5.2 Double Integrals over General Regions

General Regions of Integration

An example of a general bounded region $D$ on a plane is shown in Figure 5.12. 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 $R\left(D\subseteq R\right).$

Suppose $z=f\left(x,y\right)$ is defined on a general planar bounded region $D$ as in Figure 5.12. 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 $g\left(x,y\right)$ on $R$ as follows:

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 Double Integrals over Rectangular Regions used an integrable function $f\left(x,y\right),$ we must be careful about $g\left(x,y\right)$ and verify that $g\left(x,y\right)$ 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.

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 5.15). 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 $x\text{in}\left[0,1\right].$ Hence, as Type I, $D$ is described as the set $\left\{\left(x,y\right)|0\le x\le 1,x^3\le y\le \sqrt{x}\right\}.$

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 $\left\{\left(x,y\right)|0\le y\le 1,y^2\le x\le \sqrt[3]{y}\right\}.$

Double Integrals over Nonrectangular 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 integrations in double integrals. As a first step, let us look at the following theorem.

The right-hand side of this equation is what we have seen before, so this theorem is reasonable because $R$ is a rectangle and ${\displaystyle \underset{R}{\iint }g\left(x,y\right)}dA$ has been discussed in the preceding section. Also, the equality works because the values of $g\left(x,y\right)$ are $0$ for any point $\left(x,y\right)$ that lies outside $D,$ and hence these points do not add anything to the integral. However, it is important that the rectangle $R$ contains the region $D.$

As a matter of fact, if the region $D$ 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 $R$ containing the region.

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 $f\left(x,y\right)$ with $x$ being held constant and the limits of integration being $g_1\left(x\right)\text{and}{g}_2\left(x\right).$ In the inner integral in the second expression, we integrate $f\left(x,y\right)$ with $y$ being held constant and the limits of integration are $h_1\left(y\right)\text{and}{h}_2\left(y\right).$

Evaluating an Iterated Integral over a Type I Region

Evaluate the integral ${\displaystyle \underset{D}{\iint }{x}^2{e}^{xy}dA}$ where $D$ is shown in Figure 5.16.

First construct the region $D$ as a Type I region (Figure 5.16). Here $D=\left\{\left(x,y\right)|0\le x\le 2,\frac{1}{2}x\le y\le 1\right\}.$ Then we have

$${\displaystyle \underset{D}{\iint }{x}^2{e}^{xy}dA}={\displaystyle \underset{x=0}{\overset{x=2}{\int }}{\displaystyle \underset{y=1\text{/}2x}{\overset{y=1}{\int }}{x}^2{e}^{xy}dydx}}.$$

Therefore, we have

$$\begin{array}{ccccc}\hfill {\displaystyle \underset{x=0}{\overset{x=2}{\int }}{\displaystyle \underset{y=\frac{1}{2}x}{\overset{y=1}{\int }}{x}^2{e}^{xy}dydx}}& ={\displaystyle \underset{x=0}{\overset{x=2}{\int }}\left[{\displaystyle \underset{y=1\text{/}2x}{\overset{y=1}{\int }}{x}^2{e}^{xy}dy}\right]}dx\hfill & & & \text{Iterated integral for a Type I region.}\hfill \\ & ={\displaystyle \underset{x=0}{\overset{x=2}{\int }}{\left[x^2\frac{e^{xy}}{x}\right]|}_{y=1\text{/}2x}^{y=1}}dx\hfill & & & \begin{array}{c}\text{Integrate with respect to}y\text{using}\hfill \\ u\text{-substitution with}u=xy\text{where}x\text{is held}\hfill \\ \text{constant.}\hfill \end{array}\hfill \\ & ={\displaystyle \underset{x=0}{\overset{x=2}{\int }}\left[x{e}^x-x{e}^{x^2\text{/}2}\right]}dx\hfill & & & \begin{array}{c}\text{Integrate with respect to}x\text{using}\hfill \\ u\text{-substitution with}u=\frac{1}{2}{x}^2.\hfill \end{array}\hfill \\ & ={\left[x{e}^x-e^x-e^{\frac{1}{2}{x}^2}\right]|}_{x=0}^{x=2}=2\hfill & & & \end{array}$$

In Example 5.12, we could have looked at the region in another way, such as $D=\left\{\left(x,y\right)|0\le y\le 1,0\le x\le 2y\right\}$ (Figure 5.17).

