Math 227 Problem Set VI Solutions
1. Identify (or, failing that, describe) each of the following surfaces. In each case, find two different
parametrizations r(u, v) for the surface.
(a) x2 + y 2 + z 2 = 1, z > 0.
(b) z 2 = x2 + y 2 , z > 0.
Solution. (a) This is a hemisphere of radius one centered on the origin. Two parametrizations are
p
r(u, v) = u ı̂ + v ̂ + 1 − u2 − v 2 k̂, u2 + v 2 < 1
r(u, v) = sin v cos u ı̂ + sin v sin u ̂ + cos v k̂, 0 ≤ u < 2π, 0 ≤ v <
π
2
(b) This is a cone centered on the z–axis with vertex at the origin, though the vertex itself is excluded.
Two parametrizations are
p
r(u, v) = u ı̂ + v ̂ + u2 + v 2 k̂, u2 + v 2 < 1
r(u, v) = u cos v ı̂ + u sin v ̂ + u k̂, u > 0, 0 ≤ v < 2π
2. Identify (or, failing that, describe) each of the following surfaces and find an expression for a normal
vector at each point of the surface.
(a)
x = 3 cos θ sin φ,
y = 2 sin θ sin φ,
z = cos φ,
(b)
x = sin v,
y = u,
z = cos v,
(c)
x = (2 − cos v) cos u, y = (2 − cos v) sin u, z = sin v,
(d)
x = r cos θ,
Solution. (a)
an ellipsoid .
x2
9
+
y = r sin θ,
y2
4
z = θ,
0 ≤ θ ≤ 2π,
−1 ≤ u ≤ 3,
−π ≤ u ≤ π,
0 < r < 1,
0≤φ≤π
0 ≤ v ≤ 2π
−π ≤ v ≤ π
0 < θ < 4π
+ z 2 = cos2 θ sin2 φ + sin2 θ sin2 φ + cos2 φ = sin2 φ + cos2 φ = 1. The surface is
Tθ = − 3 sin θ, 2 cos θ, 0 sin φ
=⇒
Tφ = 3 cos θ cos φ, 2 sin θ cos φ, − sin φ
Tθ × Tφ = − (2 cos θ sin φ, 3 sin θ sin φ, 6 cos φ) sin φ
(b)
x2 + z 2 = sin2 v + cos2 v = 1
The surface is a cylinder parallel to the y–axis . Actually, the portion of the cylinder with −1 ≤ y ≤ 3.
Tu = 0, 1, 0
Tv = cos v, 0, − sin v
=⇒
Tu × Tv = − (sin v, 0, cos v)
(c) For each fixed v, x2 + y 2 = (1 − cos v)2 , z = sin v is a circle centred on the z–axis and parallel to
the xy–plane. The surface is a union (over the various possible values of v) of such circles. So it is a
surface of revolution about the z–axis. To see what curve gets revolved around the z–axis, fix u = 0.
This selects the part of the surface in the half of the xz–plane with x > 0. When u = 0,
x = 2 − cos v
y=0
z = sin v
1
=⇒
(x − 2)2 + z 2 = 1
This is a circle centred on x = 2, z = 0. When this circle is rotated about the z–axis a torus results.
Tu = (2 − cos v) − sin u, cos u, 0
Tv = sin v cos u, sin v sin u, cos v
=⇒
Tu × Tv = (2 − cos v) (cos u cos v, sin u cos v, − sin v)
(d) For each fixed θ, as r runs from 0 to 1, x = r cos θ, y = r sin θ, z = θ runs along a straight line
segment, which starts at the z–axis, has length one and is parallel to the xy–plane. As θ increases, the
height of the line segment increases and the line segment rotates about the z–axis.
=⇒
Tθ = − r sin θ, r cos θ, 1
Tr = cos θ, sin θ, 0
Tθ × Tr = (− sin θ, cos θ, −r)
A shaded image of this surface, which is called a Helicoid, is available at
http://www.math.ubc.ca/e feldman/surfaces/.
3. Given a sphere of radius 2 centred at the origin, find the equation for the plane that is tangent to it at
√
the point (1, 1, 2) by considering the sphere as
(a) a surface parametrized by (θ, φ) 7→ (2 cos θ sin φ, 2 sin θ sin φ, 2 cos φ)
(b) a level surface of f (x, y, z) = x2 + y 2 + z 2
p
(c) the graph of g(x, y) = 4 − x2 − y 2
√
Solution. We find a normal vector to the sphere at (1, 1, 2), using each of the three representations.
(a)
Tθ = − 2 sin θ sin φ, 2 cos θ sin φ, 0
Tφ = 2 cos θ cos φ, 2 sin θ cos φ, −2 sin φ
√
The point (1, 1, 2) corresponds to θ = φ = π4 . There
Tθ = − 1, 1, 0
√ Tφ = 1, 1, − 2
=⇒
Tθ × Tφ = −
√
√
2, − 2, −2
√ So one normal vector to the sphere is − √12 Tθ × Tφ = 1, 1, 2 .
√
√
√ (b) ∇f (1, 1, 2) = 2x, 2y, 2z (1,1,√2) = 2 1, 1, 2 is normal to f (x, y, z) = 4 at (1, 1, 2). So, once
√
√
again, we see that (1, 1, 2) is normal to the sphere at (1, 1, 2).
