APEX Iterated Integrals and Area

Section 14.1 Iterated Integrals and Area

In Section 13.3 we found that it was useful to differentiate functions of several variables with respect to one variable, while treating all the other variables as constants or coefficients. We can integrate functions of several variables in a similar way. For instance, if we are told that

f_{x} (x, y) = 2 x y,

we can treat

y

as staying constant and integrate to obtain

f (x, y) :

\begin{aligned} f (x, y) & = \int f_{x} (x, y) d x \\ = \int 2 x y d x \\ = x^{2} y + C . \end{aligned}

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Make a careful note about the constant of integration,

C .

This “constant” is something with a derivative of

0

with respect to

x,

so it could be any expression that contains only constants and functions of

y .

For instance, if

f (x, y) = x^{2} y + \sin (y) + y^{3} + 17,

then

f_{x} (x, y) = 2 x y .

To signify that

C

is actually a function of

y,

we write:

f (x, y) = \int f_{x} (x, y) d x = x^{2} y + C (y) .

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Using this process we can even evaluate definite integrals.

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Subsection 14.1.1 Iterated integrals

Example 14.1.1. Integrating functions of more than one variable.

Evaluate the integral

\int_{1}^{2 y} 2 x y d x .

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Solution.

We find the indefinite integral as before, then apply the Fundamental Theorem of Calculus to evaluate the definite integral:

\begin{aligned} \int_{1}^{2 y} 2 x y d x & = x^{2} y |_{1}^{2 y} \\ = (2 y)^{2} y - (1)^{2} y \\ = 4 y^{3} - y . \end{aligned}

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We can also integrate with respect to

y .

In general,

\int_{h_{1} (y)}^{h_{2} (y)} f_{x} (x, y) d x = f (x, y) |_{h_{1} (y)}^{h_{2} (y)} = f (h_{2} (y), y) - f (h_{1} (y), y),

and

\int_{g_{1} (x)}^{g_{2} (x)} f_{y} (x, y) d y = f (x, y) |_{g_{1} (x)}^{g_{2} (x)} = f (x, g_{2} (x)) - f (x, g_{1} (x)) .

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Note that when integrating with respect to

x,

the bounds are functions of

y

(of the form

x = h_{1} (y)

and

x = h_{2} (y)

) and the final result is also a function of

y .

When integrating with respect to

y,

the bounds are functions of

x

(of the form

y = g_{1} (x)

and

y = g_{2} (x)

) and the final result is a function of

x .

Another example will help us understand this.

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Example 14.1.2. Integrating functions of more than one variable.

Evaluate

\int_{1}^{x} (5 x^{3} y^{- 3} + 6 y^{2}) d y .

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Solution.

We consider

x

as staying constant and integrate with respect to

y :

\begin{aligned} \int_{1}^{x} (5 x^{3} y^{- 3} + 6 y^{2}) d y & = (\frac{5 x^{3} y^{- 2}}{- 2} + \frac{6 y^{3}}{3}) |_{1}^{x} \\ = (- \frac{5}{2} x^{3} x^{- 2} + 2 x^{3}) - (- \frac{5}{2} x^{3} + 2) \\ = \frac{9}{2} x^{3} - \frac{5}{2} x - 2 . \end{aligned}

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Note how the bounds of the integral are from

y = 1

y = x

and that the final answer is a function of

x .

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In the previous example, we integrated a function with respect to

y

and ended up with a function of

x .

We can integrate this as well. This process is known as iterated integration, or multiple integration.

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Example 14.1.3. Evaluating an integral.

Evaluate

\int_{1}^{2} (\int_{1}^{x} (5 x^{3} y^{- 3} + 6 y^{2}) d y) d x .

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Solution.

We follow a standard “order of operations” and perform the operations inside parentheses first (which is the integral evaluated in Example 14.1.2.)

\begin{aligned} \int_{1}^{2} (\int_{1}^{x} (5 x^{3} y^{- 3} + 6 y^{2}) d y) d x & = \int_{1}^{2} ([\frac{5 x^{3} y^{- 2}}{- 2} + \frac{6 y^{3}}{3}] |_{1}^{x}) d x \\ = \int_{1}^{2} (\frac{9}{2} x^{3} - \frac{5}{2} x - 2) d x \\ = (\frac{9}{8} x^{4} - \frac{5}{4} x^{2} - 2 x) |_{1}^{2} \\ = \frac{89}{8} . \end{aligned}

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Note how the bounds of

x

were

x = 1

x = 2

and the final result was a number.

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The previous example showed how we could perform something called an iterated integral; we do not yet know why we would be interested in doing so nor what the result, such as the number

89 / 8,

means. Before we investigate these questions, we offer some definitions.

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Definition 14.1.4. Iterated Integration.