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

However, if we integrate first with respect to $x,$ 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 ${\displaystyle \underset{D}{\iint }\left(3x^2+y^2\right)dA}$ where $=\left\{\left(x,y\right)|-2\le y\le 3,y^2-3\le x\le y+3\right\}.$

Notice that $D$ can be seen as either a Type I or a Type II region, as shown in Figure 5.18. However, in this case describing $D$ as Type $\text{I}$ is more complicated than describing it as Type II. Therefore, we use $D$ as a Type II region for the integration.

Choosing this order of integration, we have

$$\begin{array}{ccccc}\hfill {\displaystyle \underset{D}{\iint }\left(3x^2+y^2\right)dA}& ={\displaystyle \underset{y=\mathrm{-2}}{\overset{y=3}{\int }}{\displaystyle \underset{x=y^2-3}{\overset{x=y+3}{\int }}\left(3x^2+y^2\right)dxdy}}\hfill & & & \text{Iterated integral, Type II region.}\hfill \\ & ={{\displaystyle \underset{y=\mathrm{-2}}{\overset{y=3}{\int }}\left(x^3+x{y}^2\right)}|}_{y^2-3}^{y+3}dy\hfill & & & \text{Integrate with respect to}x.\hfill \\ & ={\displaystyle \underset{y=\mathrm{-2}}{\overset{y=3}{\int }}\left({\left(y+3\right)}^3+\left(y+3\right)y^2-{\left(y^2-3\right)}^3-\left(y^2-3\right)y^2\right)}dy\hfill & & & \\ & ={\displaystyle \underset{\mathrm{-2}}{\overset{3}{\int }}\left(54+27y-12y^2+2y^3+8y^4-y^6\right)}dy\hfill & & & \text{Integrate with respect to}y.\hfill \\ & ={\left[54y+\frac{27y^2}{2}-4y^3+\frac{y^4}{2}+\frac{8y^5}{5}-\frac{y^7}{7}\right]|}_{\mathrm{-2}}^3\hfill & & & \\ & =\frac{2375}{7}.\hfill & & & \end{array}$$

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 $3$ states:

If $R=S\cup T$ and $S\cap T=\varnothing$ except at their boundaries, then

Similarly, we have the following property of double integrals over a nonrectangular bounded region on a plane.

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 $D$ shown in Figure 5.19 as a union of regions of Type I or Type II, and evaluate the integral

$${\displaystyle \underset{D}{\iint }\left(2x+5y\right)}dA.$$

The region $D$ 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 $D_1,D_2,\text{and}{D}_3$ where, $D_1=\left\{\left(x,y\right)|-2\le x\le 0,0\le y\le {\left(x+2\right)}^2\right\},$ $D_2=\left\{\left(x,y\right)|0\le y\le 4,0\le x\le \left(y-\frac{1}{16}{y}^3\right)\right\},$ $D_3=\left\{\left(x,y\right)\overline{)-4\le y\le 0,-2\le x\le y-\frac{y^3}{16}}\right\}.$ These regions are illustrated more clearly in Figure 5.20.

Here $D_1$ is Type $\text{I}$ and $D_2$ and $D_3$ are both of Type II. Hence,

$$\begin{array}{cc}\hfill {\displaystyle \underset{D}{\iint }(2x+5y)dA}& ={\displaystyle \underset{D_1}{\iint }(2x+5y)dA}+{\displaystyle \underset{D_2}{\iint }(2x+5y)dA}+{\displaystyle \underset{D_3}{\iint }(2x+5y)dA}\hfill \\ & ={\displaystyle \underset{x=\mathrm{-2}}{\overset{x=0}{\int }}{\displaystyle \underset{y=0}{\overset{y={(x+2)}^2}{\int }}(2x+5y)dydx}}+{\displaystyle \underset{y=0}{\overset{y=4}{\int }}{\displaystyle \underset{x=0}{\overset{x=y-(1\text{/}16)y^3}{\int }}(2+5y)dxdy}}+{\displaystyle \underset{y=\mathrm{-4}}{\overset{y=0}{\int }}{\displaystyle \underset{x=\mathrm{-2}}{\overset{x=y-(1\text{/}16)y^3}{\int }}(2x+5y)dxdy}}\hfill \\ & ={\displaystyle \underset{x=\mathrm{-2}}{\overset{x=0}{\int }}\left[\frac{1}{2}{(2+x)}^2(20+24x+5x^2)\right]dx}+{\displaystyle \underset{y=0}{\overset{y=4}{\int }}\left[\frac{1}{256}{y}^6-\frac{7}{16}{y}^4+6y^2\right]dy}\hfill \\ & +{\displaystyle \underset{y=\mathrm{-4}}{\overset{y=0}{\int }}\left[\frac{1}{256}{y}^6-\frac{7}{16}{y}^4+6y^2+10y-4\right]dy}\hfill \\ & =\frac{40}{3}+\frac{1664}{35}-\frac{1696}{35}=\frac{1304}{105}.\hfill \end{array}$$

