Setting up a double integral and approximating it by double sums
Consider the function
over the rectangular region
(
[link] ).
Set up a double integral for finding the value of the signed volume of the solid
S that lies above
and “under” the graph of
Divide
R into four squares with
and choose the sample point as the upper right corner point of each square
and
(
[link] ) to approximate the signed volume of the solid
S that lies above
and “under” the graph of
Divide
R into four squares with
and choose the sample point as the midpoint of each square:
to approximate the signed volume.
The function
graphed over the rectangular region
As we can see, the function
is above the plane. To find the signed volume of
S , we need to divide the region
R into small rectangles
each with area
and with sides
and
and choose
as sample points in each
Hence, a double integral is set up as
Approximating the signed volume using a Riemann sum with
we have
Also, the sample points are (1, 1), (2, 1), (1, 2), and (2, 2) as shown in the following figure.
Subrectangles for the rectangular region
Hence,
Approximating the signed volume using a Riemann sum with
we have
In this case the sample points are (1/2, 1/2), (3/2, 1/2), (1/2, 3/2),
and (3/2, 3/2).
Hence
Divide
R into the same four squares with
and choose the sample points as the upper left corner point of each square
and
(
[link] ) to approximate the signed volume of the solid
S that lies above
and “under” the graph of
Note that we developed the concept of double integral using a rectangular region
R . This concept can be extended to any general region. However, when a region is not rectangular, the subrectangles may not all fit perfectly into
R , particularly if the base area is curved. We examine this situation in more detail in the next section, where we study regions that are not always rectangular and subrectangles may not fit perfectly in the region
R . Also, the heights may not be exact if the surface
is curved. However, the errors on the sides and the height where the pieces may not fit perfectly within the solid
S approach 0 as
m and
n approach infinity. Also, the double integral of the function
exists provided that the function
is not too discontinuous. If the function is bounded and continuous over
R except on a finite number of smooth curves, then the double integral exists and we say that
is integrable over
R .
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