
# 7.2: Double Integrals over General Regions


Previously, we studied the concept of double integrals and examined the tools needed to compute them. We learned techniques and properties to integrate functions of two variables over rectangular regions. We also discussed several applications, such as finding the volume bounded above by a function over a rectangular region, finding area by integration, and calculating the average value of a function of two variables.

In this section we consider double integrals of functions defined over a general bounded region $$D$$ on the plane. Most of the previous results hold in this situation as well, but some techniques need to be extended to cover this more general case.

## General Regions of Integration

An example of a general bounded region $$D$$ on a plane is shown in Figure $$\PageIndex{1}$$. Since $$D$$ is bounded on the plane, there must exist a rectangular region $$R$$ on the same plane that encloses the region $$D$$ that is, a rectangular region $$R$$ exists such that $$D$$ is a subset of $$R (D \subseteq R)$$.

Suppose $$z = f(x,y)$$ is defined on a general planar bounded region $$D$$ as in Figure $$\PageIndex{1}$$. In order to develop double integrals of $$f$$ over $$D$$ we extend the definition of the function to include all points on the rectangular region $$R$$ and then use the concepts and tools from the preceding section. But how do we extend the definition of $$f$$ to include all the points on $$R$$? We do this by defining a new function $$g(x,y)$$ on $$R$$ as follows:

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

Note that we might have some technical difficulties if the boundary of $$D$$ is complicated. So we assume the boundary to be a piecewise smooth and continuous simple closed curve. Also, since all the results developed in the section on Double Integrals over Rectangular Regions used an integrable function $$f(x,y)$$ we must be careful about $$g(x,y)$$ and verify that $$g(x,y)$$ is an integrable function over the rectangular region $$R$$. This happens as long as the region $$D$$ is bounded by simple closed curves. For now we will concentrate on the descriptions of the regions rather than the function and extend our theory appropriately for integration.

We consider two types of planar bounded regions.

Definition: Type I and Type II regions

A region $$D$$ in the $$(x,y)$$-plane is of Type I if it lies between two vertical lines and the graphs of two continuous functions $$g_1(x)$$ and $$g_2(x)$$. That is (Figure $$\PageIndex{2}$$),

$D = \big\{(x,y)\,|\, a \leq x \leq b, \space g_1(x) \leq y \leq g_2(x) \big\}.$

A region $$D$$ in the $$xy$$-plane is of Type II if it lies between two horizontal lines and the graphs of two continuous functions $$h_1(y)$$ and $$h_2(y)$$. That is (Figure $$\PageIndex{3}$$),

$D = \big\{(x,y)\,| \, c \leq y \leq d, \space h_1(y) \leq x \leq h_2(y) \big\}.$

Example $$\PageIndex{1}$$: Describing a Region as Type I and Also as Type II

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

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 $$D$$ is bounded above by $$y = \sqrt{x}$$ and below by $$y = x^3$$ in the interval for $$x$$ in $$[0,1]$$. Hence, as Type I, $$D$$ is described as the set $$\big\{(x,y)\,| \, 0 \leq x \leq 1, \space x^3 \leq y \leq \sqrt[3]{x}\big\}$$.

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 $$D$$ is bounded on the left by $$x = y^2$$ and on the right by $$x = \sqrt[3]{y}$$ in the interval for y in $$[0,1]$$. Hence, as Type II,