of two variables defined on a closed rectangle:
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In much the same way that our attempt to solve the area problem led to the definition of a definite integral, we now seek to find the volume of a solid and in the process we arrive at the definition of a double integral.
In a similar manner for defining a definite integrals, we consider a function
of two variables defined on a closed rectangle:
and we first suppose that
The graph of
is a surface with equation
Let S be the solid that lies above R and under the graph of
,
that is,
Our goal is to find the volume of S.
The first step of this process is to divide the region R into small rectangles. We can then compare the part of S that lies above the small rectangle, and this forms a thin box. The volume of this box is the area of the box (the base) times the height of the box. If we follow this procedure for all of the rectangles and add the volumes of the corresponding boxes, we get an approximation of the total volume of S:
Now, to find the area exactly, we simply make the boxes on the region R infinitely small (hence there are infinitely many). We do this by taking the limit of the above equation like so:
By taking the limit above, we arrive at the definition of a double integral:
The Double Integral of
over the rectangle R is
if this limit exists.
The double integral of the surface
is the volume between the rectangle and below the surface (if
).
The sum:
is called the double Riemann sum and is used as an approximation to the value of the double integral.
picture of a surface. the have a picture of a surface composed of rectangles, then have the rectangles get smaller (see bottom of page 1004)
The methods that we used for approximating single integrals ( the midpoint
rule, the trapezoidal rule, Simpson's rule) all have counterparts for double
integrals. Here we consider only the midpoint rule for double integrals. .
This means that we use a double Riemann sum to approximate the double
integral, where the sample height of the rectangle is found in the center of
the box on the region R. In other words,
is the midpoint of
![$[x_{i-1},x_{i}]$](16.1 Multiple Integrals__17.png)
and
is the midpoint of
![$[y_{i-1},y_{i}].$](16.1 Multiple Integrals__19.png)
Midpoint Rule for double integrals
where
is the midpoint of
![$[x_{i-1},x_{i}]$](16.1 Multiple Integrals__23.png)
and
is the midpoint of
![$[y_{i-1},y_{i}].$](16.1 Multiple Integrals__25.png)
Recall that the average value of a function
of one variable defined on an interval
![$[a,b]$](16.1 Multiple Integrals__27.png)
is:
In a similar fashion we define the average value of a function
of two variables defined on a rectangle R to be:
where A(R) is the area of R.
Also, if
the equation
says that the box with base R and height
has the same volume as the solid that lies under the graph of
.
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