It is now time to think about integrating functions over
some surface, S, in threedimensional
space. Let’s start off with a sketch of
the surface S since the notation can
get a little confusing once we get into it.
Here is a sketch of some surface S.
The region S will
lie above (in this case) some region D
that lies in the xyplane. We used a rectangle here, but it doesn’t have
to be of course. Also note that we could
just as easily looked at a surface S
that was in front of some region D in
the yzplane or the xzplane. Do not get so locked into the xyplane that you can’t do problems that
have regions in the other two planes.
Now, how we evaluate the surface integral will depend upon
how the surface is given to us. There
are essentially two separate methods here, although as we will see they are
really the same.
First, let’s look at the surface integral in which the
surface S is given by . In this case the surface integral is,
Now, we need to be careful here as both of these look like
standard double integrals. In fact the
integral on the right is a standard double integral. The integral on the left however is a surface
integral. The way to tell them apart is
by looking at the differentials. The
surface integral will have a dS while
the standard double integral will have a dA.
In order to evaluate a surface integral we will substitute
the equation of the surface in for z
in the integrand and then add on the often messy square root. After that the integral is a standard double
integral and by this point we should be able to deal with that.
Note as well that there are similar formulas for surfaces
given by (with D
in the xzplane) and (with D
in the yzplane). We will see one of these formulas in the
examples and we’ll leave the other to you to write down.
The second method for evaluating a surface integral is for
those surfaces that are given by the parameterization,
In these cases the surface integral is,
where D is the
range of the parameters that trace out the surface S.

Before we work some examples let’s notice that since we can
parameterize a surface given by as,
we can always use this form for these kinds of surfaces as
well. In fact it can be shown that,
for these kinds of surfaces.
You might want to verify this for the practice of computing these cross
products.
Let’s work some examples.
Example 1 Evaluate
where S
is the portion of the plane that lies in the 1^{st} octant and is in front of the yzplane.
Solution
Okay, since we are looking for the portion of the plane
that lies in front of the yzplane
we are going to need to write the equation of the surface in the form . This is easy enough to do.
Next we need to determine just what D is. Here is a sketch of
the surface S.
Here is a sketch of the region D.
Notice that the axes are labeled differently than we are
used to seeing in the sketch of D. This was to keep the sketch consistent with
the sketch of the surface. We arrived
at the equation of the hypotenuse by setting x equal to zero in the equation of the plane and solving for z.
Here are the ranges for y
and z.
Now, because the surface is not in the form we can’t use the formula above. However, as noted above we can modify this
formula to get one that will work for us.
Here it is,
The changes made to the formula should be the somewhat
obvious changes. So, let’s do the
integral.
Notice that we plugged in the equation of the plane for
the x in the integrand. At this point we’ve got a fairly simple
double integral to do. Here is that
work.

Example 2 Evaluate
where S
is the upper half of a sphere of radius 2.
Solution
We gave the parameterization of a sphere in the previous section. Here is the parameterization for this
sphere.
Since we are working on the upper half of the sphere here
are the limits on the parameters.
Next, we need to determine . Here are the two individual vectors.
Now let’s take the cross product.
Finally, we 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. The surface integral is then,
Don’t forget that we need to plug in for x, y
and/or z in these as well, although
in this case we just needed to plug in z. Here is the evaluation for the double
integral.

Example 3 Evaluate
where S
is the portion of the cylinder that lies between and .
Solution
We parameterized up a cylinder in the previous section. Here is the parameterization of this
cylinder.
The ranges of the parameters are,
Now we need . Here are the two vectors.
Here is the cross product.
The magnitude of this vector is,
The surface integral is then,

Example 4 Evaluate
where S
is the surface whose side is the cylinder ,
whose bottom is the disk in the xyplane
and whose top is the plane .
Solution
There is a lot of information that we need to keep track
of here. First, we are using pretty
much the same surface (the integrand is different however) as the previous
example. However, unlike the previous
example we are putting a top and bottom on the surface this time. Let’s first start out with a sketch of the
surface.
Actually we need to be careful here. There is more to this sketch than the
actual surface itself. We’re going to
let be the portion of the cylinder that goes
from the xyplane to the
plane. In other words, the top of the
cylinder will be at an angle. We’ll
call the portion of the plane that lies inside (i.e. the cap on the cylinder) .
Finally, the bottom of the cylinder
(not shown here) is the disk of radius in the xyplane
and is denoted by .
In order to do this integral we’ll need to note that just
like the standard double integral, if the surface is split up into pieces we
can also split up the surface integral.
So, for our example we will have,
We’re going to need to do three integrals here. However, we’ve done most of the work for
the first one in the previous example so let’s start with that.
: The Cylinder
The parameterization of the cylinder and is,
The difference between this problem and the previous one
is the limits on the parameters. Here
they are.
The upper limit for the z’s is the plane so we can just plug that in. However, since we are on the cylinder we
know what y is from the
parameterization so we will also need to plug that in.
Here is the integral for the cylinder.
: Plane on Top of the Cylinder
In this case we don’t need to do any parameterization
since it is set up to use the formula that we gave at the start of this
section. Remember that the plane is
given by . Also note that, for this surface, D is the disk of radius centered at the origin.
Here is the integral for the plane.
Don’t forget that we need to plug in for z!
Now at this point we can proceed in one of two ways. Either we can proceed with the integral or
we can recall that is nothing more than the area of D and we know that D is the disk of radius and so there is no reason to do the
integral.
Here is the remainder of the work for this problem.
: Bottom of the Cylinder
Again, this is set up to use the initial formula we gave
in this section once we realize that the equation for the bottom is given by and D
is the disk of radius centered at the origin. Also, don’t forget to plug in for z.
Here is the work for this integral.
We can now get the value of the integral that we are
after.