Iterated integration is the process of repeatedly integrating the results of previous integrations. Evaulating one integral is denoted as follows.

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Let

a,

b,

c

and

d

be numbers and let

g_{1} (x),

g_{2} (x),

h_{1} (y)

and

h_{2} (y)

be functions of

x

and

y,

respectively. Then:

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$\int_{c}^{d} \int_{h_{1} (y)}^{h_{2} (y)} f (x, y) d x d y = \int_{c}^{d} (\int_{h_{1} (y)}^{h_{2} (y)} f (x, y) d x) d y .$
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$\int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y d x = \int_{a}^{b} (\int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y) d x .$
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Again make note of the bounds of these iterated integrals.

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With

\int_{c}^{d} \int_{h_{1} (y)}^{h_{2} (y)} f (x, y) d x d y,

x

varies from

h_{1} (y)

h_{2} (y),

whereas

y

varies from

c

d .

That is, the bounds of

x

are curves, the curves

x = h_{1} (y)

and

x = h_{2} (y),

whereas the bounds of

y

are constants,

y = c

and

y = d .

It is useful to remember that when setting up and evaluating such iterated integrals, we integrate “from curve to curve, then from point to point.”

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We now begin to investigate why we are interested in iterated integrals and what they mean.

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Subsection 14.1.2 Area of a plane region

Consider the plane region

R

bounded by

a \leq x \leq b

and

g_{1} (x) \leq y \leq g_{2} (x),

shown in Figure 14.1.5. We learned in Section 7.1 that the area of

R

is given by

\int_{a}^{b} (g_{2} (x) - g_{1} (x)) d x .

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A region in the first quadrant is bounded above and below by graphs, and left and right by vertical lines. — Figure 14.1.5. Calculating the area of a plane region $R$ with an iterated integral
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We can view the expression

(g_{2} (x) - g_{1} (x))

(g_{2} (x) - g_{1} (x)) = \int_{g_{1} (x)}^{g_{2} (x)} 1 d y = \int_{g_{1} (x)}^{g_{2} (x)} d y,

meaning we can express the area of

R

as an iterated integral:

area of R = \int_{a}^{b} (g_{2} (x) - g_{1} (x)) d x = \int_{a}^{b} (\int_{g_{1} (x)}^{g_{2} (x)} d y) d x = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} d y d x .

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In short: a certain iterated integral can be viewed as giving the area of a plane region.

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A region

R

could also be defined by

c \leq y \leq d

and

h_{1} (y) \leq x \leq h_{2} (y),

as shown in Figure 14.1.6. Using a process similar to that above, we have

the area of R = \int_{c}^{d} \int_{h_{1} (y)}^{h_{2} (y)} d x d y .

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A region in the first quadrant is bounded left and right by graphs, and above and below by horizontal lines. — Figure 14.1.6. Calculating the area of a plane region $R$ with an iterated integral
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We state this formally in a theorem.

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Theorem 14.1.7. Area of a plane region.

Let $R$ be a plane region bounded by $a \leq x \leq b$ and $g_{1} (x) \leq y \leq g_{2} (x),$ where $g_{1}$ and $g_{2}$ are continuous functions on $[a, b] .$ The area $A$ of $R$ is

$A = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} d y d x .$

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Let $R$ be a plane region bounded by $c \leq y \leq d$ and $h_{1} (y) \leq x \leq h_{2} (y),$ where $h_{1}$ and $h_{2}$ are continuous functions on $[c, d] .$ The area $A$ of $R$ is

$A = \int_{c}^{d} \int_{h_{1} (y)}^{h_{2} (y)} d x d y .$

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The following examples should help us understand this theorem.

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Example 14.1.8. Area of a rectangle.

Find the area

A

of the rectangle with corners

(- 1, 1)

and

(3, 3),

as shown in Figure 14.1.9.

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A rectangle in the plane, spanning x values from -1 to 3, and y values from 1 to 3. — Figure 14.1.9. Calculating the area of a rectangle with an iterated integral in Example 14.1.8
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Solution.

Multiple integration is obviously overkill in this situation, but we proceed to establish its use.

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The region

R

is bounded by

x = - 1,

x = 3,

y = 1

and

y = 3 .

Choosing to integrate with respect to

y

first, we have

A = \int_{- 1}^{3} \int_{1}^{3} 1 d y d x = \int_{- 1}^{3} (y |_{1}^{3}) d x = \int_{- 1}^{3} 2 d x = 2 x |_{- 1}^{3} = 8 .

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We could also integrate with respect to

x

first, giving:

A = \int_{1}^{3} \int_{- 1}^{3} 1 d x d y = \int_{1}^{3} (x |_{- 1}^{3}) d y = \int_{1}^{3} 4 d y = 4 y |_{1}^{3} = 8 .