Now we could redo this example using a union of two Type II regions (see the Checkpoint).

Changing the Order of Integration

As we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation that is significantly simpler than the other order of integration. Sometimes the order of integration does not matter, but it is important to learn to recognize when a change in order will simplify our work.

Changing the Order of Integration

Reverse the order of integration in the iterated integral ${\displaystyle \underset{x=0}{\overset{x=\sqrt{2}}{\int }}{\displaystyle \underset{y=0}{\overset{y=2-x^2}{\int }}x{e}^{x^2}dydx}}.$ Then evaluate the new iterated integral.

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 5.21.

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 $\left[0,\sqrt{2}\right].$ 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 $\left[0,2\right].$ We solved $y=2-x^2$ in terms of $x$ to obtain $x=\sqrt{2-y}.$

Hence

$$\begin{array}{ccccc}\hfill {\displaystyle \underset{0}{\overset{\sqrt{2}}{\int }}{\displaystyle \underset{0}{\overset{2-x^2}{\int }}x{e}^{x^2}dydx}}& ={\displaystyle \underset{0}{\overset{2}{\int }}{\displaystyle \underset{0}{\overset{\sqrt{2-y}}{\int }}x{e}^{x^2}dxdy}}\hfill & & & \begin{array}{c}\text{Reverse the order of}\hfill \\ \text{integration then use}\hfill \\ \text{substitution.}\hfill \end{array}\hfill \\ & ={\displaystyle \underset{0}{\overset{2}{\int }}\left[\frac{1}{2}{{e^x}^2|}_0^{\sqrt{2-y}}\right]}dy={\displaystyle \underset{0}{\overset{2}{\int }}\frac{1}{2}\left(e^{2-y}-1\right)}dy={-\frac{1}{2}\left(e^{2-y}+y\right)|}_0^2\hfill & & & \\ & =\frac{1}{2}\left(e^2-3\right).\hfill & & & \end{array}$$

Evaluating an Iterated Integral by Reversing the Order of Integration

Consider the iterated integral ${\displaystyle \underset{R}{\iint }f\left(x,y\right)}dxdy$ where $z=f\left(x,y\right)=x-2y$ over a triangular region $R$ that has sides on $x=0,y=0,$ and the line $x+y=1.$ Sketch the region, and then evaluate the iterated integral by

1. integrating first with respect to $y$ and then
2. integrating first with respect to $x.$

A sketch of the region appears in Figure 5.22.

We can complete this integration in two different ways.

1. One way to look at it is by first integrating $y$ from $y=0\text{to}y=1-x$ vertically and then integrating $x$ from $x=0\text{to}x=1\text{:}$
   
   $$\begin{array}{cc}\hfill {\displaystyle \underset{R}{\iint }f\left(x,y\right)}dxdy& ={\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\displaystyle \underset{y=0}{\overset{y=1-x}{\int }}\left(x-2y\right)dydx}=}{\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\left[xy-y^2\right]}_{y=0}^{y=1-x}dx}\hfill \\ & ={\displaystyle \underset{x=0}{\overset{x=1}{\int }}\left[x\left(1-x\right)-{\left(1-x\right)}^2\right]dx={\displaystyle \underset{x=0}{\overset{x=1}{\int }}\left[\mathrm{-1}+3x-2x^2\right]dx=}{\left[\text{−}x+\frac{3}{2}{x}^2-\frac{2}{3}{x}^3\right]}_{x=0}^{x=1}=-\frac{1}{6}}.\hfill \end{array}$$
2. The other way to do this problem is by first integrating $x$ from $x=0\text{to}x=1-y$ horizontally and then integrating $y$ from $y=0\text{to}y=1\text{:}$
   
