Calculus Answers
13. r(u, v) = u cos vi + u sin vj + vk. The parametric equations for the surface are
x = u cos v y = u sin v z = v
We look at the grid curve first; if we fix v, then x and y parametrize a straight line in the plane z = v which intersects the z−axis.If u is held constant,the projection onto the xy−plane is circular;with z = v, each grid curve is a helix. The surface is a spiraling ramp,graph IV.
14. r(u, v) = u cos vi + u sin vj + sin uk. The corresponding parametric equations for the surface are
x = u cos v y = u sin v z = sin u
, where − π ≤ u ≤ π
If u = u0 is held constant, then x = u0cos v,y = u0sin v so each grid curve is a circle of radius |u0| in the horizontal plane z = sin u0. If v = v0 is constant, then
x= u cos v0, y = u sin v0 ⇒ y = (tan v0)x,
so the grid curve lie in the vertical planes y = kx through the z−axis.
In fact, since x and y are constant multiples of u and z = sin u, each of these is a sine wave. The surface is graph I.
15. r(u, v) = sin vi + cos u sin 2vj + sin u sin 2vk. Parametric equtaions for the surface are
x = sin v y = cos u sin 2v z = sin u sin 2v.
If v = v0 is fixed, then x = sin v0 is constant, and y = (sin 2v0) cos u and z = (sin 2v0) sin u describe a circle of radius | sin 2v0|, so each corre- sponding grid curve is a circle contained in the vertical plane x = sin v0
parallel to yz−plane. The only possible surface is graph II. The grid curves we see running lengthwise along the surface correspond to hold- ing u constant, in which case
y = (cos u0) sin 2v, z = sin u0sin 2v ⇒ z = (tan u0)y, so each grid curve lies in a plane z = ky that includes the x−axis.
16.
x = (1 − u)(3 + cos v) cos 4πu y = (1 − u)(3 + cos v) sin 4πu z = 3u + (1 − u) sin v.
These equations corresponding to graph V:
When u = 0, then
x = 3 + cos v y = 0
z = sin v,
which are equations of a circle with radius 1 in the xz−plane centered at (3, 0, 0).
When u = 12, then
x = 32 +12 cos v y = 0
z = 32 +12 sin v,
which are equations of a circle with radius 12 in the xz−plane centered at (32,0,32).
When u = 1, then x = y = 0 and z = 3, giving the topmost point shown in the graph.This suggests that the grid curves with u constant are the vertically oriented circles visible on the surface. The spiralling grid curves correspond to keeping v constant.
17.
x = cos3ucos3v y = sin3ucos3v z = sin3v.
If v = v0is held constant then z = sin3v0 is constant, so the correspond- ing 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 hori- zontal grid curves here are members of family x = a cos3u, y= a sin3u and are called astroids.) The vertical grid curves we see on the surface correspond to u = u0 held constant, as then we have
x = cos3u0cos3v y = sin3u0cos3v
so the corresponding grid curve lies in the vertical plane y = (tan3u0)x through the z−axis.
18.
x = (1 − |u|) cos v y = (1 − |u|) sin v z = u.
Then
x2+ y2 = (1 − |u|)2cos2v+ (1 − |u|)2sin2v
= (1 − |u|)2,
so if u is held constant, each grid curve is a circle of radius (1 − |u|) in the horizontal plane z = u. The graph then must be graph VI.
If v is held constant, so v = v0, we have x = (1 − |u|) cos v0 and y = (1 − |u|) sin v0. Then y = (tan v0)x, so the grid curve we see running vertically along the surface in the planes y = kx correspond to keeping v constant.
32. First we graph the surfaces 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 M¨obius strip is discussed in more detail in Section 17.7 [ET 16.7].)
54. (a)
ru = a cos vi + b sin vj + 2uk rv = −au sin vi + bu cos vj + 0k
ru× rv = −2bu2cos vi − 2au2sin vj + abuk.
A(S) = Z 2π
0
Z 2
0 |ru× rv|dudv
= Z 2π
0
Z 2 0
p4b2u4cos2v+ 4a2u4sin2v+ a2b2u2dudv
(b)
x2 = a2u2cos2v, y2 = b2u2sin2v, z = u2
⇒ x2 a2 + y2
b2 = u2 = z
which is an elliptic paraboloid. To find D, notice that 0 ≤ u ≤ 2 ⇒ 0 ≤ z ≤ 4 ⇒ 0 ≤ x2
+ y2
≤ 4
Therefore, using Formula 9, we have
A(S) = Z 2a
−2a
Z b
q
4−(x2a2)
−b
q
4−(x2a2)
r
1 + (2x
a2)2+ (2y
b2)2dydx (c)
(d) We substitute a = 2, b = 3 in the integral in part(a) to get A(S) =
Z 2π 0
Z 2 0
2up
9u2cos2v+ 4u2sin2v+ 9dudv.
