Before starting on double integrals let’s do a quick review
of the definition of a definite integrals for functions of single
variables. First, when working with the
we think of x’s as
coming from the interval . For these integrals we can say that we are
integrating over the interval . Note that this does assume that ,
however, if we have then we can just use the interval .
Now, when we derived
the definition of the definite integral we first thought of this as an area
problem. We first asked what the area
under the curve was and to do this we broke up the interval into n
subintervals of width and choose a point, ,
from each interval as shown below,
Each of the rectangles has height of and we could then use the area of each of
these rectangles to approximate the area as follows.
To get the exact area we then took the limit as n goes to infinity and this was also the
definition of the definite integral.
In this section we want to integrate a function of two
variables, . With functions of one variable we integrated
over an interval (i.e. a
one-dimensional space) and so it makes some sense then that when integrating a
function of two variables we will integrate over a region of (two-dimensional space).
We will start out by assuming that the region in is a rectangle which we will denote as
This means that the ranges for x and y are and .
Also, we will initially assume that although this doesn’t really have to be the
case. Let’s start out with the graph of
the surface S given by graphing over the rectangle R.
Now, just like with functions of one variable let’s not
worry about integrals quite yet. Let’s
first ask what the volume of the region under S (and above the xy-plane
of course) is.
We will first approximate the volume much as we approximated
the area above. We will first divide up into n
subintervals and divide up into m
subintervals. This will divide up R into a series of smaller rectangles
and from each of these we will choose a point . Here is a sketch of this set up.
Now, over each of these smaller rectangles we will construct
a box whose height is given by . Here is a sketch of that.
Each of the rectangles has a base area of and a height of so the volume of each of these boxes is . The volume under the surface S is then approximately,
We will have a double sum since we will need to add up
volumes in both the x and y directions.
To get a better estimation of the volume we will take n and m larger and larger and to get the exact volume we will need to
take the limit as both n and m go to infinity. In other words,
Now, this should look familiar. This looks a lot like the definition of the
integral of a function of single variable.
In fact this is also the definition of a double integral, or more
exactly an integral of a function of two variables over a rectangle.
Here is the official definition of a double integral of a
function of two variables over a rectangular region R as well as the notation that we’ll use for it.
Note the similarities and differences in the notation to
single integrals. We have two integrals
to denote the fact that we are dealing with a two dimensional region and we
have a differential here as well. Note
that the differential is dA instead
of the dx and dy that we’re used to seeing.
Note as well that we don’t have limits on the integrals in this
notation. Instead we have the R written below the two integrals to
denote the region that we are integrating over.
Note that one interpretation of the double integral of over the rectangle R is the volume under the function (and above the xy-plane). Or,
We can use this double sum in the definition to estimate the
value of a double integral if we need to.
We can do this by choosing to be the midpoint of each rectangle. When we do this we usually denote the point
as . This leads to the Midpoint Rule,
In the next section we start looking at how to actually
compute double integrals.