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In this section we will look at the lone application (aside
from the area and volume interpretations) of multiple integrals in this
material. This is not the first time
that we’ve looked at surface area We
first saw surface area in Calculus II,
however, in that setting we were looking at the surface area of a solid of
revolution. In other words we were
looking at the surface area of a solid obtained by rotating a function about
the x or y axis. In this section we
want to look at a much more general setting although you will note that the
formula here is very similar to the formula we saw back in Calculus II.
Here we want to find the surface area of the surface given
by 
where is a point from the region D in the xy-plane. In this case the
surface area is given by,
Let’s take a look at a couple of examples.
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Example 1 Find
the surface area of the part of the plane that lies in the first octant.
Solution
Remember that the first octant is the portion of the xyz-axis system in which all three
variables are positive. Let’s first
get a sketch of the part of the plane that we are interested in.
 
We’ll also need a sketch of the region D.
Remember that to get the region D we can pretend that we are standing directly over the plane and
what we see is the region D. We can get the equation for the hypotenuse
of the triangle by realizing that this is nothing more than the line where
the plane intersects the xy-plane
and we also know that on the xy-plane. Plugging into the equation of the plane will give us
the equation for the hypotenuse.
Notice that in order to use the surface area formula we
need to have the function in the form and so solving for z and taking the partial derivatives gives,
The limits defining D
are,
The surface area is then,
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Example 2 Determine
the surface area of the part of that lies in the cylinder given by .
Solution
In this case we are looking for the surface area of the
part of where comes from the disk of radius 1 centered at
the origin since that is the region that will lie inside the given cylinder.
Here are the partial derivatives,
The integral for the surface area is,
Given that D is
a disk it makes sense to do this integral in polar coordinates.
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