1
Parametric surfaces
A curve:
x = x(t) , y = y(t) , z = z(t) , t0 ≤ t ≤ t1
z
Pa
a
orientation
A surface:
b
Pb
t
x
orientation
x = x(u, v) , y = y(u, v) , z = z(u, v) , (u, v) ∈ R
1
y
Example. z = 4 − x2 − y 2
Rectangular coordinates: x = u, y = v, z = 4 − u2 − v 2
Polar coordinates: x = r cos θ, y = r sin θ, z = 4 − r2
2
z
6
Polar coordinates are useful for
surfaces of revolution
generated by revolving a curve
5
4
3
2
z = f (x), x ≥ 0 in the xz-plane,
1
or, equivalently, a curve
1
2
3
4
5
6
x
z = f (y), y ≥ 0 in the yz-plane
about the z-axis.
These surfaces are graphs of
p
functions z = f ( x2 + y 2),
and can be parametrised as
x = r cos θ, y = r sin θ, z = f (r)
Example. A right cone of height h and base radius a oriented along
the z-axis, with vertex pointing up, and with the base located at z = 0.
r
x = r cos θ , y = r sin θ , z = h(1 − ) ,
a
3
0 ≤ r ≤ a , 0 ≤ θ ≤ 2π
Cylindrical coordinates
x = r cos θ , y = r sin θ , z
r ≥ 0 , 0 ≤ θ ≤ 2π
Plane: z = const
Circular cylinder: r = const
Half-plane: θ = const
Cylindrical coordinates are useful for
z
surfaces of revolution generated
6
by revolving a curve x = f (z)
5
in the xz-plane or, equivalently,
4
a curve y = f (z) in the yz-plane
3
about the z-axis.
2
1
These surfaces are parametrised as
1
2
3
4
5
6
x
x = f (ζ) cos θ, y = f (ζ) sin θ, z = ζ
Example. A right cone of height h
and base radius a oriented along the z-axis, with vertex pointing up,
and with the base located at z = 0.
ζ
ζ
x = (1− )a cos θ , y = (1− )a sin θ , z = ζ ,
h
h
4
0 ≤ ζ ≤ h , 0 ≤ θ ≤ 2π
Spherical coordinates
x = ρ sin φ cos θ ,
y = ρ sin φ sin θ ,
z = ρ cos φ ,
ρ ≥ 0 , 0 ≤ θ ≤ 2π ,
0 ≤ φ ≤ π.
Sphere: r = const
Half-plane: θ = const
Cone: φ = const
Example. x2 + y 2 + z 2 = 9 ⇒ ρ = 3
x = 3 sin φ cos θ , y = 3 sin φ sin θ , z = 3 cos φ ,
0 ≤ θ ≤ 2π , 0 ≤ φ ≤ π .
The constant φ-curves are the lines of latitude.
The constant θ-curves are the lines of longitude.
q
2
2
π
⇒
φ
=
Example. z = x +y
3
6
x=
1
2 ρ cos θ ,
y=
1
2 ρ sin θ ,
√
z=
3
2 ρ,
ρ ≥ 0 , 0 ≤ θ ≤ 2π .
The constant ρ-curves are half-lines.
The constant θ-curves are circles.
5
Sphere is an example of a
z
surface of revolution generated by
3.0
2.5
revolving a parametric curve
2.0
x = f (t), z = g(t) or, equivalently,
1.5
a parametric curve y = f (t), z = g(t)
1.0
about the z-axis.
0.5
Such a surface is parametrised as
1
2
3
x = f (t) cos θ , y = f (t) sin θ , z = g(t) .
Example. Torus.
Similarly, we can get surfaces of revolution by
revolving a parametric curve
y = f (u), x = g(u) or, equivalently,
a parametric curve z = f (u), x = g(u)
about the x-axis.
They are parametrised as
x = g(u) , y = f (u) cos v ,
z = f (u) sin v
Finally, we can get surfaces of
revolution by revolving a parametric curve z = f (u), y = g(u) or,
equivalently, a parametric curve x = f (u), y = g(u) about the y-axis.
They are parametrised as
x = f (u) sin v , y = g(u) , z = f (u) cos v
6
4
x
Vector-valued functions of two variables
The vector form of parametric eqs for a surface
~r = ~r(u, v) = x(u, v)~i + y(u, v) ~j + z(u, v) ~k
~r(u, v) is a vector-valued functions of two variables.
Partial derivatives
~ru =
∂~r ∂x ~ ∂y ~ ∂z ~
=
i+
j+
k
∂u ∂u
∂u
∂u
~rv =
∂~r ∂x ~ ∂y ~ ∂z ~
=
i+
j+
k
∂v ∂v
∂v
∂v
Tangent planes to parametric surfaces
Let σ be a parametric surface in 3-space.
Definition. A plane is said to be tangent to σ at P0 provided a line
through P0 lies in the plane if and only if it is a tangent line at P0 to
a curve on σ.
σ : ~r(u, v) , P0(a, b, c) ∈ σ
a = x(u0, v0) , b = y(u0, v0) , c = z(u0, v0)
If
If
∂~r
∂u
∂~r
∂v
6= 0 then it is tangent to the constant v-curve.
6= 0 then it is tangent to the constant u-curve.
∂~r
∂~r
Thus, if ∂u
× ∂v
6= 0 at (u0, v0) then it is orthogonal to both tangent
vectors and is normal to the tangent plane and the surface at P0.
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~n =
∂~r
∂u
∂~r
∂u
∂~r
× ∂v
×
∂~r ∂v
is called the principle normal vector
to the surface ~r = ~r(u, v) at (u0, v0).
Thus, the tangent plane equation is
~n · (~r − ~r0) = 0
Example. Tangent plane to
x = uv , y = u , z = v 2 at (2, −1)
Figure 1:
Whitney’s umbrella.
Answer : x + y + z = 1
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Surface area
Let ~r = x(u, v)~i + y(u, v) ~j + z(u, v) ~k be a smooth parametric
surface on a region R of the uv-plane, that is
∂~r
∂u
∂~r
× ∂v
6= 0 on R, and
therefore there is a tangent plane for every (u, v) ∈ R.
Rk : ∆Ak = ∆uk ∆vk
∂~r
∂~r
∆Sk ≈ Area of parallelogram = ∆uk × ∆vk ∂u
∂v
∂~r ∂~r ∂~r ∂~r = × ∆uk ∆vk = × ∆Ak
∂u ∂v
∂u ∂v
Thus,
n X
∂~r ∂~r × ∆Ak
S≈
∂u ∂v k=1
In the limit n → ∞
¨ ∂~r ∂~r × dA
S=
∂v R ∂u
Examples. Surfaces of revolution: Sphere, cone, torus
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Surface area of surfaces of the form z = f (x, y)
x = u , y = v , z = f (u, v) ,
∂~r ~ ∂z ~
=i+
k,
∂u
∂u
~r = u~i + v ~j + z ~k
∂~r ~ ∂z ~
=j+
k
∂v
∂v
∂~r ∂~r ~ ∂z ~ ∂z ~
×
=k−
j−
i
∂u ∂v
∂v
∂u
s
2 2
∂~r ∂~r × = 1 + ∂z + ∂z
∂u ∂v ∂u
∂v
¨
s
1+
S=
R
∂z
∂x
2
+
Example. S √
of z = x2 + y 2 below z = 1.
Answer : π6 (5 5 − 1) ≈ 5.330
10
∂z
∂y
2
dA