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 \(R (D \subseteq R)\).

Figure \(\PageIndex{1}\). For a region \(D\) that is a subset of \(R\), we can define a function \(g(x,y)\) to equal \(f(x,y)\) at every point in \(D\) and \(0\) at every point of \(R\) not in \(D\).

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 \(\PageIndex{3}\)),

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

Figure \(\PageIndex{2}\). A Type I region lies between two vertical lines and the graphs of two functions of \(x\).Figure \(\PageIndex{3}\). A Type II region lies between two horizontal lines and the graphs of two functions of \(y\).

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.

Figure \(\PageIndex{4}\): Region \(D\) can be described as Type I or 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, D is described as the set \(\big\{(x,y) \,| \, 0 \leq y \leq 1, \space y^2 \leq x \leq \sqrt[3]{y}\big\}\).

Exercise \(\PageIndex{1}\)

Consider the region in the first quadrant between the functions \(y = 2x\) and \(y = x^2\). Describe the region first as Type I and then as Type II.

Hint

Graph the functions, and draw vertical and horizontal lines.

Answer

Type I and Type II are expressed as \(\big\{(x,y) \,|\, 0 \leq x \leq 2, \space x^2 \leq y \leq 2x\big\}\) and \(\big\{(x,y)|\, 0 \leq y \leq 4,

\space \frac{1}{2} y \leq x \leq \sqrt{y}\big\}\), respectively.

Double Integrals over Non-rectangular Regions

To develop the concept and tools for evaluation of 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 decide the limits of integration in double integrals.

As a first step, let us look at the following theorem.

Theorem: Double Integrals over Nonrectangular Regions

Suppose \(g(x,y)\) is the extension to the rectangle \(R\) of the function \(f(x,y)\) defined on the regions \(D\) and \(R\) as shown in

Figure \(\PageIndex{1}\) inside \(R\). Then \(g(x,y)\) is integrable and we define the double integral of \(f(x,y)\) over \(D\) by

Figure \(\PageIndex{5}\). We can express region \(D\) as a Type I region and integrate from \(y = \frac{1}{2} x\) to \(y = 1\) between the lines \(x = 0\) and \(x = 2\).

Therefore, we have

\[\begin{align*} \int_{x=0}^{x=2}\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\,dx &= \int_{x=0}^{x=2}\left[\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\right] dx &\text{Iterated integral for a Type I region.}\\ &=\int_{x=0}^{x=2} \left.\left[ x^2 \frac{e^{xy}}{x} \right] \right|_{y=1/2x}^{y=1}\,dx & \text{Integrate with respect to $y$}\\ &= \int_{x=0}^{x=2} \left[xe^x – xe^{x^2/2}\right]dx &\text{Integrate with respect to $x$} \\ &=\left[xe^x – e^x – e^{\frac{1}{2}x^2} \right] \Big|_{x=0}^{x=2} = 2. \end{align*}\]

In Example \(\PageIndex{2}\), we could have looked at the region in another way, such as \(D = \big\{(x,y)\,|\,0 \leq y \leq 1, \space 0 \leq x \leq 2y\big\}\)

(Figure \(\PageIndex{6}\)).

Figure \(\PageIndex{6}\).

This is a Type II region and the integral would then look like

The region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II. Refer to Figure \(\PageIndex{10}\).

Figure \(\PageIndex{10}\). Converting a region from Type I to Type II.

We can see from the limits of integration that the region is bounded above by \(y = 2 – x^2\) and below by \(y = 0\) where \(x\) is in the interval

\([0, \sqrt{2}]\). By reversing the order, we have the region bounded on the left by \(x = 0\) and on the right by \(x = \sqrt{2 – y}\) where \(y\)

is in the interval \([0, 2]\). We solved \(y = 2 – x^2\) in terms of \(x\) to obtain \(x = \sqrt{2 – y}\).

Hence \[\begin{align*} \int_0^{\sqrt{2}} \int_0^{2-x^2} xe^{x^2} dy \space dx&= \int_0^2 \int_0^{\sqrt{2-y}} xe^{x^2}\,dx \space dy &\text{Reverse the order of integration then use substitution.} \\&= \int_0^2 \left[\left.\frac{1}{2}e^{x^2}\right|_0^{\sqrt{2-y}}\right] dy = \int_0^2\frac{1}{2}(e^{2-y} – 1)\,dy \\&= -\left.\frac{1}{2}(e^{2-y} + y)\right|_0^2 = \frac{1}{2}(e^2 – 3). \end{align*}\]

Example \(\PageIndex{6}\): Evaluating an Iterated Integral by Reversing the Order of Integration

Consider the iterated integral \[\iint\limits_R f(x,y)\,dx \space dy\] where \(z = f(x,y) = x – 2y\) over a triangular region \(R\) that has sides on

\(x = 0, \space y = 0\), and the line \(x + y = 1\). Sketch the region, and then evaluate the iterated integral by

integrating first with respect to \(y\) and then

integrating first with respect to \(x\).

Solution

A sketch of the region appears in Figure \(\PageIndex{11}\).

Figure \(\PageIndex{11}\). A triangular region \(R\) for integrating in two ways.

We can complete this integration in two different ways.

a. One way to look at it is by first integrating \(y\) from \(y = 0\) to \(y = 1 – x\) vertically and then integrating \(x\) from \(x = 0\) to \(x = 1\):

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