General Regions of Integration

So far we have computed double integrals over rectangular regions. However, most regions we encounter in applications are not rectangular.

Consider a general bounded region [latex]D[/latex] on a plane as shown in Figure 1. Since [latex]D[/latex] is bounded, there must exist a rectangular region [latex]R[/latex] that encloses [latex]D[/latex]. In other words, we can find a rectangle [latex]R[/latex] such that [latex]D[/latex] is a subset of [latex]R[/latex], written as [latex]D \subseteq R[/latex].

Now suppose [latex]z=f(x, y)[/latex] is defined on this general bounded region [latex]D[/latex]. To develop double integrals of [latex]f[/latex] over [latex]D[/latex], we extend the definition of the function to include all points on a rectangular region [latex]R[/latex] that contains [latex]D[/latex]. We define a new function [latex]g(x, y)[/latex] on [latex]R[/latex] as follows:

[latex]g(x,y)= \begin{cases} f(x,y) & \text{if } (x,y) \text{ is in } D \\ 0 & \text{if } (x,y) \text{ is in } R \text{ but not in } D \end{cases}[/latex]

This extension allows us to use the integration techniques from rectangular regions. The function [latex]g(x,y)[/latex] equals [latex]f(x,y)[/latex] inside [latex]D[/latex] and equals zero outside [latex]D[/latex] but within [latex]R[/latex].

We assume the boundary of [latex]D[/latex] is a piecewise smooth and continuous simple closed curve to avoid technical difficulties. Under this assumption, [latex]g(x, y)[/latex] is integrable over the rectangular region [latex]R[/latex] as long as [latex]f(x,y)[/latex] is integrable over [latex]D[/latex]. For now we focus on describing different types of regions rather than the technical details of the function.

We consider two types of planar bounded regions.

type I and type II regions

A region [latex]D[/latex] in the [latex](x, y)[/latex]-plane is of Type I if it lies between two vertical lines and the graphs of two continuous functions [latex]g_1(x)[/latex] and [latex]g_2(x)[/latex]. That is,,

[latex]{D} = \left \{(x,y) \mid a \leq x \leq \ b,g_1 \leq y \leq g_2 \ (x) \right \}[/latex].

A region [latex]D[/latex] in the [latex]xy[/latex] plane is of Type II if it lies between two horizontal lines and the graphs of two continuous functions [latex]h_1(y)[/latex] and [latex]h_2(y)[/latex]. That is,

[latex]{D} = \left \{(x,y) \mid c \leq y \leq \ d,h_1 \leq x \leq h_2 \ (y) \right \}[/latex].

Figures 2 and 3 illustrate the geometric interpretation of Type I and Type II regions.

 

Identifying Region Types

For a Type I region, think vertically: identify the functions that form the top and bottom boundaries, then determine the [latex]x[/latex]-interval.

For a Type II region, think horizontally: identify the functions that form the left and right boundaries, then determine the [latex]y[/latex]-interval.

Consider the region in the first quadrant between the functions [latex]{y} = {\sqrt{x}}[/latex] and [latex]y=x^{3}[/latex] (Figure 4). Describe the region first as Type I and then as Type II.

Show Solution

When describing a region as Type I, we need to identify the function that lies above the region and the function that lies below the region. Here, region [latex]D[/latex] is bounded above by [latex]{y} = {\sqrt{x}}[/latex] and below by [latex]y=x^{3}[/latex] in the interval for [latex]x[/latex] in [latex][0,1][/latex]. Hence, as Type I, [latex]D[/latex] is described as the set [latex]{\left \{(x,y) \mid 0 \leq x \leq 1,x^3 \leq y \leq \sqrt{x} \right \}}[/latex].

However, when describing a region as Type II, we need to identify the function that lies on the left of the region and the function that lies on the right of the region. Here, the region [latex]D[/latex] is bounded on the left by [latex]x=y^{2}[/latex] and on the right by [latex]x = \sqrt[3]{y}[/latex] in the interval for [latex]y[/latex] in [latex][0,1][/latex]. Hence, as Type II, [latex]D[/latex] is described as the set [latex]{\left \{(x,y) \mid 0 \leq y \leq 1,y^2 \leq x \leq \sqrt[3]{y} \right \}}[/latex].