+ ln 2 +√
2 + 422 0
Use trigonometric substitution or Formula 21 in the Table of Integrals
= 6√ 2 + ln
4 + 3√ 2
− ln√ 2 or 6√
2 + ln4 + 3√
√ 2
2 = 6√ 2 + ln
2√ 2 + 3
Note: In general, if = ( ) then r× r= i −
j−
kand () =
1 +
2
+
2
.
47. A parametric representation of the surface is = , = 2+ 2, = with 0 ≤ 2+ 2 ≤ 16.
Hence r× r= (i + 2 j) × (2 j + k) = 2 i − j + 2 k.
Note: In general, if = ( ) then r× r=
i− j +
k, and () =
1 +
2
+
2
. Then
() =
0≤ 2+ 2≤ 16
√1 + 42+ 42 =2
0
4 0
√1 + 42
=2
0 4 0 √
1 + 42 = 2
1
12(1 + 42)324 0=6
6532− 1 48. A parametric representation of the surface is = 2+ 2, = , = with 0 ≤ 2+ 2 ≤ 9.
Hence r× r = (2 i + j) × (2 i + k) = i − 2 j − 2 k.
Note: In general, if = ( ) then r× r= i −
j−
k, and () =
1 +
2
+
2
. Then
() =
0≤ 2+ 2≤ 9
1 + 42+ 42 =2
0
3 0
√1 + 42
=2
0 3 0 √
1 + 42 = 2
1
12(1 + 42)323 0=6
37√ 37 − 1 49. r= h2 0i, r= h0 i, and r× r=
2 −2 22. Then
() =
|r× r| =1 0
2 0
√4+ 422+ 44 =1 0
2 0
(2+ 22)2
=1 0
2
0 (2+ 22) =1 0
1
33+ 22=2
=0 =1 0
8
3 + 42
=8
3 +4331 0= 4 50. The cylinder encloses separate portions of the sphere in the upper and lower halves. The top half of the sphere is
= ( ) =
2− 2− 2and is given by
( ) 2+ 2≤ 2. By Formula 9, the surface area of the upper enclosed portion is
=
1 +
−
2− 2− 2
2
+
−
2− 2− 2
2
=
1 + 2+ 2
2− 2− 2
=
2
2− 2− 2 =
2
0
0
√
2− 2 =
2
0
0
√
2− 2
=
2
0
−√
2− 2
0 = 2
−√
2− 2+√
2− 0
= 2
−√
2− 2 The lower portion of the sphere enclosed by the cylinder has identical shape, so the total area is 2 = 4
−√
2− 2.
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SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS ¤ 681
(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 area is given by () =2
0
0 |r× r| =2
0
0
36 sin4 cos2 + 9 sin4 sin2 + 4 cos2 sin2
60. (a) = cosh cos , = cosh sin , = sinh ⇒
2
2 +2
2 −2
2 = cosh2 cos2 + cosh2 sin2 − sinh2
= cosh2 − sinh2 = 1
and the parametric equations represent a hyperboloid of one sheet.
(b)
(c) r= sinh cos i + 2 sinh sin j + 3 cosh kand
r= − cosh sin i + 2 cosh cos j, so r× r= −6 cosh2 cos i − 3 cosh2 sin j + 2 cosh sinh k.
We integrate between = sinh−1(−1) = − ln 1 +√
2and = sinh−11 = ln 1 +√
2, since then varies between
−3 and 3, as desired. So the surface area is
() =
2
0
ln(1 +√2)
− ln(1 +√2)|r× r|
=
2
0
ln(1 +√2)
− ln(1 +√2)
36 cosh4 cos2 + 9 cosh4 sin2 + 4 cosh2 sinh2
61. To find the region : = 2+ 2implies + 2= 4or 2− 3 = 0. Thus = 0 or = 3 are the planes where the surfaces intersect. But 2+ 2+ 2= 4implies 2+ 2+ ( − 2)2= 4, so = 3 intersects the upper hemisphere.
Thus ( − 2)2= 4 − 2− 2or = 2 +
4 − 2− 2. Therefore is the region inside the circle 2+ 2+ (3 − 2)2= 4, that is, =
( ) | 2+ 2≤ 3.
() =
1 + [(−)(4 − 2− 2)−12]2+ [(−)(4 − 2− 2)−12]2
=
2
0
√3 0
1 + 2
4 − 2 =
2
0
√3 0
2
√4 − 2 =
2
0
−2(4 − 2)12=√ 3
=0
=2
0 (−2 + 4) = 22
0 = 4
62. We first find the area of the face of the surface that intersects the positive -axis. A parametric representation of the surface is
= , =√
1 − 2, = with 2+ 2≤ 1. Then r( ) =
√
1 − 2
⇒ r= h1 0 0i,
r= 0 −√
1 − 2 1
and r× r=
0 −1 −√ 1 − 2
⇒ | r× r| =
1 + 2
1 − 2 = 1
√1 − 2.
() =
2+2≤1
| r× r| =
1
−1
√1−2
−√
1−2
√ 1
1 − 2 = 4
1 0
√1−2 0
√ 1
1 − 2
by the symmetry of the surface
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682 ¤ CHAPTER 16 VECTOR CALCULUS This integral is improper [when = 1], so
() = lim
→1−4
0
√1−2 0
√ 1
1 − 2 = lim
→1−4
0
√1 − 2
√1 − 2 = lim
→1−4
0
= lim
→1−4 = 4 Since the complete surface consists of four congruent faces, the total surface area is 4(4) = 16.
Alternate solution: The face of the surface that intersects the positive -axis can also be parametrized as
r( ) = h cos sin i for −2 ≤ ≤2 and 2+ 2≤ 1 ⇔ 2+ sin2 ≤ 1 ⇔
−
1 − sin2 ≤ ≤
1 − sin2 ⇔ − cos ≤ ≤ cos . Then r= h1 0 0i, r = h0 − sin cos i and r× r= h0 − cos − sin i ⇒ | r× r| = 1, so
() =2
−2
cos
− cos 1 =2
−22 cos = 2 sin 2
−2= 4. Again, the area of the complete surface is 4(4) = 16.
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 1is = sin cos , = sin sin ,
= cos and |r× r| = 2sin . For , 0 ≤ ≤2 and for each fixed ,
−122
+ 2≤1 22
or
sin cos −122
+ 2sin2 sin2 ≤ (2)2implies 2sin2 − 2sin cos ≤ 0 or sin (sin − cos ) ≤ 0. But 0 ≤ ≤2, so cos ≥ sin or sin
2 +
≥ sin or −2 ≤ ≤ 2 − .
Hence =
( ) | 0 ≤ ≤2, −2 ≤ ≤ 2 − . Then
(1) =2 0
(2)−
− (2)2sin = 22
0 ( − 2) sin
= 2[(− cos ) − 2(− cos + sin )]20 = 2( − 2) Thus () = 22( − 2).
Alternate solution: Working on 1we could parametrize the portion of the sphere by = , = , =
2− 2− 2.
Then |r× r| =
1 + 2
2− 2− 2 + 2
2− 2− 2 =
2− 2− 2 and
(1) =
0≤ ( − (2))2+ 2≤ (2)2
2− 2− 2 =
2
−2
cos 0
√
2− 2
=2
−2 −(2− 2)12 = cos
= 0 =2
−22[1 − (1 − cos2)12]
=2
−22(1 − |sin |) = 222
0 (1 − sin ) = 22 2 − 1 Thus () = 42
2 − 1
= 22( − 2).
Notes:
(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 1as above and arrived at (1) = 2, you now see your error.
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