We use a CAS to estimate the integral accurate to four decimal places. To speed up the calculation, we can set ”Digit:= 7;”(in Maple) or use the approximation command N (in Mathematica).
We find that A(S) ≈ 115.6596
58. We first find the area of the face of the surface that intersects the positive y-axis. A parametric representation of the surface is
x = x
y = √
1 − z2 z = z, with x2+ z2 =≤ 1. Then
r(x, z) = hx,√
1 − z2, zi ⇒ rx = h1, 0, 0i, rz= h0, − z
√1 − z2,1i and
rx× rz = h0, −1, − z
√1 − z2i ⇒ |rx× rz| = r
1 + z2
1 − z2 = 1
√1 − z2.
A(S) = Z Z
x2+z2≤1
|rx× rz|dA = Z 1
−1
Z −√1−z2
√1−z2
√ 1
1 − z2dxdz
= 4 Z 1
0
Z √1−z2
0
√ 1
1 − z2dxdz
by the symmetry of the surface. The integral is improper[when z = 1], so
A(S) = lim
t→1−4 Z t
0
Z √1−z2
0
√ 1
1 − z2dxdz
= lim
t→1−4 Z t
0
√1 − z2
√1 − z2dz
= lim
t→1−4 Z t
0
dz = lim
t→1−4t = 4
Since the complete surface consists of four congruent faces, the total surface is (4(4)) = 16.
Alternate solution: The face of the surface that intersects the positive y-axis can also be parametrized as
r(x, θ) = hx, cos θ, sin θi for − π
2 ≤ θ ≤ π
2 and x2+ z2 ≤ 1
⇔ x2+ sin2θ ≤ 1
⇔ −p
1 − sin2θ ≤ x ≤p
1 − sin2θ
⇔ − cos θ ≤ x ≤ cos θ.
Then rx = h1, 0, 0i, rθ = h0, − sin θ, cos θi and rx×rz = h0, − cos θ, − sin θi ⇒
|rx× rz| = 1, so
A(S) = Z π2
−π2
Z cos θ
− cos θ
1dxdθ
= Z π2
−π2
2 cos θdθ = 2 sin θ]
π 2
−π2 = 4.
Again, the area of the complete surface is 4(4) = 16.
59. Let A(S1) be the surface area of that portion of the surface which lies above the plane z = 0. Then A(S) = 2A(S1). Following Example 10, a parametric representation of S1 is
x = a sin φ cos θ y = a sin φ sin θ z = a cos φ, and |rφ× rθ| = a2sin φ.
For D,0 ≤ φ ≤ π2 and for each fixed φ, (x − a
2)2+ y2 ≤ (a
2)2 or [a sin φ cos θ −a
2]2+ a2sin2φsin2θ ≤ (a 2)2
⇒ a2sin2φ− a2sin φ cos θ ≤ 0 or sin φ(sin φ − cos θ) ≤ 0.
But 0 ≤ φ ≤ π2, so cos θ ≥ sin φ or sin(π2 + θ) ≥ sin φ or φ − π2 ≤ θ ≤
π
2 − φ. Hence D = {(φ, θ)|0 ≤ φ ≤ π2, φ− π2 ≤ θ ≤ π2 − φ}. Then A(S1) =
Z π2
0
Z π2−φ
φ−π2
a2sin φdθdφ
= a2[(−π cos φ) − 2(−φ cos φ + sin φ)]
π
02 = a2(π − 2).
Thus A(S) = 2a2(π − 2).
Alternate solution: Working on S1 we could parametrize the portion of the sphere by
x = x
y = y
z = pa2− x2 − y2. Then
|rx× ry| = s
1 + x2
a2− x2− y2 + y2
a2− x2− y2 = a pa2− x2− y2 and
A(S1) =
Z Z
0≤(x−(a2)2)2+y2≤(a2)2
a
pa2− x2− y2dA
= Z π2
−π2
Z a cos θ 0
√ a
a2− r2drdθ
= Z π2
−π2
−a(a2 − r2)1/2]r=a cos θr=0 dθ
= Z π2
−π2
a2[1 − (1 − cos2θ)1/2]dθ
= Z π2
−π2
a2(1 − | sin θ|)dθ = 2a2 Z π2
0 (1 − sin θ)dθ = 2a2(π 2 − 1).
Thus A(S) = 4a2(π2 − 1) = 2a2(π − 2).
Notes:
(1) Perhaps working in spherical coordinates is the most obvious ap- proch here. However, you must be careful in setting up D.
(2) In the alternate solution, you can avoid having to use | sin θ| by working in the first octant and then multiplying by 4. However, if you set up S1 as above and arrived at A(S1) = a2π, you now see your error.