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Clearly there are simpler ways to find this area, but it is interesting to note that this method works.

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Example 14.1.10. Area of a triangle.

Find the area

A

of the triangle with vertices at

(1, 1),

(3, 1)

and

(5, 5),

as shown in Figure 14.1.11.

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An obtuse triangle in the plane. The base is horizontal, and the other sides are lines with positive slope. — Figure 14.1.11. Calculating the area of a triangle with iterated integrals in Example 14.1.10
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Solution.

The triangle is bounded by the lines as shown in the figure. Choosing to integrate with respect to

x

first gives that

x

is bounded by

x = y

x = \frac{y + 5}{2},

while

y

is bounded by

y = 1

y = 5 .

(Recall that since

x

-values increase from left to right, the leftmost curve,

x = y,

is the lower bound and the rightmost curve,

x = (y + 5) / 2,

is the upper bound.) The area is

\begin{aligned} A & = \int_{1}^{5} \int_{y}^{\frac{y + 5}{2}} d x d y \\ = \int_{1}^{5} (x |_{y}^{\frac{y + 5}{2}}) d y \\ = \int_{1}^{5} (- \frac{1}{2} y + \frac{5}{2}) d y \\ = (- \frac{1}{4} y^{2} + \frac{5}{2} y) |_{1}^{5} \\ = 4 . \end{aligned}

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We can also find the area by integrating with respect to

y

first. In this situation, though, we have two functions that act as the lower bound for the region

R,

y = 1

and

y = 2 x - 5 .

This requires us to use two iterated integrals. Note how the

x

-bounds are different for each integral:

\begin{aligned} A & = \int_{1}^{3} \int_{1}^{x} 1 d y d x & + & \int_{3}^{5} \int_{2 x - 5}^{x} 1 d y d x \\ = \int_{1}^{3} (y) |_{1}^{x} d x & + & \int_{3}^{5} (y) |_{2 x - 5}^{x} d x \\ = \int_{1}^{3} (x - 1) d x & + & \int_{3}^{5} (- x + 5) d x \\ = 2 & + & 2 \\ = 4 . \end{aligned}

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As expected, we get the same answer both ways.

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Example 14.1.12. Area of a plane region.

Find the area of the region enclosed by

y = 2 x

and

y = x^{2},

as shown in Figure 14.1.13.

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A region in the first quadrant of the plane, bounded by a line and a parabola. — Figure 14.1.13. Calculating the area of a plane region with iterated integrals in Example 14.1.12
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Solution.

Once again we’ll find the area of the region using both orders of integration.

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Using

d y d x :

\int_{0}^{2} \int_{x^{2}}^{2 x} 1 d y d x = \int_{0}^{2} (2 x - x^{2}) d x = (x^{2} - \frac{1}{3} x^{3}) |_{0}^{2} = \frac{4}{3} .

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Using

d x d y :

\int_{0}^{4} \int_{y / 2}^{\sqrt{y}} 1 d x d y = \int_{0}^{4} (\sqrt{y} - y / 2) d y = (\frac{2}{3} y^{3 / 2} - \frac{1}{4} y^{2}) |_{0}^{4} = \frac{4}{3} .

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Subsection 14.1.3 Changing Order of Integration

In each of the previous examples, we have been given a region

R

and found the bounds needed to find the area of

R

using both orders of integration. We integrated using both orders of integration to demonstrate their equality.

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We now approach the skill of describing a region using both orders of integration from a different perspective. Instead of starting with a region and creating iterated integrals, we will start with an iterated integral and rewrite it in the other integration order. To do so, we’ll need to understand the region over which we are integrating.

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The simplest of all cases is when both integrals are bound by constants. The region described by these bounds is a rectangle (see Example 14.1.8), and so:

\int_{a}^{b} \int_{c}^{d} 1 d y d x = \int_{c}^{d} \int_{a}^{b} 1 d x d y .

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When the inner integral’s bounds are not constants, it is generally very useful to sketch the bounds to determine what the region we are integrating over looks like. From the sketch we can then rewrite the integral with the other order of integration.

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Examples will help us develop this skill.

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Example 14.1.14. Changing the order of integration.

Rewrite the iterated integral

\int_{0}^{6} \int_{0}^{x / 3} 1 d y d x

with the order of integration

d x d y .

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Solution.

We need to use the bounds of integration to determine the region we are integrating over.

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The bounds tell us that

y

is bounded by

0

and

x / 3;

x

is bounded by 0 and 6. We plot these four curves:

y = 0,

y = x / 3,

x = 0

and

x = 6

to find the region described by the bounds. Figure 14.1.15 shows these curves, indicating that

R

is a triangle.