   $$\begin{array}{cc}\hfill {\displaystyle \underset{R}{\iint }f\left(x,y\right)}dxdy& ={\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{x=0}{\overset{x=1-y}{\int }}\left(x-2y\right)dxdy}=}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\left[\frac{1}{2}{x}^2-2xy\right]}_{x=0}^{x=1-y}dy}\hfill \\ & ={\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left[\frac{1}{2}{\left(1-y\right)}^2-2y\left(1-y\right)\right]dy={\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left[\frac{1}{2}-3y+\frac{5}{2}{y}^2\right]dy}}\hfill \\ & ={\left[\frac{1}{2}y-\frac{3}{2}{y}^2+\frac{5}{6}{y}^3\right]}_{y=0}^{y=1}=-\frac{1}{6}.\hfill \end{array}$$

Calculating Volumes, Areas, and Average Values

We can use double integrals over general regions to compute volumes, areas, and average values. The methods are the same as those in Double Integrals over Rectangular Regions, but without the restriction to a rectangular region, we can now solve a wider variety of problems.

Finding the Volume of a Tetrahedron

Find the volume of the solid bounded by the planes $x=0,y=0,z=0,$ and $2x+3y+z=6.$

The solid is a tetrahedron with the base on the $xy$-plane and a height $z=6-2x-3y.$ The base is the region $D$ bounded by the lines, $x=0,y=0$ and $2x+3y=6$ where $z=0$ (Figure 5.23). Note that we can consider the region $D$ as Type I or as Type II, and we can integrate in both ways.

First, consider $D$ as a Type I region, and hence $D=\left\{\left(x,y\right)|0\le x\le 3,0\le y\le 2-\frac{2}{3}x\right\}.$

Therefore, the volume is

$$\begin{array}{cc}\hfill V& ={\displaystyle \underset{x=0}{\overset{x=3}{\int }}{\displaystyle \underset{y=0}{\overset{y=2-(2x\text{/}3)}{\int }}(6-2x-3y)dydx}}={\displaystyle \underset{x=0}{\overset{x=3}{\int }}\left[{\left(6y-2xy-\frac{3}{2}{y}^2\right)|}_{y=0}^{y=2-(2x\text{/}3)}\right]}dx\hfill \\ & ={\displaystyle \underset{x=0}{\overset{x=3}{\int }}\left[\frac{2}{3}{(x-3)}^2\right]}dx=6.\hfill \end{array}$$

Now consider $D$ as a Type II region, so $D=\left\{(x,y)|0\le y\le 2,0\le x\le 3-\frac{3}{2}y\right\}.$ In this calculation, the volume is

$$\begin{array}{cc}\hfill V& ={\displaystyle \underset{y=0}{\overset{y=2}{\int }}{\displaystyle \underset{x=0}{\overset{x=3-(3y\text{/}2)}{\int }}(6-2x-3y)dxdy}}={\displaystyle \underset{y=0}{\overset{y=2}{\int }}\left[{\left(6x-x^2-3xy\right)|}_{x=0}^{x=3-(3y\text{/}2)}\right]}dy\hfill \\ & ={\displaystyle \underset{y=0}{\overset{y=2}{\int }}\left[\frac{9}{4}{(y-2)}^2\right]}dy=6.\hfill \end{array}$$

Therefore, the volume is $6$ cubic units.

Finding the area of a rectangular region is easy, but finding the area of a nonrectangular region is not so easy. As we have seen, we can use double integrals to find a rectangular area. As a matter of fact, this comes in very handy for finding the area of a general nonrectangular region, as stated in the next definition.

We have already seen how to find areas in terms of single integration. Here we are seeing another way of finding areas by using double integrals, which can be very useful, as we will see in the later sections of this chapter.

Finding the Area of a Region

Find the area of the region bounded below by the curve $y=x^2$ and above by the line $y=2x$ in the first quadrant (Figure 5.24).

We just have to integrate the constant function $f\left(x,y\right)=1$ over the region. Thus, the area $A$ of the bounded region is ${\displaystyle \underset{x=0}{\overset{x=2}{\int }}{\displaystyle \underset{y=x^2}{\overset{y=2x}{\int }}dydx}}$ or ${\displaystyle \underset{y=0}{\overset{x=4}{\int }}{\displaystyle \underset{x=y\text{/}2}{\overset{x=\sqrt{y}}{\int }}dxdy}}\text{:}$

$$A={\displaystyle \underset{D}{\iint }1dxdy={\displaystyle \underset{x=0}{\overset{x=2}{\int }}{\displaystyle \underset{y=x^2}{\overset{y=2x}{\int }}1dydx=}}}{\displaystyle \underset{x=0}{\overset{x=2}{\int }}\left[{y|}_{y=x^2}^{y=2x}\right]}dx={\displaystyle \underset{x=0}{\overset{x=2}{\int }}\left(2x-x^2\right)}dx={x^2-\frac{x^3}{3}|}_0^2=\frac{4}{3}.$$

We can also use a double integral to find the average value of a function over a general region. The definition is a direct extension of the earlier formula.

Finding an Average Value

Find the average value of the function $f\left(x,y\right)=7x{y}^2$ on the region bounded by the line $x=y$ and the curve $x=\sqrt{y}$ (Figure 5.25).

First find the area $A\left(D\right)$ where the region $D$ is given by the figure. We have

$$A\left(D\right)={\displaystyle \underset{D}{\iint }1dA={\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{x=y}{\overset{x=\sqrt{y}}{\int }}1dxdy={\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left[{x|}_{x=y}^{x=\sqrt{y}}\right]}}}}dy={\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left(\sqrt{y}-y\right)dy=\frac{2}{3}}{y}^{3\text{/}2}-{\frac{y^2}{2}|}_0^1=\frac{1}{6}.$$

Then the average value of the given function over this region is

$$\begin{array}{cc}\hfill f_{ave}& =\frac{1}{A\left(D\right)}{\displaystyle \underset{D}{\iint }f\left(x,y\right)dA=\frac{1}{A\left(D\right)}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{x=y}{\overset{x=\sqrt{y}}{\int }}7x{y}^{2}dxdy=\frac{1}{1\text{/}6}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left[{\frac{7}{2}{x}^2{y}^2|}_{x=y}^{x=\sqrt{y}}\right]}}}}dy\hfill \\ & =6{\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left[\frac{7}{2}{y}^2\left(y-y^2\right)\right]dy=6{\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left[\frac{7}{2}\left(y^3-y^4\right)\right]dy}=\frac{42}{2}}{\left(\frac{y^4}{4}-\frac{y^5}{5}\right)|}_0^1=\frac{42}{40}=\frac{21}{20}.\hfill \end{array}$$

Improper Double Integrals

An improper double integral is an integral ${\displaystyle \underset{D}{\iint }fdA}$ where either $D$ is an unbounded region or $f$ is an unbounded function. For example, $D=\left\{\left(x,y\right)|\left|x-y\right|\ge 2\right\}$ is an unbounded region, and the function $f\left(x,y\right)=1\text{/}\left(1-x^2-2y^2\right)$ over the ellipse $x^2+2y^2\le 1$ is an unbounded function. Hence, both of the following integrals are improper integrals:

1. ${\displaystyle \underset{D}{\iint }xydA}$ where $D=\left\{\left(x,y\right)|\left|x-y\right|\ge 2\right\};$
2. ${\displaystyle \underset{D}{\iint }\frac{1}{1-x^2-2y^2}dA}$ where $D=\left\{\left(x,y\right)|x^2+2y^2\le 1\right\}.$

In this section we would like to deal with improper integrals of functions over rectangles or simple regions such that $f$ has only finitely many discontinuities. Not all such improper integrals can be evaluated; however, a form of Fubini’s theorem does apply for some types of improper integrals.

It is very important to note that we required that the function be nonnegative on $D$ for the theorem to work. We consider only the case where the function has finitely many discontinuities inside $D.$

Evaluating a Double Improper Integral

Consider the function $f\left(x,y\right)=\frac{e^y}{y}$ over the region $D=\{(x,y)\text{:}0\le x\le 1,x\le y\le \sqrt{x}\}.$

Notice that the function is nonnegative and continuous at all points on $D$ except $\left(0,0\right).$ Use Fubini’s theorem to evaluate the improper integral.

First we plot the region $D$ (Figure 5.26); then we express it in another way.

The other way to express the same region $D$ is

$$D=\left\{\left(x,y\right)\text{:}0\le y\le 1,y^2\le x\le y\right\}.$$

Thus we can use Fubini’s theorem for improper integrals and evaluate the integral as

$${\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{x=y^2}{\overset{x=y}{\int }}\frac{e^y}{y}}}dxdy.$$

Therefore, we have

$$\underset{y=0}{\overset{y=1}{\int }}\underset{x=y^2}{\overset{x=y}{\int }}\frac{e^y}{y}dxdy=\underset{y=0}{\overset{y=1}{\int }}\frac{e^y}{y}{x|}_{x=y^2}^{x=y}dy=\underset{y=0}{\overset{y=1}{\int }}\frac{e^y}{y}(y-y^2)dy=\underset{0}{\overset{1}{\int }}(e^y-y{e}^y)dy=e-2.$$

As mentioned before, we also have an improper integral if the region of integration is unbounded. Suppose now that the function $f$ is continuous in an unbounded rectangle $R.$

The following example shows how this theorem can be used in certain cases of improper integrals.

In some situations in probability theory, we can gain insight into a problem when we are able to use double integrals over general regions. Before we go over an example with a double integral, we need to set a few definitions and become familiar with some important properties.

Application to Probability

At Sydney’s Restaurant, customers must wait an average of $15$ minutes for a table. From the time they are seated until they have finished their meal requires an additional $40$ minutes, on average. What is the probability that a customer spends less than an hour and a half at the diner, assuming that waiting for a table and completing the meal are independent events?

Waiting times are mathematically modeled by exponential density functions, with $m$ being the average waiting time, as

$$f(t)=\{\begin{array}{cc}0\hfill & \text{if}t<0,\hfill \\ \frac{1}{m}{e}^{\text{−}t\text{/}m}\hfill & \text{if}t\ge 0.\hfill \end{array}$$

If $X$ and $Y$ are random variables for ‘waiting for a table’ and ‘completing the meal,’ then the probability density functions are, respectively,

$$f_1(x)=\{\begin{array}{cc}0\hfill & \text{if}x<0,\hfill \\ \frac{1}{15}{e}^{\text{−}x\text{/}15}\hfill & \text{if}x\ge 0.\hfill \end{array}\text{and}{f}_2(y)=\{\begin{array}{cc}0\hfill & \text{if}y<0,\hfill \\ \frac{1}{40}{e}^{\text{−}y\text{/}40}\hfill & \text{if}y\ge 0.\hfill \end{array}$$

Clearly, the events are independent and hence the joint density function is the product of the individual functions

$$f(x,y)=f_1(x)f_2(y)=\{\begin{array}{cc}0\hfill & \text{if}x<0\text{or}y<0,\hfill \\ \frac{1}{600}{e}^{\text{−}x\text{/}15}{e}^{\text{−}y\text{/}60}\hfill & \text{if}x,y\ge 0.\hfill \end{array}$$

We want to find the probability that the combined time $X+Y$ is less than $90$ minutes. In terms of geometry, it means that the region $D$ is in the first quadrant bounded by the line $x+y=90$ (Figure 5.27).

Hence, the probability that $\left(X,Y\right)$ is in the region $D$ is

$$P\left(X+Y\le 90\right)=P\left(\left(X,Y\right)\in D\right)={\displaystyle \underset{D}{\iint }f\left(x,y\right)dA=}{\displaystyle \underset{D}{\iint }\frac{1}{600}{e}^{\text{−}x\text{/}15}{e}^{\text{−}y\text{/}40}dA}.$$

Since $x+y=90$ is the same as $y=90-x,$ we have a region of Type I, so

$$\begin{array}{rcl}D& =& \left\{(x,y)\right|0\le x\le 90,0\le y\le 90-x\},\\ P(X+Y\le 90)& =& \begin{array}{lc}\frac{1}{600}& \underset{x=0}{\overset{x=90}{\int }}\underset{y=0}{\overset{y=90-x}{\int }}{e}^{\text{−}x\text{/}15}{e}^{\text{−}y\text{/}40}dy dx\end{array}\\ & =& \begin{array}{lc}\frac{1}{600}& \underset{x=0}{\overset{x=90}{\int }}\underset{y=0}{\overset{y=90-x}{\int }}{e}^{\text{−}(x\text{/}15+y\text{/}40)}dy dx=0.8328.\end{array}\end{array}$$

Thus, there is an $83.28\text{\%}$ chance that a customer spends less than an hour and a half at the restaurant.

Another important application in probability that can involve improper double integrals is the calculation of expected values. First we define this concept and then show an example of a calculation.