In
[link] , we could have looked at the region in another way, such as
(
[link] ).
This is a Type II region and the integral would then look like
However, if we integrate first with respect to
this integral is lengthy to compute because we have to use integration by parts twice.
Evaluating an iterated integral over a type ii region
Evaluate the integral
where
Notice that
can be seen as either a Type I or a Type II region, as shown in
[link] . However, in this case describing
as Type
is more complicated than describing it as Type II. Therefore, we use
as a Type II region for the integration.
Recall from
Double Integrals over Rectangular Regions the properties of double integrals. As we have seen from the examples here, all these properties are also valid for a function defined on a nonrectangular bounded region on a plane. In particular, property
states:
If
and
except at their boundaries, then
Similarly, we have the following property of double integrals over a nonrectangular bounded region on a plane.
Decomposing regions into smaller regions
Suppose the region
can be expressed as
where
and
do not overlap except at their boundaries. Then
This theorem is particularly useful for nonrectangular regions because it allows us to split a region into a union of regions of Type I and Type II. Then we can compute the double integral on each piece in a convenient way, as in the next example.
Decomposing regions
Express the region
shown in
[link] as a union of regions of Type I or Type II, and evaluate the integral
The region
is not easy to decompose into any one type; it is actually a combination of different types. So we can write it as a union of three regions
where,
These regions are illustrated more clearly in
[link] .
Here
is Type
and
and
are both of Type II. Hence,
Now we could redo this example using a union of two Type II regions (see the Checkpoint).
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