Double Integrals over General Regions

In Double Integrals over Rectangular Regions, 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

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

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),

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.

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

has been discussed in the preceding section. Also, the equality works because the values of

g (x, y)

and hence these points do not add anything to the integral. However, it is important that the rectangle

R

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

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)

being held constant and the limits of integration being

g_{1} (x) and g_{2} (x) .

being held constant and the limits of integration are

h_{1} (x) and h_{2} (x) .

In [link], we could have looked at the region in another way, such as

D = {(x, y) \| 0 \leq y \leq 1, 0 \leq x \leq 2 y}

this integral is lengthy to compute because we have to use integration by parts twice.

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

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.

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.

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 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.

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.

Improper Double Integrals

is an unbounded function. Hence, both of the following integrals are improper integrals:

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 .

As mentioned before, we also have an improper integral if the region of integration is unbounded. Suppose now that the function

f

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.

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.

Key Concepts

Key Equations

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

The region $D$

bounded by $y = x^{3},$

y = x^{3} + 1,

x = 0,

and $x = 1$

as given in the following figure.

A region is bounded by y = 1 + x cubed, y = x cubed, x = 0, and x = 1.

Find the average value of the function $f (x, y) = 3 x y$

on the region graphed in the previous exercise.

\frac{27}{20}

Find the area of the region $D$

given in the previous exercise.

The region $D$

bounded by $y = sin x, y = 1 + sin x, x = 0, and x = \frac{π}{2}$

as given in the following figure.

A region is bounded by y = 1 + sin x, y = sin x, x = 0, and x = pi/2.

Type I but not Type II

Find the average value of the function $f (x, y) = cos x$

on the region graphed in the previous exercise.

Find the area of the region $D$

given in the previous exercise.

\frac{π}{2}

The region $D$

bounded by $x = y^{2} - 1$

and $x = \sqrt{1 - y^{2}}$

as given in the following figure.

A region is bounded by x = negative 1 + y squared and x = the square root of the quantity (1 minus y squared).

Find the volume of the solid under the graph of the function $f (x, y) = x y + 1$

and above the region in the figure in the previous exercise.

\frac{1}{6} (8 + 3 π)

The region $D$

bounded by $y = 0, x = −10 + y, and x = 10 - y$

as given in the following figure.

A region is bounded by x = negative 10 + y, x = 10 minus y, and y = 0.

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.

\frac{1000}{3}

The region $D$

bounded by $y = 0, x = y - 1,$

x = \frac{π}{2}

as given in the following figure.

A region is bounded by x = pi/2, y = 0, and x = negative 1 + y.

The region $D$

bounded by $y = 0$

and $y = x^{2} - 1$

as given in the following figure.

A region is bounded by y = 0 and y = negative 1 + x squared.

Type I and Type II

Let $D$

be the region bounded by the curves of equations $y = x, y = − x,$

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 $y = cos x$

and $y = 4 - x^{2}$

and the $x$

-axis. Explain why $D$

is neither of Type I nor II.

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 $D$

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 $\iint_{D} f (x, y) d A$

over the region $D .$

f (x, y) = 2 x + 5 y

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

f (x, y) = 1

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

\frac{π}{2}

f (x, y) = 2

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

f (x, y) = x y

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

0

f (x, y) = sin y

and $D$

is the triangular region with vertices $(0, 0), (0, 3), and (3, 0)$

f (x, y) = − x + 1

and $D$

is the triangular region with vertices $(0, 0), (0, 2), and (2, 2)$

\frac{2}{3}

Evaluate the iterated integrals.

\int_{0}^{1} \int_{2 x}^{3 x} (x + y^{2}) d y d x

\int_{0}^{1} \int_{2 \sqrt{x}}^{2 \sqrt{x} + 1} (x y + 1) d y d x

\frac{41}{20}

\int_{e}^{e^{2}} \int_{ln u}^{2} (v + ln u) d v d u

\int_{1}^{2} \int_{− u^{2} - 1}^{− u} (8 u v) d v d u

−63

\int_{0}^{1} \int_{− \sqrt{1 - y^{2}}}^{\sqrt{1 - y^{2}}} (2 x + 4 x^{3}) d x d y

\int_{0}^{1 / 2} \int_{− \sqrt{1 - 4 y^{2}}}^{\sqrt{1 - 4 y^{2}}} 4 d x d y

π

Let $D$

be the region bounded by $y = 1 - x^{2}, y = 4 - x^{2},$

and the $x$

and $y$

-axes.

Show that $\iint_{D} x d A = \int_{0}^{1} \int_{1 - x^{2}}^{4 - x^{2}} x d y d x + \int_{1}^{2} \int_{0}^{4 - x^{2}} x d y d x$
by dividing the region
$D$
into two regions of Type I.
Evaluate the integral $\iint_{D} x d A .$

Let $D$

be the region bounded by $y = 1,$

y = x,

y = ln x,

and the $x$

-axis.

