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Before we get into surface integrals we first need to talk
about how to parameterize a surface.
When we parameterized a curve we took values of t from some interval 
and plugged them into
and the resulting set of vectors will be the position
vectors for the points on the curve.
With surfaces we’ll do something similar. We will take points, ,
out of some two-dimensional space D
and plug them into
and 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.
We will sometimes need to write the parametric equations for a surface.
There are really nothing more than the components of the parametric
representation explicitly written down.
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Example 1 Determine
the surface given by the parametric representation
Solution
Let’s first write down the parametric equations.
Now if we square y
and z and then add them together we
get,
So, we were able to eliminate the parameters and the
equation in x, y, and z is given by,
From the Quadric Surfaces
section notes we can see that this is a cone that opens along the x-axis.
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We are much more likely to need to be able to write down the
parametric equations of a surface than identify the surface from the parametric
representation so let’s take a look at some examples of this.
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Example 2 Give
parametric representations for each of the following surfaces.
(a) The
elliptic paraboloid . [Solution]
(b) The
elliptic paraboloid that is in front of the yz-plane.
[Solution]
(c) The
sphere . [Solution]
(d) The
cylinder . [Solution]
Solution
(a) The elliptic
paraboloid .
This one is probably the easiest one of the four to see
how to do. Since the surface is in the
form we can quickly write down a set of
parametric equations as follows,
The last two equations are just there to acknowledge that
we can choose y and z to be anything we want them to
be. The parametric representation is
then,
[Return to Problems]
(b) The elliptic
paraboloid that is in front of the yz-plane.
This is really a restriction on the previous parametric
representation. The parametric
representation stays the same.
However, since we only want the surface that lies in front
of the yz-plane we also need to
require that . This is equivalent to requiring,
[Return to Problems]
(c) The sphere .
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,
and so the equation of this sphere (in spherical
coordinates) is . Now, we also have the following conversion
formulas for converting Cartesian coordinates into spherical coordinates.
However, we know what is for our sphere and so if we plug this
into these conversion formulas we will arrive at a parametric representation
for the sphere. Therefore, the
parametric representation is,
All we need to do now is come up with some restriction on
the variables. First we know that we
have the following restriction.
This is enforced upon us by choosing to use spherical
coordinates. Also, to make sure that
we only trace out the sphere once we will also have the following
restriction.
[Return to Problems]
(d) The
cylinder .
As with the last one this can be tricky until you see how
to do it. In this case it makes some sense
to use cylindrical coordinates since they can be easily used to write down
the equation of a cylinder.
In cylindrical coordinates the equation of a cylinder of
radius a is given by
and so the equation of the cylinder in this problem is .
Next, we have the following conversion formulas.
Notice that they are slightly different from those that we
are used to seeing. We needed to change
them up here since the cylinder was centered upon the x-axis.
Finally, we know what r
is so we can easily write down a parametric representation for this cylinder.
We will also need the restriction to make sure that we don’t retrace any
portion of the cylinder. Since we
haven’t put any restrictions on the “height” of the cylinder there won’t be
any restriction on x.
[Return to Problems]
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In the first part of this example we used the fact that the
function was in the form to quickly write down a parametric
representation. This can always be done
for functions that are in this basic form.
Okay, now that we have practice writing down some parametric
representations for some surfaces let’s take a quick look at a couple of
applications.
Let’s take a look at finding the tangent plane to the
parametric surface S given by,
First, define
Now, provided it can be shown that the vector will be orthogonal to the surface S.
This means that it can be used for the normal vector that we need in
order to write down the equation of a tangent plane. This is an important idea that will be used
many times throughout the next couple of sections.
Let’s take a look at an example.
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Example 3 Find
the equation of the tangent plane to the surface given by
at the point .
Solution
Let’s first compute . Here are the two individual vectors.
Now the cross product (which will give us the normal
vector ) is,
Now, this is all fine, but in order to use it we will need
to determine the value of u and v that will give us the point in
question. We can easily do this by
setting the individual components of the parametric representation equal to
the coordinates of the point in question.
Doing this gives,
Now, as shown, we have the value of u, but there are two possible values of v. To determine the
correct value of v let’s plug u into the third equation and solve
for v. This should tell us what the correct value
is.
Okay so we now know that we’ll be at the point in question
when and . At this point the normal vector is,
The tangent plane is then,
You do remember how to
write down the equation of a plane, right?
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The second application that we want to take a quick look at
is the surface area of the parametric surface S given by,
and as we will see it again comes down to needing the vector
.
So, provided S is
traced out exactly once as ranges over the points in D the surface area of S
is given by,
Let’s take a look at an example.
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Example 4 Find
the surface area of the portion of the sphere of radius 4 that lies inside
the cylinder 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.
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 .
Now, we need to determine a range for . This will take a little work, although it’s
not too bad. First, let’s start with the
equation of the sphere.
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.
Now, since we also specified that we only want the portion
of the sphere that lies above the xy-plane
we know that we need . We also know that . Plugging this into the following conversion
formula we get,
So, it looks like the range of will be,
Finally, we need to determine . Here are the two individual vectors.
Now let’s take the cross product.
We now need the magnitude of this,
We can drop the absolute value bars in the sine because
sine is positive in the range of that we are working with.
We can finally get the surface area.
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