Parametric Surfaces
Group 4
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Boateng Freda Owusu
Alornyo Maxwell Mawunyo
Sefe Sewoenam Kwasi
Amoah Naphtali
Kwarteng Elvis Kwadwo
Kumah David Jnr.
Gyimah Ntim Gyakari Obed
Awudu Seidu Rashid Tobrazune
Dadzie Dennis
July 24, 2023
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Parametric Surfaces
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Introduction - Definition
A parametric surface is a surface in Euclidean space defined by a
parametric equation with two parameters.
Parametric representation is a general way to specify a surface,
frequently used along with implicit representation.
Surfaces in vector calculus theorems like Stokes’ theorem and the
divergence theorem are often given in parametric form.
Various properties of the surface, such as curvature, arc length,
surface area, and differential geometric invariants, can be computed
from a given parametrization.
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Parametric Surfaces
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Parametric Representation
Given the function r (t) = x(t) + y (t)j + z(t)k on the interval [a, b], for
points u and v out of some two-dimensional space D, we can plug them
into:
r (u, v ) = x(u, v )i + y (u, v )j + z(u, v )k
The resulting set of vectors will be the position vectors for the points on
the surface S that we are trying to parameterize. This is often called the
parametric representation of the parametric surface S.
Parametric Equations for a Surface:
x = x(u, v )
y = y (u, v )
z = z(u, v )
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Example 1
Determine the surface given by the parametric representation:
r(u, v ) = ui + u cos v j + u sin v k
Solution: Let’s first write down the parametric equations.
x(u, v ) = u
y (u, v ) = u cos v
z(u, v ) = u sin v
Now if we square y and z and then add them together, we get:
y 2 + z 2 = u 2 (cos2 v + sin2 v ) = u 2 (cos2 v + sin2 v ) = u 2 = x 2
So, we were able to eliminate the parameters, and the equation in x, y ,
and z is given by:
x2 = y2 + z2
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Example 2
Give the parametric representations for the elliptic paraboloid surface
x = 5y 2 + 2z 2 − 10
Solution
The elliptic paraboloid x 2 + y 2 − z = 10
Since the surface is in the form x = f (y , z), we can quickly write down a
set of parametric equations as follows:
x = y 2 + z 2,
y = y,
z =z
The parametric representation is then:
r(y , z) = (5y 2 + 2z 2 − 10)i + y j + zk
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Parametric Surfaces
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Example 3
Write down the parametric representations for the elliptic paraboloid
x = 5y 2 + 2z 2 − 10 that is in front of the yz-plane.
Solution
The elliptic paraboloid x 2 + y 2 − z = 10 that is in front of the yz plane.
This is really a restriction on the previous parametric representation. The
parametric representation stays the same:
r(y , z) = (5y 2 + 2z 2 − 10)i + y j + zk
However, since we only want the surface that lies in front of the yz-plane,
we also need to require that x ≥ 0.
This is equivalent to requiring:
5y 2 + 2z 2 − 10 ≥ 10
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Example 4
Give the parametric representations for the sphere x 2 + y 2 + z 2 = 30
Solution
The sphere x 2 + y 2 + z 2 = 30.
This one can be a little tricky until you see how to do it. In spherical
coordinates, we know that the equation of a sphere of radius a is given by
ρ=√
a, and so the equation of this sphere (in spherical coordinates) is
ρ = 30. Now, we also have the following conversion formulas for
converting Cartesian coordinates into spherical coordinates:
x = ρ sin ϕ cos θ,
Group 4 (KNUST)
y = ρ sin ϕ sin θ,
Parametric Surfaces
z = ρ cos ϕ
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Cont....
However, we know what ρ is for our sphere, and if we plug this into these
conversion formulas, we will arrive at a parametric representation for the
sphere.
Therefore, the parametric representation is:
r (θ, ϕ) =
Group 4 (KNUST)
√
30ϕ cos θi +
√
30 sin ϕ sin θj +
Parametric Surfaces
√
30 cos ϕk
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Example 5
Find the surface area of the portion of the sphere of radius 4 that lies
inside the cylinder x 2 + y 2 = 12 and above the xy − plane.
Solution
Okay, we’ve got a couple of things to do here. First, we need the
parameterization of the sphere. We parameterized a sphere earlier in this
section, so there isn’t too much to do at this point. Here is the
parameterization.


4 sin θ cos ϕ
r(θ, ϕ) =  4 sin θ sin ϕ 
4 cos θ
Next, we need to determine D. Since we are not restricting how far around
the z-axis we are rotating with the sphere, we can take the following range
for θ.
0 ≤ θ ≤ 2π
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Solution Cont...
Now, we need to determine a range for ϕ. This will take a little work,
although it’s not too bad. First, let’s start with the equation of the sphere.
x 2 + y 2 + z 2 = 16
Now, if we substitute the equation for the cylinder into this equation, we
can find the value of z where the sphere and the cylinder intersect.
x 2 + y 2 + z 2 = 16
12 + z 2 = 16
z2 = 4
z = ±2
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Sol Cont....
Since we only want the portion of the sphere that lies above the xy -plane,
we know that we need z = 2. We also know that ρ = 4. Plugging this into
the following conversion formula, we get,
cos ϕ =
z
2
1
= =
ρ
4
2
⇒
ϕ=
π
3
So, it looks like the range of ϕ will be,
π
3
Finally, we need to determine rρ × rϕ . Here are the two individual vectors.




4 sin θ sin ϕ
4 cos θ cos ϕ
rθ = −4 sin θ cos ϕ , rϕ =  4 cos θ sin ϕ 
0
−4 sin θ
0≤ϕ≤
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Sol Cont...
Now let’s take the cross product.
i
j
k
0
rθ × rϕ = −4 sin θ sin ϕ 4 sin θ cos ϕ
4 cos θ cos ϕ 4 cos θ sin ϕ −4 sin θ
= −16 sin2 ϕ cos θi − 16 sin2 ϕ cos θj − 16 sin ϕ cos ϕk
We now need the magnitude of this:
p
||r θ × r ϕ|| 256 sin4 ϕ cos2 θ + 256 sin4 ϕ sin2 θ + 256 sin2 ϕ cos2 ϕ
q
= 256 sin4 ϕ(cos2 θ + sin2 θ) + 256 sin2 θ + 256 sin2 ϕ cos2 ϕ
q
= 256 sin2 ϕ(sin2 ϕ + cos2 ϕ)
p
= 16 sin2 ϕ
= 16|ϕ|
= 16ϕ
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Sol. Cont.....
We can drop the absolute value bars in the sine because sine is positive in
the range of ϕ that we are working with.
We can finally get the surface area
Z Z
A=
16 sin ϕdA
Z
2π
π
3
Z
=
0
16 sin ϕ dϕ dθ
0
= 16π
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Conclusion
Parametric surfaces are a fundamental concept in mathematics and
computer graphics used to describe three-dimensional shapes in a
parameterized manner. Instead of representing a surface directly as an
equation in the form of
F (x, y , z) = 0,
parametric surfaces define each point on the surface using a set of
continuous parameters.
In general, a parametric surface can be represented by two functions,
typically denoted as X (u, v ), Y (u, v ), and Z (u, v ). Here, u and v are
the parameters that define the position of a point on the surface. As
the parameters u and v vary within their respective domains, the
functions X (u, v ), Y (u, v ), and Z (u, v ) generate the coordinates of
the points lying on the surface.
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