
# 15.2: 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 15.2.1. 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 15.2.1. 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 15.2.1. 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 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:

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 15.2.2),

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

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 15.2.3),

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

Figure 15.2.2. A Type I region lies between two vertical lines and the graphs of two functions of $$x$$.

Figure 15.2.3. 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 15.2.4). Describe the region first as Type I and then as Type II.

Figure 15.2.4. 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 $${(x,y)|0 \leq x \leq 1, \space x^3 \leq y \leq \sqrt[3]{x}}$$.

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, is described as the set $${(x,y) |0 \leq y \leq 1, \space y^2 \leq x \leq \sqrt[3]{y}}$$.

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.

Solution

Type I and Type II are expressed as $${(x,y) |0 \leq x \leq 2, \space x^2 \leq y \leq 2x}$$ and $${(x,y)| 0 \leq y \leq 4, \space \frac{1}{2} y \leq x \leq \sqrt{y}}$$, respectively.

# 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.

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 15.2.1 inside $$R$$. Then $$g(x,y)$$ is integrable and we define the double integral of $$f(x,y)$$ over $$D$$ by

$\iint\limits_D f(x,y) dA = \iint\limits_R g(x,y) dA.$

The right-hand side of this equation is what we have seen before, so this theorem is reasonable because $$R$$ is a rectangle and $$\iint\limits_R g(x,y)dA$$ has been discussed in the preceding section. Also, the equality works because the values of $$g(x,y)$$ are $$0$$ for any point $$(x,y)$$ 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.

Theorem: Fubini’s Theorem (Strong Form)

For a function $$f(x,y)$$ that is continuous on a region $$D$$ of Type I, we have

$\iint\limits_D f(x,y)dA = \iint\limits_D f(x,y)dy \space dx = \int_a^b \left[\int_{g_1(x)}^{g_2(x)} f(x,y)dy \right] dx.$

Similarly, for a function $$f(x,y)$$ that is continuous on a region $$D$$ of Type II, we have

$\iint\limits_D f(x,y)dA = \iint\limits_D f(x,y)dx \space dy = \int_c^d \left[\int_{h_1(y)}^{h_2(y)} f(x,y)dy \right] dy.$

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 $$f(x,y)$$ with $$x$$ being held constant and the limits of integration being $$g_1(x)$$ and $$g_2(x)$$. In the inner integral in the second expression, we integrate $$f(x,y)$$ with $$y$$ being held constant and the limits of integration are $$h_1(x)$$ and $$h_2(x)$$.

Example $$\PageIndex{2}$$: Evaluating an Iterated Integral over a Type I Region

Evaluate the integral $$\iint \limits _D x^2 e^{xy} dA$$ where $$D$$ is shown in Figure 15.2.5.

Solution

First construct the region as a Type I region (Figure 15.2.5). Here $$D = {(x,y) |0 \leq x \leq 2, \space \frac{1}{2} x \leq y \leq 1}$$. Then we have

$\iint \limits _D x^2e^{xy} dA = \int_{x=0}^{x=2} \int_{y=1/2x}^{y=1} x^2e^{xy}dydx.$

Figure 15.2.5. 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}dydx &= \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, w could have looked at the region in another way, such as $$D = {(x,y)|0 \leq y \leq 1, \space 0 \leq x \leq 2y}$$ (Figure 15.2.6).

Figure 15.2.6.

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

$\iint \limits _D x^2e^{xy}dA = \int_{y=0}^{y=1} \int_{x=0}^{x=2y} x^2 e^{xy}dx \space dy.$

However, if we integrate first with respect to $$x$$ this integral is lengthy to compute because we have to use integration by parts twice.

Example $$\PageIndex{3}$$: Evaluating an Iterated Integral over a Type II Region

Evaluate the integral

$\iint \limits _D (3x^2 + y^2) dA$

where $$D = {(x,y)| -2 \leq y \leq 3, \space y^2 - 3 \leq x \leq y + 3}$$.

Solution

Notice that $$D$$ can be seen as either a Type I or a Type II region, as shown in Figure 15.2.7. However, in this case describing $$D$$ as Type I is more complicated than describing it as Type II. Therefore, we use $$D$$ as a Type II region for the integration.

Figure 15.2.7. The region $$D$$ in this example can be either (a) Type I or (b) Type II.

Choosing this order of integration, we have

\begin{align} \iint \limits _D (3x^2 + y^2)dA &= \int_{y=-2}^{y=3} \int_{x=y^2-3}^{x=y+3} (3x^2 + y^2) dx \space dy \\ &=\int_{y=-2}^{y=3} (x^3 + xy^2) |_{y^2-3}^{y+3} dy &\text{Iterated integral, Type II region}\\ &=\int_{y=-2}^{y=3} \left((y + 3)^3 + (y + 3)y^2 - (y^2 - 3)y^2\right)dy \\ &=\int_{-2}^3 (54 + 27y - 12y^2 + 2y^3 + 8y^4 - y^6)dy&\text{Integrate with repsect to x.} \\ &= \left[ 54y + \frac{27y^2}{2} - 4y^3 + \frac{y^4}{2} + \frac{8y^5}{5} - \frac{y^7}{7} \right]_{-2}^3 \\ &=\frac{2375}{7}. \end{align}

Exercise $$\PageIndex{2}$$

Sketch the region and evaluate the iterated integral $\iint \limits _D xy \space dy \space dx$ where is the region bounded by the curves $$y = \cos \space x$$ and $$y = \sin \space x$$ in the interval $$[-3\pi/4, \space \pi/4]$$.

Hint:

Express $$D$$ as a Type I region, and integrate with respect to $$y$$ first.

Solution

$$\frac{\pi}{4}$$

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 $$S \cap T = 0$$ except at their boundaries, then

$\iint \limits _R f(x,y)fA = \iint\limits _S f(x,y)dA + \iint_T f(x,y) dA.$

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

Theorem: Decomposing Regions into Smaller Regions

Suppose the region $$D$$ can be expressed as $$D = D_1 \cup D_2$$ where $$D_1$$ and $$D_2$$ do not overlap except at their boundaries. Then

$\iint \limits _D f(x,y) dA = \iint \limits _{D_1} f(x,y) dA + \iint \limits _{D_2} f(x,y)dA.$

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.

Example $$\PageIndex{4}$$: Decomposing Regions

Express the region $$D$$ shown in Figure 15.2.8 as a union of regions of Type I or Type II, and evaluate the integral

$\iint \limits _D (2x + 5y)dA.$

Figure 15.2.8. This region can be decomposed into a union of three regions of Type I or Type II.

Solution

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 $$D_1$$, $$D_2$$, and $$D_3$$ where, $$D_1 = {(x,y)| -2 \leq x \leq 0, \space 0 \leq y \leq (x + 2)^2 }$$, $$D_2 = {(x,y)|0 \leq y \leq 4, \space 0 \leq x \leq (y - \frac{1}{10} y^3)}$$. These regions are illustrated more clearly in Figure 15.2.9.

Figure 15.2.9. Breaking the region into three subregions makes it easier to set up the integration.

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

\begin{align} \iint\limits_D (2x + 5y)dA &= \iint\limits_{D_1} (2x + 5y)dA + \iint\limits_{D_2} (2x + 5y)dA + \iint\limits_{D_3} (2x + 5y)dA \\ &= \int_{x=-2}^{x=0} \int_{y=0}^{y=(x+2)^2} (2x + 5y) dy \space dx + \int_{y=0}^{y=4} \int_{x=0}^{x=y-(1/16)y^3} (2 + 5y)dx \space dy + \int_{y=-4}^{y=0} \int_{x=-2}^{x=y-(1/16)y^3} (2x + 5y)dx \space dy \\ &= \int_{x=-2}^{x=0} \left[\frac{1}{2}(2 + x)^2 (20 + 24x + 5x^2)\right] + \int_{y=0}^{y=4} \left[\frac{1}{256}y^6 - \frac{7}{16}y^4 + 6y^2 \right] \\ \;\;&+\int{y=-4}^{y=0} \left[\frac{1}{256}y^6 - \frac{7}{16}y^4 + 6y^2 + 10y - 4\right] \\ &= \frac{40}{3} + \frac{1664}{35} - \frac{1696}{35} = \frac{1304}{105}.\end{align}

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

Exercise $$\PageIndex{3}$$

Consider the region bounded by the curves $$y = ln \space x$$ and $$y = e^x$$ in the interval $$[1,2]$$. Decompose the region into smaller regions of Type II.

Hint:

Sketch the region, and split it into three regions to set it up.

Solution

${(x,y)|0 \leq y \leq 1, \space 1 \leq x \leq e^y} \cup {(x,y)|1 \leq y \leq e, \space 1 \leq x \leq 2} \cup {(x,y)|e \leq y \leq e^2, \space ln \space y \leq x \leq 2}$

Exercise $$\PageIndex{4}$$

Redo Example using a union of two Type II regions.

Hint:
${(x,y)|0 \leq y \leq 4, \space 2 + \sqrt{y} \leq x \leq (y - \frac{1}{16}y^3)} \cup {(x,y)| - 4 \leq y \leq 0, \space -2 \leq x \leq (y - \frac{1}{16}y^{13})}$

Solution

Same as in the example shown.

# 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.

Example $$\PageIndex{5}$$: Changing the Order of Integration

Reverse the order of integration in the iterated integral

$\int_{x=0}^{x=\sqrt{2}} \int_{y=0}^{y=2-x^2} xe^{x^2} dy \space dx.$

Then evaluate the new iterated integral.

Solution

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 15.2.10.

Figure 15.2.10. 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}}dx \space dy &\text{Reverse the order of integration then use substitution.} \\ &= \int_0^2 \left[\left.\frac{1}{2}ex^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

a. integrating first with respect to $$y$$ and then

b. integrating first with respect to $$x$$.

Solution

A sketch of the region appears in Figure 15.2.11.

Figure 15.2.11. 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 $$x = 1$$:

\begin{align} \iint\limits_R f(x,y) dx \space dy &= \int_{x=0}^{x=1} \int_{y=0}^{y=1-x} (x - 2y) dy \space dx = \int_{x=0}^{x=1}\left[xy - 2y^2\right]_{y=0}^{y=1-x} dx \\ &\int_{x=0}^{x=1} \left[ x(1 - x) - (1 - x)^2\right] dx = \int_{x=0}^{x=1} [ -1 + 3x - 2x^2] dx = \left[ -x + \frac{3}{2}x^2 - \frac{2}{3} x^3 \right]_{x=0}^{x=1} = -\frac{1}{6}. \end{align}

b. The other way to do this problem is by first integrating $$x$$ from $$x = 0$$ to $$x = 1 - y$$ horizontally and then integrating $$y$$ from $$y = 0$$ to $$y = 1$$:

\begin{align} \iint\limits_R f(x,y) dx \space dy &= \int_{y=0}^{y=1} \int_{x=0}^{x=1-y} (x - 2y) dx \space dy = \int_{y=0}^{y=1} \left[\frac{1}{2} x^2 = 2xy \right]_{x=0}^{x=1-y} dy \\&= \int_{y=0}^{y=1} \left[\frac{1}{2} (1 - y)^2 - 2y(1 - y)\right] dy = \int_{y=0}^{y=1}\left[\frac{1}{2} - 3y + \frac{5}{2} y^2\right] dy \\ &= \left[\frac{1}{2} y - \frac{3}{2} y^2 + \frac{5}{6}y^3 \right]_{y=0}^{y=1} = -\frac{1}{6}. \end{align}

Exercise $$\PageIndex{5}$$:

Evaluate the iterated integral $$\iint\limits_D (x^2 + y^2)dA$$ over the region $$D$$ in the first quadrant between the functions $$y = 2x$$ and $$y = x^2$$. Evaluate the iterated integral by integrating first with respect to $$y$$ and then integrating first with resect to $$x$$.

Hint:

Sketch the region and follow Example.

Solution

$$\frac{216}{35}$$

# 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.

Example $$\PageIndex{7}$$: Finding the Volume of a Tetrahedron

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

Solution

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 15.2.12). Note that we can consider the region as Type I or as Type II, and we can integrate in both ways.

Figure 15.2.12. A tetrahedron consisting of the three coordinate planes and the plane $$z = 6 - 2x - 3y$$, with the base bound by $$x = 0, \space y = 0$$, and $$2x + 3y = 6$$.

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

Therefore, the volume is

\begin{align} V &= \int_{x=0}^{x=3} \int_{y=0}^{y=2-(2x/3)} (6 - 2x - 3y) dy \space dx = \int_{x=0}^{x=3} \left[ \left.\left( 6y - 2xy - \frac{3}{2}y^2\right)\right|_{y=0}^{y=2-(2x/3)} \right] dx\\&= \int_{x=0}^{x=3} \left[\frac{2}{3} (x - 3)^2 \right] dx = 6. \end{align}

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

\begin{align} V &= \int_{y=0}^{y=2} \int_{x=0}^{x=3-(3y/2)} (6 - 2x - 3y)dx \space dy = \int_{y=0}^{y=2} \left[\left.(6x - x^2 - 3xy)\right|_{x=0}^{x=3-(3y/2)} \right] dy \\ &= \int_{y=0}^{y=2} \left[\frac{9}{4}(y - 2)^2 \right] dy = 6.\end{align}

Therefore, the volume is 6 cubic units.

Exercise $$\PageIndex{6}$$:

Find the volume of the solid bounded above by $$f(x,y) = 10 - 2x + y$$ over the region enclosed by the curves $$y = 0$$ and $$y = e^x$$ where $$x$$ is in the interval $$[0,1]$$.

Hint:

Sketch the region, and describe it as Type I.

Solution

$$\frac{e^2}{4} + 10e - \frac{49}{4}$$ 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.

Definition

The area of a plane-bounded region $$D$$ is defined as the double integral $\iint\limits_D 1dA.$

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.

Example $$\PageIndex{8}$$: 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 15.2.13).

Figure 15.2.13. The region bounded by $$y = x^2$$ and $$y = 2x$$.

Solution

We just have to integrate the constant function $$f(x,y) = 1$$ over the region. Thus, the area $$A$$ of the bounded region is $$\int_{x=0}^{x=2} \int_{y=x^2}^{y=2x} dy \space dx \space or \space \int_{y=0}^{y=4} \int_{x=y/2}^{x=\sqrt{y}} dx \space dy:$$

$A = \iint\limits_D 1dx \space dy = \int_{x=0}^{x=2} \int_{y=x^2}^{y=2x} 1dy \space dx = \int_{x=0}^{x=2} \left[\left.y\right|_{y=x^2}^{y=2x} \right] dx = \int_{x=0}^{x=2} (2x - x^2)dx = \left.x^2 - \frac{x^3}{3} \right|_0^2 = \frac{4}{3}.$

Exercise $$\PageIndex{7}$$:

Find the area of a region bounded above by the curve $$y = x^3$$ and below by $$y = 0$$ over the interval $$[0,3]$$.

Hint:

Sketch the region.

Solution

$$\frac{81}{4}$$ square units

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.

Definition

If $$f (x,y)$$ is integrable over a plane-bounded region $$D$$ with positive area $$A(D)$$, then the average value of the function is

$f_{ave} = \frac{1}{A(D)} \iint\limits_D f(x,y) dA.$

Note that the area is $$A(D) = \iint\limits_D 1dA$$.

Example $$\PageIndex{9}$$: Finding an Average Value

Find the average value of the function $$f(x,y) = 7xy^2$$ on the region bounded by the line $$x = y$$ and the curve $$x = \sqrt{y}$$ (Figure 15.2.14).

Figure 15.2.14. The region bounded by $$x = y$$ and $$x = \sqrt{y}$$.

Solution

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

$A(D) = \iint\limits_D 1dA = \int_{y=0}^{y=1} \int_{x=y}^{x=\sqrt{y}} 1dx \space dy = \int_{y=0}^{y=1} \left[\left. x \right|_{x=y}^{x=\sqrt{y}} \right] dy = \int_{y=0}^{y=1} (\sqrt{y} - y) dy = \frac{2}{3}\left. y^{2/3} - \frac{y^2}{2} \right|_0^1 = \frac{1}{6}$

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

\begin{align} f_{ave} &= \frac{1}{A(D)} \iint\limits_D f(x,y) dA = \frac{1}{A(D)} \int_{y=0}^{y=1}\int_{x=y}^{x=\sqrt{y}} 7xy^2 dx \space dy = \frac{1}{1/6} \int_{y=0}^{y=1} \left[ \left. \frac{7}{2} x^2y^2 \right|_{x=y}^{x=\sqrt{y}} \right] dy \\ &= 6 \int_{y=0}^{y=1} \left[ \frac{7}{2} y^2 (y - y^2)\right] dy = 6\int_{y=0}^{y=1} \left[ \frac{7}{2} (y^3 -y^4) \right] dy = \frac{42}{2} \left. \left( \frac{y^4}{4} - \frac{y^5}{5}\right) \right|_0^1 = \frac{42}{40} = \frac{21}{20}. \end{align}

Exercise $$\PageIndex{8}$$

Find the average value of the function $$f(x,y) = xy$$ over the triangle with vertices $$(0,0), \space (1,0)$$ and $$(1,3)$$.

Hint:

Express the line joining $$(0,0)$$ and $$(1,3)$$ as a function $$y = g(x)$$.

Solution

$$\frac{3}{4}$$

# Improper Double Integrals

An improper double integral is an integral $$\iint\limits_D fdA$$  where either $$D$$ is an unbounded region or $$f$$ is an unbounded function. For example, $$D = {(x,y) ||x - y| \geq 2}$$ is an unbounded region, and the function $$f(x,y) = 1/(1 - x^2 - 2y^2)$$ over the ellipse $$x^2 + 3y^2 \geq 1$$ is an unbounded function. Hence, both of the following integrals are improper integrals:

i.    $\iint\limits_D xy \space dA \space \text{where} \space D = {(x,y)| |x - y| \geq 2};$

ii.    $\iint\limits_D \frac{1}{1 - x^2 -2y^2}dA \space \text{where} \space D = {(x,y)|x^2 + 3y^2 \leq 1}.$

In this section we would like to deal with improper integrals of functions over rectangles or simple regions such that 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.

Theorem: Fubini’s Theorem for Improper Integrals

If $$D$$ is a bounded rectangle or simple region in the plane defined by

$${(x,y): a \leq x \leq b, \space g(x) \leq y \leq h(x) }$$ and also by

$${(x,y): c \leq y \leq d, \space j(y) \leq x \leq k(y)}$$ and $$f$$ is a nonnegative function on $$D$$ with finitely many discontinuities in the interior of $$D$$ then

$\iint\limits_D f \space dA = \int_{x=a}^{x=b} \int_{y=g(x)}^{y=h(x)} f(x,y) dy \space dx = \int_{y=c}^{y=d} \int_{x=j(y)}^{x=k(y)} f(x,y) dx \space dy$

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 $$D$$.

Example $$\PageIndex{10}$$: Evaluating a Double Improper Integral

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

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

Solution

First we plot the region $$D$$ (Figure 15.2.15); then we express it in another way.

Figure 15.2.15. The function $$f$$ is continuous at all points of the region $$D$$ except $$(0,0)$$.

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

$D = {(x,y): 0 \leq y \leq 1, \space y^2 \leq x \leq y}.$

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

$\int_{y=0}^{y=1} \int_{x=y^2}^{x=y} \frac{e^y}{y} dx \space dy.$

Therefore, we have

$\int_{y=0}^{y=1} \int_{x=y^2}^{x=y} \frac{e^y}{y} dx \space dy = \int_{y=0}^{y=1} \left. \frac{e^y}{y}x\right|_{x=y^2}^{x=y} dy = \int_{y=0}^{y=1} \frac{ey}{y} (y - y^2) dy = \int_0^1 (ey - ye^y)dy = e - 2.$

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 $$R$$.

Theorem: Improper Integrals on an Unbounded Region

If $$R$$ is an unbounded rectangle such as $$R = {(x,y): a \leq x \leq \infty, \space c \leq y \leq \infty }$$, then when the limit exists, we have

$\iint\limits_R f(x,y) dA = \lim_{(b,d) \rightarrow (\infty, \infty)} \int_a^b \left(\int_c^d f (x,y) dy \right) dx = \lim_{(b,d) \rightarrow (\infty, \infty)} \int_c^d \left(\int_a^b f(x,y) dy \right) dy.$

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

Example $$\PageIndex{11}$$

Evaluate the integral $$\iint\limits_R xye^{-x^2-y^2}dA$$ where $$R$$ is the first quadrant of the plane.

Solution

The region $$R$$ is the first quadrant of the plane, which is unbounded. So

\begin{align} \iint\limits_R xye^{-x^2-y^2} dA &= \lim_{(b,d) \rightarrow (\infty, \infty)} \int_{x=0}^{x=b} \left(\int_{y=0}^{y=d} xye^{-x^2-y^2} dy\right) dx \\ &= \lim_{(b,d) \rightarrow (\infty, \infty)} \int_{y=0}^{x=b} xye^{-x^2-y^2} dy \\&= \lim_{(b,d) \rightarrow (\infty, \infty)} \frac{1}{4} \left(1 - e^{-b^2}\right) \left( 1 - e^{-d^2}\right) = \frac{1}{4} \end{align}

Thus,

$\iint\limits_R xye^{-x^2-y^2}dA$

is convergent and the value is $$\frac{1}{4}$$.

Exercise $$\PageIndex{9}$$: Evaluate the improper integral

$\iint\limits_D \frac{y}{\sqrt{1 - x^2 - y^2}}dA$ where $$D = {(x,y) x \geq 0, \space y \geq 0, \space x^2 + y^2 \leq 1}.$$

Hint:

Notice that the integral is nonnegative and discontinuous on $$x^2 + y^2 = 1$$. Express the region $$D$$ as $$D = {(x,y): 0 \leq x \leq 1, \space 0 \leq y \leq \sqrt{1 - x^2}}$$ and integrate using the method of substitution.

Solution

$$\frac{\pi}{4}$$

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.

Definition

Consider a pair of continuous random variables $$X$$ and $$Y$$ such as the birthdays of two people or the number of sunny and rainy days in a month. The joint density function $$f$$ of $$X$$ and $$Y$$ satisfies the probability that $$(X,Y)$$ lies in a certain region $$D$$: $P((X,Y) \in D) = \iint\limits_D f(x,y) dA.$

Since the probabilities can never be negative and must lie between 0 and 1 the joint density function satisfies the following inequality and equation:

$f(x,y) \geq 0 \space and \space \iint\limits_R f(x,y) dA = 1.$

Definition

The variables $$X$$ and $$Y$$ are said to be independent random variables if their joint density function is the product of their individual density functions:

$f(x,y) = f_1(x) f_2(y).$

Example $$\PageIndex{12}$$: 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?

Solution

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

$f(t) = \begin{cases} 0 &\text{if}\; t<0 \\ \dfrac{1}{m}e^{-t/m} &\text{if} \; t\geq 0.\end{cases}$

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

$f_1(x) = \begin{cases} 0 &\text{if}\; x<0. \\ \dfrac{1}{15} e^{-x/15} &\text{if} \; x\geq 0. \end{cases} \; \text{and} \; f_2(y) = \begin{cases} 0 &\text{if}\; y<0 \\ \dfrac{1}{40} e^{-y/40} &\text{if}\; y\geq 0. \end{cases}$

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

$f(x,y) = f_1(x)f_2(y) = \begin{cases} 0 &\text{if} \; x<0 \; \text{or} \; y<0, \\ \dfrac{1}{600} e^{-x/15} &\text{if} \; x,y\geq 0 \end{cases}$

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 15.2.16).

Figure 15.2.16. The region of integration for a joint probability density function.

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

$$P(X + Y \leq 90) = P((X,Y) \in D) = \iint\limits_D f(x,y) fA = \iint\limits_D \frac{1}{600}e^{-x/15} e^{-y/40} dA.\] Since \(x + y = 90$$ is the same as $$y = 90 - x$$, we have a region of Type I, so

\begin{align} D &= {(x,y)|0 \leq x \leq 90, \space 0 \leq y \leq 90 - x}, \\ P(X + Y \leq 90) &= \frac{1}{600} \int_{x=0}^{x=90} \int_{y=0}^{y=90-x} e^{-(/15}e^{-y/40}dx \space dy = \frac{1}{600} \int_{x=0}^{x=90} \int_{y=0}^{y=90-x}e^{-x/15}e^{-y/40} dx \space dy \\ &= \frac{1}{600} \int_{x=0}^{x=90} \int_{y=0}^{y=90-x} e^{-(x/15+y/40)}dx \space dy = 0.8328 \end{align}

Thus, there is an $$83.2\%$$ 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.

Note: Definition

In probability theory, we denote the expected values $$E(X)$$ and $$E(Y)$$ respectively, as the most likely outcomes of the events. The expected values $$E(X)$$ and $$E(Y)$$ are given by

$E(X) = \iint\limits_S xf(x,y) dA \space and \space E(Y) = \iint\limits_S yf (x,y) dA,$

where S is the sample space of the random variables $$X$$ and $$Y$$.

Example $$\PageIndex{13}$$:

Finding Expected Value

Find the expected time for the events ‘waiting for a table’ and ‘completing the meal’ in Example.

Solution

Using the first quadrant of the rectangular coordinate plane as the sample space, we have improper integrals for $$E(X)$$ and $$E(Y)$$. The expected time for a table is

\begin{align} E(X) &= \iint\limits_S x\frac{1}{600} e^{-x/15}e^{-y/40} dA = \frac{1}{600} \int_{x=0}^{x=\infty} \int_{y=0}^{y=\infty} xe^{-x/15} e^{-y/40}dA\\&= \frac{1}{600} \lim_{(a,b) \rightarrow (\infty, \infty)} \int_{x=0}^{x=a} \int_{y=0}^{y=b} xe^{-x/15} e^{-y/40} dx \space dy \\ &= \frac{1}{600} \left(\lim_{a\rightarrow \infty}\int_{x=0}^{x=a} xe^{-x/15} dx \right) \left( \lim_{b\rightarrow \infty} \int_{y=0}^{y=b} e^{-y/40} dy \right) \\ & \frac{1}{600}\left(\left. (\lim_{a\rightarrow \infty} (-15e^{-x/15}(x + 15)))\right|_{x=0}^{x=a} \right) \left( \left. ( \lim_{b\rightarrow \infty}(-40e^{-y/40})) \right|_{y=0}^{y=b}\right) \\&=\frac{1}{600} \left( \lim_{a \rightarrow \infty} (-15e^{-a/15} (x + 15) + 225) \right)\left(\lim_{b\rightarrow \infty} (-40e^{-b/40} + 40) \right) \\ &= \frac{1}{600} (225)(40) = 15. \end{align}

A similar calculation shows that $$E(Y) = 40$$. This means that the expected values of the two random events are the average waiting time and the average dining time, respectively.

Exercise $$\PageIndex{10}$$:

The joint density function for two random variables $$X$$ and $$Y$$ is given by

$f(x,y) =\begin{cases}\frac{1}{600} (x^2 + y^2)\; &\text{if} \; \leq x \leq 15, \; 0 \leq y \leq 10 \\ 0 &\text{otherwise} \end{cases}$

Find the probability that $$X$$ is at most 10 and $$Y$$ is at least 5.

Hint:

Compute the probability

$P(X \leq 10, \space Y \geq 5) = \int_{x=-\infty}^{10} \int_{y=5}^{y=10} \frac{1}{6000} (x^2 + y^2) dy \space dx.$

Solution

$$\frac{55}{72} \approx 0.7638$$

# Key Concepts

• A general bounded region $$D$$  on the plane is a region that can be enclosed inside a rectangular region. We can use this idea to define a double integral over a general bounded region.
• To evaluate an iterated integral of a function over a general nonrectangular region, we sketch the region and express it as a Type I or as a Type II region or as a union of several Type I or Type II regions that overlap only on their boundaries.
• We can use double integrals to find volumes, areas, and average values of a function over general regions, similarly to calculations over rectangular regions.
• We can use Fubini’s theorem for improper integrals to evaluate some types of improper integrals.

# Key Equations

Iterated integral over a Type I region

$\iint\limits_D f(x,y) dA = \iint\limits_D f(x,y) dy \space dx = \int_a^b \left[\int_{g_1(x)}^{g_2(x)} f(x,y) dy \right] dx$

Iterated integral over a Type II region

$\iint\limits_D f(x,y) dA = \iint\limits_D (x,y)dx \space dy = \int_c^d \left[ \int_{h_1(y)}^{h_2(y)} f(x,y) dx \right] dy$

### Contributors

In the following exercises, specify whether the region is of Type I or Type II.

The region $$D$$ bounded by $$y = x^3, \space y = x^3 + 1, \space x = 0,$$ and $$x = 1$$ as given in the following figure.

Find the average value of the function $$f(x,y) = 3xy$$ on the region graphed in the previous exercise.

[Hide Solution]

$$\frac{27}{20}$$

Find the area of the region $$D$$ given in the previous exercise.

The region $$D$$ bounded by $$y = sin \space x, \space y = 1 + sin \space x, \space x = 0$$, and $$x = \frac{\pi}{2}$$ as given in the following figure.

[Hide Solution]

Type I but not Type II

Find the average value of the function $$f(x,y) = cos \space x$$ on the region graphed in the previous exercise.

Find the area of the region $$D$$ given in the previous exercise.

[Hide Solution]

$$\frac{\pi}{2}$$

The region $$D$$ bounded by $$x = y^2 - 1$$ and $$x = \sqrt{1 - y^2}$$ as given in the following figure.

Find the volume of the solid under the graph of the function $$f(x,y) = xy + 1$$ and above the region in the figure in the previous exercise.

[Hide Solution]

$$\frac{1}{6}(8 + 3\pi)$$

The region $$D$$ bounded by $$y = 0, \space x = -10 + y,$$ and $$x = 10 - y$$ as given in the following figure.

Find the volume of the solid under the graph of the function $$f(x,y) = x + y$$ and above the region in the figure from the previous exercise.

[Hide Solution]

$$\frac{1000}{3}$$

The region $$D$$ bounded by $$y = 0, \space x = y - 1, \space x = \frac{\pi}{2}$$ as given in the following figure.

The region $$D$$ bounded by $$y = 0$$ and $$y = x^2 - 1$$  as given in the following figure.

Type I and Type II

Let $$D$$ be the region bounded by the curves of equations $$y = x, \space y = -x$$ and $$y = 2 - x^2$$. Explain why $$D$$ is neither of Type I nor II.

Let $$D$$ be the region bounded by the curves of equations $$y = cos \space x$$ and $$y = 4 - x^2$$ and the $$x$$-axis. Explain why $$D$$ is neither of Type I nor II.

[Hide Solution]

The region $$D$$ is not of Type I: it does not lie between two vertical lines and the graphs of two continuous functions $$g_1(x)$$ and $$g_2(x)$$. The region is not of Type II: it does not lie between two horizontal lines and the graphs of two continuous functions $$h_1(y)$$ and $$h_2(y)$$.

In the following exercises, evaluate the double integral $$\iint_D f(x,y) dA$$ over the region $$D$$.

$$f(x,y) = 2x + 4y$$ and

$$D = {(x,y)|0 \leq x \leq 1, \space x^3 \leq y \leq x^3 + 1}$$

$$f(x,y) = 1$$ and

$$D = {(x,y)|0 \leq x \leq \frac{\pi}{2}, \space sin \space x \leq y \leq 1 + sin \space x}$$

[Hide Solution]

$$\frac{\pi}{2}$$

$$f(x,y) = 2$$ and

$$D = {(x,y)|0 \leq y \leq 1, \space y - 1 \leq x \leq arccos \space y}$$

$$f(x,y) = xy$$ and

$$D = {(x,y)| -1 \leq y \leq 1, \space y^2 - 1 \leq x \leq \sqrt{1 - y^2}}$$

[Hide Solution]

0

$$f(x,y) = sin \space y$$ and $$D$$ is the triangular region with vertices $$(0,0), \space (0,3)$$, and $$(3,0)$$

$$f(x,y) = -x + 1$$ and $$D$$ is the triangular region with vertices  $$(0,0), \space (0,2)$$, and $$(2,2)$$

[Hide Solution]

$$\frac{2}{3}$$

Evaluate the iterated integrals.

$\int_0^3 \int_{2x}^{3x} (x + y^2) dy \space dx$

$\int_0^1 \int_{2\sqrt{x}}^{2\sqrt{x}+1} (xy + 1) dy \space dx$

[Hide Solution]

$$\frac{41}{20}$$

$\int_e^{e^2} \int_{ln \space u}^2 (v + ln \space u) dv \space du$

$\int_1^2 \int_{-u^2-1}^{-u} (8 uv) dv \space du$

[Hide Solution]

$$-63$$

$\int_0^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (2x + 4y^3) dx \space dy$

$\int_0^1 \int_{-\sqrt{1-4y^2}}^{\sqrt{1-4y^2}} 4dx \space dy$

[Hide Solution]

$$\pi$$

Let $$D$$ be the region bounded by $$y = 1 - x^2, \space y = 4 - x^2$$, and the $$x$$- and $$y$$-axes.

a.    Show that

$\iint_D xdA = \int_0^1 \int_{1-x^2}^{4-x^2} x \space dy \space dx + \int_1^2 \int_0^{4-x^2} x \space dy \space dx$ by dividing the region $$D$$ into two regions of Type I.

b.     Evaluate the integral $\iint_D s \space dA.$

Let $$D$$ be the region bounded by $$y = 1, \space y = x, \space y = ln \space x$$, and the $$x$$-axis.

a.    Show that

$\iint_D y^2 dA = \int_{-1}^0 \int_{-x}^{2-x^2} y^2 dy \space dx + \int_0^1 \int_x^{2-x^2} y^2 dy \space dx$ by dividing the region $$D$$ into two regions of Type I, where $$D = {(x,y)|y \geq x, y \geq -x, \space y \leq 2-x^2}$$.

b.     Evaluate the integral $\iint_D y^2 dA.$

Let $$D$$ be the region bounded by $$y = x^2$$, $$y = x + 2$$, and $$y = -x$$.

a. Show that $\iint_D x \space dA = \int_0^1 \int_{-y}^{\sqrt{y}} x \space dx \space dy + \int_1^2 \int_{y-2}^{\sqrt{y}} x \space dx \space dy$ by dividing the region $$D$$ into two regions of Type II, where $$D = {(x,y)|y \geq x^2, \space y \geq -x, \space y \leq x + 2}. b. Evaluate the integral $\iint_D x \space dA.$ [Hide Solution] a. Answers may vary; b. \(\frac{8}{12}$$

The region $$D$$ bounded by $$x = 0, y = x^5 + 1$$, and $$y = 3 - x^2$$ is shown in the following figure. Find the area $$A(D)$$ of the region $$D$$.

The region $$D$$ bounded by $$y = cos \space x, \space y = 4 \space cos \space x$$, and $$x = \pm \frac{\pi}{3}$$ is shown in the following figure. Find the area $$A(D)$$ of the region $$D$$.

[Hide Solution]

$$\frac{8\pi}{3}$$

Find the area $$A(D)$$ of the region $$D = {(x,y)|y \geq 1 - x^2, y \leq 4 - x^2, \space y \geq 0, \space x \geq 0}.$$

Let $$D$$ be the region bounded by $$y = 1, \space y = x, \space y = ln \space x$$, and the $$x$$-axis. Find the area $$A(D)$$ of the region $$D$$.

[Hide Solution]

$$e - \frac{3}{2}$$

Find the average value of the function $$f(x,y) = sin \space y$$ on the triangular region with vertices $$(0,0), \space (0,3)$$, and $$(3,0)$$.

Find the average value of the function $$f(x,y) = -x + 1$$ on the triangular region with vertices $$(0,0), \space (0,2)$$, and $$(2,2)$$.

[Hide Solution]

$$\frac{2}{3}$$

In the following exercises, change the order of integration and evaluate the integral. $\int_{-1}^{\pi/2} \int_0^{x+1} sin \space x \space dy \space dx$

$\int_0^1 \int_{x-1}^{1-x} x \space dy \space dx$

[Hide Solution]

$\int_0^1 \int_{x-1}^{1-x} x \space dy \space dx = \int_{-1}^0 \int_0^{y+1} x \space dx \space dy + \int_0^1 \int_-^{1-y} x \space dx \space dy = \frac{1}{3}$

$\int_{-1}^0 \int_{-\sqrt{y+1}}^{\sqrt{y+1}} y^2 dx \space dy$

$\int_{-1/2}^{1/2} \int_{-\sqrt{y^2+1}}^{\sqrt{y^2+1}} y \space dx \space dy$

[Hide Solution]

$\int_{-1/2}^{1/2} \int_{-\sqrt{y^2+1}}^{\sqrt{y^2+1}} y \space dx \space dy = \int_1^2 \int_{-\sqrt{x^2-1}}^{\sqrt{x^2-1}} y \space dy \space dx = 0$

The region $$D$$ is shown in the following figure. Evaluate the double integral $\iint_D (x^2 + y) dA$ by using the easier order of integration.

The region $$D$$ is shown in the following figure. Evaluate the double integral $\iint_D (x^2 - y^2) dA$ by using the easier order of integration.

[Hide Solution]

$\iint_D (x^2 - y^2) dA = \int_{-1}^1 \int_{y^4-1}^{1-y^4} (x^2 - y^2)dx \space dy = \frac{464}{4095}$

Find the volume of the solid under the surface $$z = 2x + y^2$$ and above the region bounded by $$y = x^5$$ and $$y = x$$.

Find the volume of the solid under the plane $$z = 3x + y$$ and above the region determined by $$y = x^7$$ and $$y = x$$.

[Hide Solution]

$$\frac{4}{5}$$

Find the volume of the solid under the plane $$z = 3x + y$$ and above the region bounded by $$x = tan \space y, \space x = -tan \space y$$, and $$x = 1$$.

Find the volume of the solid under the surface $$z = x^3$$ and above the plane region bounded by $$x = sin \space y, \space x = -sin \space y$$, and $$x = 1$$.

[Hide Solution]

$$\frac{5\pi}{32}$$

Let $$g$$ be a positive, increasing, and differentiable function on the interval $$[a,b]$$. Show that the volume of the solid under the surface $$z = g'(x)$$ and above the region bounded by $$y = 0, \space y = g(x), \space x = a$$, and $$x = b$$ is given by $$\frac{1}{2}(g^2 (b) - g^2 (a))$$.

Let $$g$$ be a positive, increasing, and differentiable function on the interval $$[a,b]$$ and let $$k$$ be a positive real number. Show that the volume of the solid under the surface $$z = g'(x)$$ and above the region bounded by $$y = g(x), \space y = g(x) + k, \space x = a$$, and $$x = b$$ is given by $$k(g(b) - g(a)).$$

Find the volume of the solid situated in the first octant and determined by the planes $$z = 2$$, $$z = 0, \space x + y = 1, \space x = 0$$, and $$y = 0$$.

Find the volume of the solid situated in the first octant and bounded by the planes $$x + 2y = 1$$, $$x = 0, \space z = 4$$, and $$z = 0$$.

[Hide Solution]

1

Find the volume of the solid bounded by the planes $$x + y = 1, \space x - y = 1, \space x = 0, \space z = 0$$, and $$z = 10$$.

Find the volume of the solid bounded by the planes $$x + y = 1, \space x - y = 1, \space x - y = -1, \space z = 1$$, and $$z = 0$$

[Hide Solution]

2

Let $$S_1$$ and $$S_2$$ be the solids situated in the first octant under the planes $$x + y + z = 1$$ and $$x + y + 2z = 1$$ respectively, and let $$S$$ be the solid situated between $$S_1, \space S_2, \space x = 0$$, and $$y = 0$$.

