text: # 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.
  ```