# Section 14.6: Parametric Surfaces and Their Areas

```Section 14.6: Parametric Surfaces and Their Areas
~
A space curve can be described by a vector function R(t)
of one parameter. Similarly, a
~
surface can be described by a vector function R(u,
v) of two parameters.
Definition: A surface S described by a vector function
~
R(u,
v) = hx(u, v), y(u, v), z(u, v)i
(u, v) ∈ D
is called a parametric surface.
Example: Identify the surface defined by
~
R(u,
v) = hu cos(v), u sin(v), ui.
Note that the parametric equations satisfy z 2 = x2 + y 2 or z =
described by this vector function is a cone.
p
x2 + y 2 . The surface
Example: Find a parametric representation of the cylinder x2 + y 2 = 9, 0 ≤ z ≤ 5.
The cylinder has a simple representation of r = 3 in cylindrical coordinates. This cylinder
can be parameterized by
~ z) = h3 cos θ, 3 sin θ, zi
R(θ,
for 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 5.
Example: Find a parametric representation of the part of the sphere x2 + y 2 + z 2 = 36 that
p
lies above the cone z = x2 + y 2 .
The sphere has a simple representation of ρ = 6 in spherical coordinates. Moreover, the
π
equation of the cone in spherical coordinates is φ = . This part of the sphere can be
4
parameterized by
~ φ) = h6 sin φ cos θ, 6 sin φ sin θ, 6 cos φi
R(θ,
π
for 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ .
4
~
Example: Find an equation of the tangent plane to the parametric surface given by R(u,
v) =
2
hu + v, 3u , u − vi at the point (2, 3, 0).
The tangent vectors are
~ u = h1, 6u, 1i,
R
~ v = h1, 0, −1i.
R
Thus, the normal vector is
~i ~j ~k
~u &times; R
~ v = 1 6u 1
R
1 0 −1
= h−6u, 2, −6ui.
Note that the point (2, 3, 0) corresponds to the parameter values u = v = 1. So the normal
vector at (2, 3, 0) is
~ = h−6, 2, −6i.
N
Therefore, the tangent plane is
−6(x − 2) + 2(y − 3) − 6(z − 0) = 0
−6x + 2y − 6z = −6
3x − y + 3z = 3.
Definition: If a smooth parametric surface S is given by the equation
~
R(u,
v) = hx(u, v), y(u, v), z(u, v)i
(x, y) ∈ D
and S is covered once as (u, v) ranges throughout D, then the surface area of S is
ZZ
~u &times; R
~ v ||dA.
A=
||R
D
~
v) = huv, u + v, u − vi, for
Example: Find the area of the parametric surface given by R(u,
2
2
u + v ≤ 1.
The tangent vectors are
~ u = hv, 1, 1i,
R
~ v = hu, 1, −1i.
R
The cross product of these vectors is
~i
~u &times; R
~v = v
R
u
~j ~k
1 1
1 −1
= h−2, u + v, v − ui.
Thus
~u &times; R
~ v || =
||R
p
4 + (u + v)2 + (v − u)2 =
√
4 + 2u2 + 2v 2 =
Therefore, the area of this surface is
Z
2π
Z
A =
0
0
1
1
√
r 2r2 + 4drdθ
√
r 2r2 + 4dr
0
1
π2 2
3/2 =
(2r + 4) 23
0
√
π √
(6 6 − 2 2).
=
3
Z
= 2π
√
2r2 + 4.
```
The text passage contains several examples and definitions related to parametric surfaces and their areas, including a few mathematical questions and their solutions.
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system
# Section 14.6: Parametric Surfaces and Their Areas

```Section 14.6: Parametric Surfaces and Their Areas
~
A space curve can be described by a vector function R(t)
of one parameter. Similarly, a
~
surface can be described by a vector function R(u,
v) of two parameters.
Definition: A surface S described by a vector function
~
R(u,
v) = hx(u, v), y(u, v), z(u, v)i
(u, v) ∈ D
is called a parametric surface.
Example: Identify the surface defined by
~
R(u,
v) = hu cos(v), u sin(v), ui.
Note that the parametric equations satisfy z 2 = x2 + y 2 or z =
described by this vector function is a cone.
p
x2 + y 2 . The surface
Example: Find a parametric representation of the cylinder x2 + y 2 = 9, 0 ≤ z ≤ 5.
The cylinder has a simple representation of r = 3 in cylindrical coordinates. This cylinder
can be parameterized by
~ z) = h3 cos θ, 3 sin θ, zi
R(θ,
for 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 5.
Example: Find a parametric representation of the part of the sphere x2 + y 2 + z 2 = 36 that
p
lies above the cone z = x2 + y 2 .
The sphere has a simple representation of ρ = 6 in spherical coordinates. Moreover, the
π
equation of the cone in spherical coordinates is φ = . This part of the sphere can be
4
parameterized by
~ φ) = h6 sin φ cos θ, 6 sin φ sin θ, 6 cos φi
R(θ,
π
for 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ .
4
~
Example: Find an equation of the tangent plane to the parametric surface given by R(u,
v) =
2
hu + v, 3u , u − vi at the point (2, 3, 0).
The tangent vectors are
~ u = h1, 6u, 1i,
R
~ v = h1, 0, −1i.
R
Thus, the normal vector is
~i ~j ~k
~u &times; R
~ v = 1 6u 1
R
1 0 −1
= h−6u, 2, −6ui.
Note that the point (2, 3, 0) corresponds to the parameter values u = v = 1. So the normal
vector at (2, 3, 0) is
~ = h−6, 2, −6i.
N
Therefore, the tangent plane is
−6(x − 2) + 2(y − 3) − 6(z − 0) = 0
−6x + 2y − 6z = −6
3x − y + 3z = 3.
Definition: If a smooth parametric surface S is given by the equation
~
R(u,
v) = hx(u, v), y(u, v), z(u, v)i
(x, y) ∈ D
and S is covered once as (u, v) ranges throughout D, then the surface area of S is
ZZ
~u &times; R
~ v ||dA.
A=
||R
D
~
v) = huv, u + v, u − vi, for
Example: Find the area of the parametric surface given by R(u,
2
2
u + v ≤ 1.
The tangent vectors are
~ u = hv, 1, 1i,
R
~ v = hu, 1, −1i.
R
The cross product of these vectors is
~i
~u &times; R
~v = v
R
u
~j ~k
1 1
1 −1
= h−2, u + v, v − ui.
Thus
~u &times; R
~ v || =
||R
p
4 + (u + v)2 + (v − u)2 =
√
4 + 2u2 + 2v 2 =
Therefore, the area of this surface is
Z
2π
Z
A =
0
0
1
1
√
r 2r2 + 4drdθ
√
r 2r2 + 4dr
0
1
π2 2
3/2 =
(2r + 4) 23
0
√
π √
(6 6 − 2 2).
=
3
Z
= 2π
√
2r2 + 4.
```
user
The text passage contains several examples and definitions related to parametric surfaces and their areas, including a few mathematical questions and their solutions.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS