Double Integrals over Nonrectangular Regions
To evaluate a double integral over a general nonrectangular region, we need to first understand the region and be able to express it as Type I or Type II or a combination of both. Without understanding the regions, we will not be able to determine the limits of integration in double integrals. As a first step, let us look at the following theorem.
double integrals over nonrectangular regions
Suppose [latex]g(x, y)[/latex] is the extension to the rectangle [latex]R[/latex] of the function [latex]f(x, y)[/latex] defined on the regions [latex]D[/latex] and [latex]R[/latex] inside [latex]R[/latex]. Then [latex]g(x, y)[/latex] is integrable and we define the double integral of [latex]f(x, y)[/latex] over [latex]D[/latex] by:
[latex]\underset{D}{\displaystyle\iint} {f(x,y)dA} = \underset{R}{\displaystyle\iint} {g(x,y)dA}[/latex].
The right-hand side of this equation is what we have seen before, so this theorem is reasonable because [latex]R[/latex] is a rectangle and [latex]\underset{R}{\displaystyle\iint}g(x,y)dA[/latex] has been discussed in the preceding section. Also, the equality works because the values of [latex]g(x, y)[/latex] are [latex]0[/latex] for any point [latex](x, y)[/latex] that lies outside [latex]D[/latex], and hence these points do not add anything to the integral. However, it is important that the rectangle [latex]R[/latex] contains the region [latex]D[/latex].
If the region [latex]D[/latex] is bounded by smooth curves on a plane and we are able to describe it as Type I or Type II or a mix of both, then we can use the following theorem and not have to find a rectangle [latex]R[/latex] containing the region.
Fubini’s theorem (strong form)
For a function [latex]f(x, y)[/latex] that is continuous on a region [latex]D[/latex] of Type I, we have:
[latex]\underset{D}{\displaystyle\iint}{f(x,y)dA} = \underset{D}{\displaystyle\iint}{f(x,y)dy \ dx}=\displaystyle\int_a^b\left[\displaystyle\int_{g_1(x)}^{g_2(x)}f(x,y)dy\right]dx.[/latex]
Similarly, for a function [latex]f(x, y)[/latex] that is continuous on a region [latex]D[/latex] of Type II, we have:
[latex]\underset{D}{\displaystyle\iint}{f(x,y)dA} = \underset{D}{\displaystyle\iint}{f(x,y)dx \ dy}=\displaystyle\int_c^d\left[\displaystyle\int_{h_1(y)}^{h_2(y)}f(x,y)dx\right]dy.[/latex]
The integral in each of these expressions is an iterated integral, similar to those we have seen before. Notice that, in the inner integral in the first expression, we integrate [latex]f(x, y)[/latex] with [latex]x[/latex] being held constant and the limits of integration being [latex]g_1(x)[/latex] and [latex]g_2(x)[/latex]. In the inner integral in the second expression, we integrate [latex]f(x, y)[/latex] with [latex]y[/latex] being held constant and the limits of integration are [latex]h_1(x)[/latex] and [latex]h_2(x)[/latex].
In the previous example, we could have looked at the region in another way, such as [latex]{D} = {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {y} \ {\leq} \ {1,0} \ {\leq} \ {x} \ {\leq} \ {2y} \right \}}[/latex] (Figure 6).
This is a Type II region and the integral would then look like:
[latex]\large{\underset{D}{\displaystyle\iint}{x^2}{e^{xy}}{dA} = \displaystyle\int_{y=0}^{y=1}\displaystyle\int_{x=0}^{x=2u}{x^2}{e^{xy}}{dx} \ {dy}}[/latex]
However, if we integrate first with respect to [latex]x[/latex], this integral is lengthy to compute because we have to use integration by parts twice.
As these examples show, choosing the right region type can significantly simplify your calculations. Another useful technique is breaking complex regions into simpler pieces.
Property of Double Integrals
Recall from Double Integrals over Rectangular Regions that all properties of double integrals are also valid for a function defined on a nonrectangular bounded region on a plane. In particular, property 3 states:
If [latex]R = S \cup T[/latex] and [latex]S \cap T = \emptyset[/latex] except at their boundaries, then:
[latex]\underset{R}{\displaystyle\iint}f(x,y)dA=\underset{S}{\displaystyle\iint}f(x,y)dA+\underset{T}{\displaystyle\iint}f(x,y)dA[/latex]
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 [latex]D[/latex] can be expressed as [latex]{D} = {D_1} \ {\cup} \ {D_2}[/latex] where [latex]D_1[/latex] and [latex]D_2[/latex] do not overlap except at their boundaries. Then:
[latex]\underset{D}{\displaystyle\iint}f(x,y)dA=\underset{D_1}{\displaystyle\iint}f(x,y)dA=\underset{D_2}{\displaystyle\iint}f(x,y)dA[/latex].
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.
[latex]\underset{D}{\displaystyle\iint}(2x+5y)dA[/latex].