(c) A vector normal to z = g(x, y) at (x0 , y0 ) is given by the formula − gx (x0 , y0 ), −gy (x0 , y0 ), 1 .
p
When g(x, y) = 4 − x2 − y 2
−x
gx = p
4 − x2 − y 2
−y
gy = p
4 − x2 − y 2
=⇒
gx (1, 1) = − √12
=⇒
gx (1, 1) = − √12
√
So one vector normal to the sphere at (1, 1, 2) is
in parts a and b. The tangent plane is
√1 , √1 , 1
2
2
√ √ 1, 1, 2 · x − 1, y − 1, z − 2 = 0
2
or
. This is parallel to the vector found
x+y+
√
2z = 4
4. Find the surface area of the torus obtained by rotating the circle (x − R)2 + z 2 = r2 , y = 0 (the circle
is contained in the xz–plane) about the z–axis.
Solution. The torus may be parametrized by
x = (R + r cos φ) cos θ
y = (R + r cos φ) sin θ
z = r sin φ
0 ≤ φ, θ ≤ 2π
Then
Tθ = (R + r cos φ) − sin θı̂ı + cos θ̂
Tφ = r − sin φ cos θı̂ı − sin φ sin θ̂ + cos φk̂
As − sin θı̂ı + cos θ̂ and − sin φ cos θı̂ı − sin φ sin θ̂ + cos φk̂ are mutually perpendicular unit vectors, their
cross product is a unit vector and
|Tθ × Tφ | = r(R + r cos φ)
=⇒
dS = r(R + r cos φ) dθ dφ
The total surface area of the torus is
r
Z
2π
dφ
0
Z
Z
2π
dθ (R + r cos φ) = 2πr
0
2π
dφ (R + r cos φ) = (2π)2 Rr
0
1
,
y 2 +z 2
5. Show that the surface x = √
where 1 ≤ x < ∞ can be filled but not painted!
Solution. We must show that the volume bounded by the surface is finite, but the surface area is
infinite. The cross–section x = x0 is a circle of radius x10 . So the volume contained in the surface is
V =
Z
∞
π
1
1 2
x
∞
dx = π − x1 1 = π < ∞
q
2
++
On the other hand, using our formula dS = 1 + ∂h
∂y
whose equation is of the form x = h(y, z), the surface area is
A=
Z
1+
y 2 +z 2 ≤1
=
Z
y 2 +z 2 ≤1
≥
=
r
Z
q
1+
q
y 2 +z 2 ≤1
Z 1 Z 2π
dr
0
0
−
y
(y 2 +z 2 )3/2
2
y 2 +z 2
(y 2 +z 2 )3
1
(y 2 +z 2 )2
dy dz =
Z
dy dz =
dθ r r12 = 2π
Z
−
dr
1
r
z
(y 2 +z 2 )3/2
y 2 +z 2 ≤1
y 2 +z 2 ≤1
1
0
+
Z
∂h 2
dydz
∂z
q
1+
1
y 2 +z 2
for the area element of a surface
2
dy dz
1
(y 2 +z 2 )2
dy dz
dy dz
1
= 2π ln r 0 = ∞
6. Find the area of that part of the cylinder x2 + y 2 = 2ay lying outside z 2 = x2 + y 2 , by parametrizing
the cylinder using the cylindrical coordinates θ and z.
Solution. In cylindrical coordinates the cylinder x2 + y 2 = 2ay is r = 2a sin θ. Parametrize the cylinder
x = 2a sin θ cos θ = a sin 2θ
y = 2a sin θ sin θ = a(1 − cos 2θ)
3
z=z
Then
Tθ = 2a cos 2θı̂ı+2a sin 2θ̂
Tz = k̂
=⇒
Tθ ×Tz = −2a cos 2θ̂+2a sin 2θı̂ı
=⇒
dS = 2a dθ dz
The cone z 2 = x2 + y 2 = r2 and the cylinder r = 2a sin θ intersect at z 2 = 4a2 sin2 θ. Here is a side view.
z
y
Area =
Z
2a dθ dz = 2
Z
π
dθ
0
|z|≤2a sin θ
Z
2a sin θ
dz 2a = 4a
Z
π
0
0
π
dθ 2a sin θ = 8a2 − cos θ 0 = 16a2
7. A thin spherical shell of radius a is centred at the origin. Find the centroid, (x̄, ȳ, z̄), of the part of the
sphere that lies in the first octant, by parametrizing the sphere using the spherical coordinates θ and φ.
Here x̄, for example, is the average value of x.
Solution. By symmetry x̄ = ȳ = z̄. Parametrize the sphere using spherical coordinates.
x = a sin φ cos θ
y = a sin φ sin θ
z = a cos φ
Then
Tθ = −a sin φ sin θı̂ı + a sin φ cos θ̂
=⇒
2
Tφ = a cos φ cos θı̂ı + a cos φ sin θ̂ − a sin φk̂
Tθ × Tφ = −a sin φ cos θı̂ı − a2 sin2 φ sin θ̂ − a2 sin φ cos φk̂
2
dS = a2 sin φ dθ dφ
=⇒
As the surface area of the part of the sphere in the first octant is 18 4πa2
x̄ = ȳ = z̄ =
1
πa2 /2
Z
0
π/2
dθ
Z
π/2
dφ (a2 sin φ)(a cos φ) =
0
4
2a π
π 2
Z
0
π/2
dφ sin φ cos φ = a
1
2
π/2
sin2 φ 0 =
a
2