Show that $\iint_{D} y d A = \int_{0}^{1} \int_{0}^{x} y d y d x + \int_{1}^{e} \int_{ln x}^{1} y d y d x$
by dividing
$D$
into two regions of Type I.
Evaluate the integral $\iint_{D} y d A .$

a. Answers may vary; b. $\frac{2}{3}$

Show that $\iint_{D} y^{2} d A = \int_{−1}^{0} \int_{− x}^{2 - x^{2}} y^{2} d y d x + \int_{0}^{1} \int_{x}^{2 - x^{2}} y^{2} d y d x$
by dividing the region
$D$
into two regions of Type I, where
$D = {(x, y) \| y \geq x, y \geq - x, y \leq 2 - x^{2}} .$
Evaluate the integral $\iint_{D} y^{2} d A .$

Let $D$

be the region bounded by $y = x^{2}, y = x + 2,$

and $y = − x .$

Show that $\iint_{D} x d A = \int_{0}^{1} \int_{− y}^{\sqrt{y}} x d x d y + \int_{1}^{2} \int_{y - 2}^{\sqrt{y}} x d x d y$
by dividing the region
$D$
into two regions of Type II, where
$D = {(x, y) \| y \geq x^{2}, y \geq - x, y \leq x + 2} .$
Evaluate the integral $\iint_{D} x d A .$

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 $D .$

A region is bounded by y = 1 + x to the fifth power, y = 3 minus x squared, and x = 0.

The region $D$

bounded by $y = cos x, y = 4 cos x,$

and $x = \pm \frac{π}{3}$

is shown in the following figure. Find the area $A (D)$

of the region $D .$

A region is bounded by y = cos x, y = 4 + cos x, x = negative 1, and x = 1.

\frac{8 π}{3}

Find the area $A (D)$

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

Let $D$

be the region bounded by $y = 1, y = x, y = ln x,$

and the $x$

-axis. Find the area $A (D)$

of the region $D .$

e - \frac{3}{2}

Find the average value of the function $f (x, y) = sin y$

on the triangular region with vertices $(0, 0), (0, 3),$

and $(3, 0) .$

Find the average value of the function $f (x, y) = − x + 1$

on the triangular region with vertices $(0, 0), (0, 2),$

and $(2, 2) .$

\frac{2}{3}

In the following exercises, change the order of integration and evaluate the integral.

\int_{−1}^{π / 2} \int_{0}^{x + 1} sin x d y d x

\int_{0}^{1} \int_{x - 1}^{1 - x} x d y d x

\int_{0}^{1} \int_{x - 1}^{1 - x} x d y d x = \int_{−1}^{0} \int_{0}^{y + 1} x d x d y + \int_{0}^{1} \int_{0}^{1 - y} x d x d y = \frac{1}{3}

\int_{−1}^{0} \int_{− \sqrt{y + 1}}^{\sqrt{y + 1}} y^{2} d x d y

\int_{− 1 / 2}^{1 / 2} \int_{− \sqrt{y^{2} + 1}}^{\sqrt{y^{2} + 1}} y d x d y

\int_{− 1 / 2}^{1 / 2} \int_{− \sqrt{y^{2} + 1}}^{\sqrt{y^{2} + 1}} y d x d y = \int_{1}^{2} \int_{− \sqrt{x^{2} - 1}}^{\sqrt{x^{2} - 1}} y d y d x = 0

The region $D$

is shown in the following figure. Evaluate the double integral $\iint_{D} (x^{2} + y) d A$

by using the easier order of integration.

A region is bounded by y = negative 4 + x squared and y = 4 minus x squared.

The region $D$

is given in the following figure. Evaluate the double integral $\iint_{D} (x^{2} - y^{2}) d A$

by using the easier order of integration.

A region is bounded by y to the fourth power = 1 minus x and y to the fourth power = 1 + x.

\iint_{D} (x^{2} - y^{2}) d A = \int_{−1}^{1} \int_{y^{4} - 1}^{1 - y^{4}} (x^{2} - y^{2}) d x d y = \frac{464}{4095}

Find the volume of the solid under the surface $z = 2 x + y^{2}$

and above the region bounded by $y = x^{5}$

and $y = x .$

Find the volume of the solid under the plane $z = 3 x + y$

and above the region determined by $y = x^{7}$

and $y = x .$

\frac{4}{5}

Find the volume of the solid under the plane $z = x - y$

and above the region bounded by $x = tan y, x = − tan y,$

and $x = 1 .$

Find the volume of the solid under the surface $z = x^{3}$

and above the plane region bounded by $x = sin y, x = − sin y,$

and $x = 1 .$

\frac{5 π}{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 $y = 0,$

y = g (x),

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 $y = g (x), y = g (x) + k, x = a,$

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, x + y = 1, x = 0, and y = 0 .