1. Find the volume of the solid $$S_1$$.
2. Find the volume of the solid $$S_2$$.
3. Find the volume of the solid $$S$$ by subtracting the volumes of the solids $$S_1$$ and $$S_2$$.

Let $$S_1$$ and $$S_2$$ be the solids situated in the first octant under the planes $$2x + 2y + z = 2$$ and $$x + y + z = 1$$ respectively, and let $$S$$ be the solid situated between $$S_1, \space S_2, \space x = 0$$, and $$y = 0$$.

1. Find the volume of the solid $$S_1$$.
2. Find the volume of the solid $$S_2$$.
3. Find the volume of the solid $$S$$ by subtracting the volumes of the solids $$S_1$$ and $$S_2$$.

[Hide Solution]

a. $$\frac{1}{3}$$; b. $$\frac{1}{6}$$; c. $$\frac{1}{6}$$.

Let $$S_1$$ and $$S_2$$ be the solids situated in the first octant under the plane $$x + y + z = 2$$ and under the sphere $$x^2 + y^2 + z^2 = 4$$, respectively. If the volume of the solid $$S_2$$ is $$\frac{4\pi}{3}$$ determine the volume of the solid $$S$$ situated between $$S_1$$ and $$S_2$$ by subtracting the volumes of these solids.

Let $$S_1$$ and $$S_2$$ be the solids situated in the first octant under the plane $$x + y + z = 2$$ and under the sphere $$x^2 + y^2 = 4$$, respectively.

1. Find the volume of the solid $$S_1$$.
2. Find the volume of the solid $$S_2$$.
3. Find the volume of the solid $$S$$ situated between $$S_1$$ and $$S_2$$ by subtracting the volumes of the solids $$S_1$$ and $$S_2$$.

[Hide Solution]

a. $$\frac{4}{3}$$; b. $$2\pi$$; c. $$\frac{6\pi - 4}{3}$$

[T] The following figure shows the region $$D$$ bounded by the curves $$y = sin \space x, \space x = 0$$, and $$y = x^4$$. Use a graphing calculator or CAS to find the -coordinates of the intersection points of the curves and to determine the area of the region $$D$$. Round your answers to six decimal places.

[T] The region $$D$$ bounded by the curves $$y = cos \space x, \space x = 0$$, and $$y = x^3$$ is shown in the following figure. Use a graphing calculator or CAS to find the x-coordinates of the intersection points of the curves and to determine the area of the region $$D$$. Round your answers to six decimal places.

[Hide Solution]

0 and 0.865474; $$A(D) = 0.621135$$

Suppose that $$(X,Y)$$ is the outcome of an experiment that must occur in a particular region in the $$xy$$-plane. In this context, the region $$S$$ is called the sample space of the experiment and are random variables. If is a region included in then the probability of (X,Y) being in is defined as $$P[(X,Y) \in D] = \iint_D p(x,y)dx \space dy$$, where $$p(x,y)$$ is the joint probability density of the experiment. Here, is a nonnegative function for which $$\iint_S p(x,y) dx \space dy = 1$$. Assume that a point $$(X,Y)$$ is chosen arbitrarily in the square $$[0,3] \times [0,3]$$ with the probability density

$p(x,y) = \frac{1}{9} (x,y) \in [0,3] \times [0,3],$

$p(x,y) = 0 \space otherwise$

Find the probability that the point $$(X,Y)$$ is inside the unit square and interpret the result.

Consider $$X$$ and $$Y$$ two random variables of probability densities $$p_1(x)$$ and $$p_2(x)$$, respectively. The random variables $$X$$ and $$Y$$ are said to be independent if their joint density function is given by $$p_(x,y) = p_1(x)p_2(y)$$. At a drive-thru restaurant, customers spend, on average, 3 minutes placing their orders and an additional 5 minutes paying for and picking up their meals. Assume that placing the order and paying for/picking up the meal are two independent events $$X$$ and $$Y$$. If the waiting times are modeled by the exponential probability densities

$p_1(x) = \frac{1}{3}e^{-x/3} \space x\geq 0,$

$p_1(x) = 0 \space ptherwise$

$p_2(y) = \frac{1}{5} e^{-y/5} \space y \geq 0$

$p_2(y) = 0 \space otherwise$

respectively, the probability that a customer will spend less than 6 minutes in the drive-thru line is given by $$P[X + Y \leq 6] = \iint_D p(x,y) dx \space dy$$, where $$D = {(x,y)|x \geq 0, \space y \geq 0, \space x + y \leq 6}$$. Find $$P[X + Y \leq 6]$$ and interpret the result.

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$$P[X + Y \leq 6] = 1 + \frac{3}{2e^2} - \frac{5}{e^{6/5}} \approx 0.45$$; there is a $$45\%$$ chance that a customer will spend minutes in the drive-thru line.

[T] The Reuleaux triangle consists of an equilateral triangle and three regions, each of them bounded by a side of the triangle and an arc of a circle of radius s centered at the opposite vertex of the triangle. Show that the area of the Reuleaux triangle in the following figure of side length $$s$$ is $$\frac{s^2}{2}(\pi - \sqrt{3})$$.

[T] Show that the area of the lunes of Alhazen, the two blue lunes in the following figure, is the same as the area of the right triangle ABC. The outer boundaries of the lunes are semicircles of diameters $$AB$$ and $$AC$$ respectively, and the inner boundaries are formed by the circumcircle of the triangle $$ABC$$.

## Glossary

improper double integral
a double integral over an unbounded region or of an unbounded function
Type I
a region $$D$$ in the $$xy$$- plane is Type I if it lies between two vertical lines and the graphs of two continuous functions $$g_1(x)$$ and $$g_2(x)$$
Type II
a region $$D$$ in the $$xy$$-plane is Type II if it lies between two horizontal lines and the graphs of two continuous functions $$h_1(y)$$ and $$h_2(h)$$