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A triangular region in the plane, bounded by lines y=0, x=6, and y=x/3. — Figure 14.1.15. Sketching the region $R$ described by the iterated integral in Example 14.1.14
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To change the order of integration, we need to consider the curves that bound the

x

-values. We see that the lower bound is

x = 3 y

and the upper bound is

x = 6 .

The bounds on

y

are

0

2 .

Thus we can rewrite the integral as

\int_{0}^{2} \int_{3 y}^{6} 1 d x d y .

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Example 14.1.16. Changing the order of integration.

Change the order of integration of

\int_{0}^{4} \int_{y^{2} / 4}^{(y + 4) / 2} 1 d x d y .

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Solution.

We sketch the region described by the bounds to help us change the integration order.

x

is bounded below and above (i.e., to the left and right) by

x = y^{2} / 4

and

x = (y + 4) / 2

respectively, and

y

is bounded between 0 and 4. Graphing the previous curves, we find the region

R

to be that shown in Figure 14.1.17.

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A sketch of the region of integration for this example. — Figure 14.1.17. Drawing the region determined by the bounds of integration in Example 14.1.16
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To change the order of integration, we need to establish curves that bound

y .

The figure makes it clear that there are two lower bounds for

y :

y = 0

0 \leq x \leq 2,

and

y = 2 x - 4

2 \leq x \leq 4 .

Thus we need two double integrals. The upper bound for each is

y = 2 \sqrt{x} .

Thus we have

\int_{0}^{4} \int_{y^{2} / 4}^{(y + 4) / 2} 1 d x d y = \int_{0}^{2} \int_{0}^{2 \sqrt{x}} 1 d y d x + \int_{2}^{4} \int_{2 x - 4}^{2 \sqrt{x}} 1 d y d x .

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This section has introduced a new concept, the iterated integral. We developed one application for iterated integration: area between curves. However, this is not new, for we already know how to find areas bounded by curves.

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In the next section we apply iterated integration to solve problems we currently do not know how to handle. The “real” goal of this section was not to learn a new way of computing area. Rather, our goal was to learn how to define a region in the plane using the bounds of an iterated integral. That skill is very important in the following sections.

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Exercises 14.1.4 Exercises

Terms and Concepts

1.

When integrating

f_{x} (x, y)

with respect to

x,

the constant of integration

C

is really which:

C (x)

C (y) ?

What does this mean?

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Problems

(b)

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

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Exercise Group.

In the following exercises, a graph of a planar region

R

is given. Give the iterated integrals, with both orders of integration

d y d x

and

d x d y,

that give the area of

R .

Evaluate one of the iterated integrals to find the area.

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11.

$The region R is a rectangle, with x from 1 to 4, and y from -2 to 1.$

A sketch of the region

R

for this exercise.

R

is a rectangle, with bounds

1 \leq x \leq 4

and

- 2 \leq y \leq 11 .

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12.

$The region R is a right triangle with vertices (1,1), (4,1), and (4,3)$

A sketch of the region

R

for this exercise. The boundary of

R

is a right-angle triangle, with vertices at

(1, 1),

(4, 1),

and

(4, 3) .

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13.

$A triangular region R. The vertices are (2,5), (2,1), and (4,3).$

The region

R

for this exercise is a triangle. The left side of the triangle is vertical, from

(2, 1)

(2, 5) .

The other vertex is at

(4, 3)

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14.

$The region R is bounded to the left by a parabola opening along the x axis, and the line x=12.$

The region

R

for this exercise is bounded by a parabola and a line.

The parabola is $x = y^{2} / 3,$ which has vertex $(0, 0)$ and opens to the right.
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The line is the vertical line $x = 12 .$
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15.

$A region in the first quadrant bounded by two curves that intersect at the points (0,0) and (1,1)$

The region

R

is bounded by two intersecting curves.

The curve that lies below (and to the right) of the region is $y = x^{4} .$ This curve is increasing and concave up.
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The curve that lies above (and to the left) of the region is $y = \sqrt{x} .$ This curve is increasing and concave down.
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The two curves intersect at the origin, and again at the point $(1, 1) .$
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16.

$A region R bounded above by a line, and below by a cubic curve.$

The region

R

lies in the first quadrant. It is bounded by two curves that intersect twice.

Above (and to the left of) the region is the line $y = 4 x .$
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Below (and to the right of) the region is the curve $y = x^{3} .$
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The two curves meet at the origin. The curve $y = x^{3}$ begins below the line $y = 4 x,$ but bends upward to meet the line again at the point $(2, 8) .$
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Exercise Group.

In the following exercises, iterated integrals are given that compute the area of a region

R

in the

x y

-plane. Sketch the region

R,

and give the iterated integral(s) that give the area of

R

\int_{- 1}^{1} \int_{(x - 1) / 2}^{(1 - x) / 2} d y d x

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