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Parametric Surfaces

Before we get into surface integrals we first need to talk about how to parameterize a surface
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With surfaces we’ll do something similar
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This is often called the parametric representation of the parametric
surface S
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There are really
nothing more than the components of the parametric representation explicitly written down
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x=u

y = u cos v

z = u sin v

Now if we square y and z and then add them together we get,

y 2 + z 2 = u 2 cos 2 v + u 2 sin 2 v = u 2 ( cos 2 v + sin 2 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 = y 2 + z 2
From the Quadric Surfaces section notes we can see that this is a cone that opens along the x-axis
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Example 2 Give parametric representations for each of the following surfaces
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[Solution]
(b) The elliptic paraboloid x = 5 y 2 + 2 z 2 − 10 that is in front of the yz-plane
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[Solution]
(d) The cylinder y 2 + z 2 = 25
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This one is probably the easiest one of the four to see how to do
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The parametric representation is then,

r ( y, z ) = ( 5 y 2 + 2 z 2 − 10 ) i + y j + z k

[Return to Problems]

(b) The elliptic paraboloid x = 5 y 2 + 2 z 2 − 10 that is in front of the yz-plane
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The parametric
representation stays the same
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This is equivalent to requiring,

5 y 2 + 2 z 2 − 10 ≥ 0

or

5 y 2 + 2 z 2 ≥ 10
[Return to Problems]

(c) The sphere x 2 + y 2 + z 2 = 30
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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
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x = ρ sin ϕ cos θ

y = ρ sin ϕ sin θ

z = ρ cos ϕ

However, we know what ρ is for our sphere and so if we plug this into these conversion
formulas we will arrive at a parametric representation for the sphere
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First we know that we
have the following restriction
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Also, to make sure that we
only trace out the sphere once we will also have the following restriction
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As with the last one this can be tricky until you see how to do it
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In cylindrical coordinates the equation of a cylinder of radius a is given by

r=a

and so the equation of the cylinder in this problem is r = 5
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y = r sin θ

x=x

z = r cos θ

Notice that they are slightly different from those that we are used to seeing
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Finally, we know what r is so we can easily write down a parametric representation for this
cylinder
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Since we haven’t put any restrictions on the “height” of the cylinder there won’t be
any restriction on x
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This can always be done for functions that are in
this basic form
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Let’s take a look at finding the tangent plane to the parametric surface S given by,

r ( u, v ) = x ( u, v ) i + y ( u, v ) j + z ( u, v ) k

First, define

∂x
∂y
∂z
ru ( u, v ) = ( u , v ) i + ( u , v ) j + ( u , v ) k
∂u
∂u
∂u
∂x
∂y
∂z
rv ( u , v ) = ( u , v ) i + ( u , v ) j + ( u , v ) k
∂v
∂v
∂v

Now, provided ru × rv ≠ 0 it can be shown that the vector ru × rv will be orthogonal to the surface
S
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This is an important idea that will be used many times throughout
the next couple of sections
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Example 3 Find the equation of the tangent plane to the surface given by
r ( u, v ) = u i + 2v 2 j + ( u 2 + v ) k
at the point ( 2, 2,3)
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Here are the two individual vectors
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We can easily do this by setting the individual components of the
parametric representation equal to the coordinates of the point in question
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To determine the
correct value of v let’s plug u into the third equation and solve for v
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3= 4+v

⇒

v = −1

Okay so we now know that we’ll be at the point in question when u = 2 and v = −1
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So, provided S is traced out exactly once as ( u , v ) ranges over the points in D the surface area of
S is given by,

A = ∫∫ ru × rv dA
D

Let’s take a look at an example
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Solution
Okay we’ve got a couple of things to do here
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We parameterized a sphere earlier in this section so there isn’t too much to do at this point
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r (θ , ϕ ) = 4sin ϕ cos θ i + 4sin ϕ sin θ j + 4 cos ϕ k

Next we need to determine D
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0 ≤ θ ≤ 2π

Now, we need to determine a range for ϕ
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First, let’s start with the equation of the sphere
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x + y + z = 16
2

2

2

12 + z 2 = 16

z2 = 4

⇒

z = ±2

Now, since we also specified that we only want the portion of the sphere that lies above the xyplane we know that we need z = 2
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Plugging this into the following
conversion formula we get,

z = ρ cos ϕ
2 = 4 cos ϕ
1
⇒
cos ϕ =
2

ϕ=

So, it looks like the range of ϕ will be,

0≤ϕ ≤

π
3

π
3

Finally, we need to determine rθ × rϕ
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rθ (θ , ϕ ) = −4sin ϕ sin θ i + 4sin ϕ cos θ j
rϕ (θ , ϕ ) = 4 cos ϕ cos θ i + 4 cos ϕ sin θ j − 4sin ϕ k
Now let’s take the cross product
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We can finally get the surface area

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