we have
x
2
+
y
2
=
u
2
cos
2
v
+
u
2
sin
2
v
=
u
2
=
z
2
.
The surface is thus the part of the
hyperboloid
x
2
+
y
2
=
z
2
between the planes
z
=

1 and
z
= 1.
15.1, Problem 5
If the surface is parametrized by
r
(
u, v
) =
u
cos
v, u
sin
v, u
2
,
0
≤
v
≤
2
π,
1
≤
u
≤
2
,
we have
x
2
+
y
2
=
u
2
cos
2
v
+
u
2
sin
2
v
=
u
2
=
z
.
The surface is thus the part of the
paraboloid
z
=
x
2
+
y
2
between the planes
z
= 1 and
z
= 2.
15.1, Problem 6
The equation for the sphere of radius 3 centered at the point (1
,
2
,
3) is
(
x

1)
2
+ (
y

2)
2
+ (
z

3)
2
= 3
2
.
If we let
x
=
x

1,
y
=
y

2 and
z
=
z

3, we get the equation (
x
)
2
+ (
y
)
2
+ (
z
)
2
= 3
2
,
which is a sphere centered at the origin with radius 3 in the
x , y , z
coordinates. We can
use spherical coordinates to parametrize the sphere:
x
= 3 cos
θ
sin
φ,
y
= 3 sin
θ
sin
φ,
z
= 3 cos
φ,
0
≤
θ
≤
2
π,
0
≤
φ
≤
π.
Reverting back to the
x, y, z
coordinates, we have thus
x
= 1 + 3 cos
θ
sin
φ,
y
= 2 + 3 sin
θ
sin
φ,
z
= 3 + 3 cos
φ,
0
≤
θ
≤
2
π,
0
≤
φ
≤
π.
or, letting
u
=
θ
and
v
=
φ
,
r
(
u, v
) =
1 + 3 cos
u
sin
v,
2 + 3 sin
u
sin
v,
3 + 3 cos
v ,
0
≤
u
≤
2
π,
0
≤
v
≤
π.
15.1, Problem 7
We use the parametrization
x
=
a
cos
θ
sin
φ,
y
=
a
sin
θ
sin
φ,
z
=
a
cos
φ,
0
≤
θ
≤
2
π,
0
≤
φ
≤
π.
or
r
(
φ, θ
) =
a
cos
θ
sin
φ, a
sin
θ
sin
φ, a
cos
φ ,
0
≤
θ
≤
2
π,
0
≤
φ
≤
π.
We have
r
φ
=
a
cos
θ
cos
φ, a
sin
θ
cos
φ,

a
sin
φ ,
and
r
θ
=

a
sin
θ
sin
φ, a
cos
θ
sin
φ,
0
.
Thus,
r
φ
×
r
θ
=
a
2
cos
θ
sin
2
φ, a
2
sin
θ
sin
2
φ, a
2
cos
φ
sin
φ
=
a
sin
φ
r
(
φ, θ
)
.
2