Find the volume of the solid situated in the first octant and bounded by the planes $x + 2 y = 1,$

x = 0, y = 0, z = 4, and z = 0 .

1

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

and $z = 10 .$

Find the volume of the solid bounded by the planes $x + y = 1, x - y = 1, x + y = −1,$

x - y = −1, z = 1 and z = 0 .

2

Let $S_{1}$

and $S_{2}$

be the solids situated in the first octant under the planes $x + y + z = 1$

and $x + y + 2 z = 1,$

respectively, and let $S$

be the solid situated between $S_{1}, S_{2}, x = 0, and y = 0 .$

Find the volume of the solid $S_{1} .$
Find the volume of the solid $S_{2} .$
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 $2 x + 2 y + z = 2$

and $x + y + z = 1,$

respectively, and let $S$

be the solid situated between $S_{1}, S_{2}, x = 0, and y = 0 .$

Find the volume of the solid $S_{1} .$
Find the volume of the solid $S_{2} .$
Find the volume of the solid $S$
by subtracting the volumes of the solids
$S_{1} and S_{2} .$

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 $\frac{4 π}{3},$

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 bounded by the cylinder $x^{2} + y^{2} = 4,$

respectively.

Find the volume of the solid $S_{1} .$
Find the volume of the solid $S_{2} .$
Find the volume of the solid $S$
situated between
$S_{1}$
and
$S_{2}$
by subtracting the volumes of the solids
$S_{1}$
and
$S_{2} .$

a. $\frac{4}{3};$

b. $2 π;$

c. $\frac{6 π - 4}{3}$

[T] The following figure shows the region $D$

bounded by the curves $y = sin x,$

x = 0,

and $y = x^{4} .$

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.

A region is bounded by y = sin x, y = x to the fourth power, and x = 0.

[T] The region $D$

bounded by the curves $y = cos x, x = 0, and y = x^{3}$

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.

A region is bounded by y = cos x, y = x cubed, and x = 0.

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 $S$

in the $x y$

-plane. In this context, the region $S$

is called the sample space of the experiment and $X and Y$

are random variables. If $D$

is a region included in $S,$

then the probability of $(X, Y)$

being in $D$

is defined as $P [(X, Y) \in D] = \iint_{D} p (x, y) d x d y,$

where $p (x, y)$

is the joint probability density of the experiment. Here, $p (x, y)$

is a nonnegative function for which $\iint_{S} p (x, y) d x d y = 1 .$

Assume that a point $(X, Y)$

is chosen arbitrarily in the square $[0, 3] \times [0, 3]$

with the probability density

p (x, y) = {\begin{matrix} \frac{1}{9} & (x, y) \in [0, 3] \times [0, 3], \\ 0 & otherwise . \end{matrix}

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

\begin{matrix} p_{1} (x) = {\begin{matrix} \frac{1}{3} e^{− x / 3} & x \geq 0, \\ 0 & otherwise, \end{matrix} & and & p_{2} (y) = {\begin{matrix} \frac{1}{5} e^{− y / 5} & y \geq 0, \\ 0 & otherwise, \end{matrix} \end{matrix}

respectively, the probability that a customer will spend less than 6 minutes in the drive-thru line is given by $P [X + Y \leq 6] = \iint_{D} p (x, y) d x d y,$

where $D = {(x, y)} \| x \geq 0, y \geq 0, x + y \leq 6} .$

Find $P [X + Y \leq 6]$

and interpret the result.

P [X + Y \leq 6] = 1 + \frac{3}{2 e^{2}} - \frac{5}{e^{6 / 5}} \approx 0.45;

there is a $45 %$

chance that a customer will spend $6$

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 $\frac{s^{2}}{2} (π - \sqrt{3}) .$

An equilateral triangle with additional regions consisting of three arcs of a circle with radius equal to the length of the side of the triangle. These arcs connect two adjacent vertices, and the radius is taken from the opposite vertex.

[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 $A B and A C,$

respectively, and the inner boundaries are formed by the circumcircle of the triangle $A B C .$

A right triangle with points A, B, and C. Point B has the right angle. There are two lunes drawn from A to B and from B to C with outer diameters AB and AC, respectively, and with the inner boundaries formed by the circumcircle of the triangle ABC.

Double Integrals over General Regions

General Regions of Integration

Double Integrals over Nonrectangular Regions

Changing the Order of Integration

Calculating Volumes, Areas, and Average Values

Improper Double Integrals

Key Concepts

Key Equations

Glossary