CC Using Definite Integrals to Find Volume by Rotation and Arc Length

Section 6.2 Using Definite Integrals to Find Volume by Rotation and Arc Length

Motivating Questions

How can we use a definite integral to find the volume of a three-dimensional solid?
How can we use a definite integral to find the volume of a three-dimensional solid of revolution that results from revolving a two-dimensional region about a particular axis?
In what circumstances do we integrate with respect to $y$ instead of integrating with respect to $x ?$
What adjustments do we need to make if we revolve about a line other than the $x$ - or $y$ -axis?
How can a definite integral be used to measure the length of a curve?

Just as we can use definite integrals to add the areas of rectangular slices to find the exact area that lies between two curves, we can also use integrals to find the volume of regions whose cross-sections have a particular shape.

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In particular, we can determine the volume of solids whose cross-sections are all thin cylinders (or washers) by adding up the volumes of these individual slices, in particular when these shapes arise by revolving one or more curves around an axis. We first consider a familiar shape in Example 6.16: a circular cone.

Example 6.16.

Consider a circular cone of radius 3 and height 5, which we view horizontally as pictured in Figure 6.17. Our goal in this example is to use a definite integral to determine the volume of the cone.

Projecting the cone onto the $x y$ -plane yields a triangle with vertices $(0, 3),$ $(0, - 3),$ and $(5, 0),$ as shown in Figure 6.17. Find a formula for the linear function $y = f (x)$ corresponding to the side of the triangle joining vertices $(0, 3)$ and $(5, 0) .$
For the representative slice of thickness $Δ x$ that is located horizontally at a location $x$ (somewhere between $x = 0$ and $x = 5$ ), what is the radius of the representative slice? Note that the radius depends on the value of $x .$
What is the volume of the representative slice you found in (b)?
What definite integral will sum the volumes of the thin slices across the full horizontal span of the cone? What is the exact value of this definite integral?
Compare the result of your work in (d) to the volume of the cone that comes from using the formula $V_{cone} = \frac{1}{3} π r^{2} h .$

Hint.

The line passes through the points $(0, 3)$ and $(5, 0) .$
Use your answer to the previous question.
A cylinder of radius $r$ and height $h$ will have volume $π r^{2} h,$ so use the answer you found in part (b.) for the radius.
Your answer to part (c.) is the volume of one cylinder, so this will become the integrand (replacing $Δ x$ with $d x$ ).
The two methods for finding the volume should agree.

Answer.

$f (x) = - \frac{3}{5} x + 3 .$
The radius at some $x$ value will be $f (x) = - \frac{3}{5} x + 3 .$
$V_{c y l i n d e r} = π (- \frac{3}{5} x + 3)^{2} Δ x$
$15 π$
The two methods for finding the volume agree: the volume formula for the cone gives $V_{c o n e} = \frac{1}{3} π (3)^{2} (5) = 15 π,$ the same as the integral method.

Solution.

We are trying to find the formula for a linear function $y = f (x)$ which passes through the points $(0, 3)$ and $(5, 0) .$ The $y$ -intercept is therefore 3, and the slope is $\frac{0 - 3}{5 - 0} = - \frac{3}{5},$ so the linear equation is $f (x) = - \frac{3}{5} x + 3 .$
The radius at some $x$ value will be $f (x) = - \frac{3}{5} x + 3 .$
A cylinder of radius $r$ and height $h$ will have volume $π r^{2} h .$ For this cylinder, the radius is $f (x) = - \frac{3}{5} x + 3$ and the height is $Δ x,$ so the volume of this cylinder is

$V_{c y l i n d e r} = π (- \frac{3}{5} x + 3)^{2} Δ x$
To find the volume of the cone, we will sum up a bunch of cylinders, and take the limit as the width of the cylinders $Δ x$ goes to 0. When we do this, the sum turns into an integral and $Δ x$ turns into $d x .$ Thus the volume of the cone is

$V_{c o n e} = \int_{0}^{5} π (- \frac{3}{5} x + 3)^{2} d x = π \int_{0}^{5} \frac{9}{25} x^{2} - \frac{18}{5} x + 9 d x = π {(\frac{3}{25} x^{3} - \frac{9}{5} x^{2} + 9 x) |}_{0}^{5} = π (\frac{3}{25} (125) - \frac{9}{5} (25) + 9 (5)) - 0 = 15 π$
The two methods for finding the volume agree: the volume formula for the cone gives $V_{c o n e} = \frac{1}{3} π (3)^{2} (5) = 15 π,$ the same as the integral method.

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We will also investigate one more application of definite integrals, to find the length of curves by again splitting the curve into small slices and adding up the smaller lengths.

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Subsection 6.2.1 The Volume of a Solid of Revolution

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A solid of revolution is a three dimensional solid that can be generated by revolving one or more curves around a fixed axis. For example, the circular cone in Figure 6.17 is the solid of revolution generated by revolving the portion of the line

y = 3 - \frac{3}{5} x

from

x = 0

x = 5

about the

x

-axis. Notice that if we slice a solid of revolution perpendicular to the axis of revolution, the resulting cross-section is a circle.

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We first consider solids whose slices are thin cylinders. Recall that the volume of a cylinder is given by

V = π r^{2} h .

Example 6.18.

Find the volume of the solid of revolution generated when the region

R

bounded by

y = 4 - x^{2}

and the

x

-axis is revolved about the

x

-axis.

Solution.

First, we observe that

y = 4 - x^{2}

intersects the

x

-axis at the points

(- 2, 0)

and

(2, 0) .

When we revolve the region

R

about the

x

-axis, we get the three-dimensional solid pictured in Figure 6.19.

Figure 6.19. The solid of revolution in Example 6.18.

We slice the solid into vertical slices of thickness

Δ x

between

x = - 2

and

x = 2 .

A representative slice is a cylinder of height

Δ x

and radius

4 - x^{2} .

Hence, the volume of the slice is

V_{slice} = π (4 - x^{2})^{2} Δ x .

Using a definite integral to sum the volumes of the representative slices, it follows that

V = \int_{- 2}^{2} π (4 - x^{2})^{2} d x .

It is straightforward to evaluate the integral and find that the volume is

V = \frac{512}{15} π .

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For a solid such as the one in Example 6.18, where each slice is a cylindrical disk, we first find the volume of a typical slice (noting particularly how this volume depends on

x

), and then integrate over the range of

x

-values that bound the solid. Often, we will be content with simply finding the integral that represents the volume; if we desire a numeric value for the integral, we typically use a calculator or computer algebra system to find that value.

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This method for finding the volume of a solid of revolution is often called the disk method.

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The Disk Method.

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y = r (x)

is a nonnegative continuous function on

[a, b],

then the volume of the solid of revolution generated by revolving the curve about the

x

-axis over this interval is given by

V = \int_{a}^{b} π r (x)^{2} d x .

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A different type of solid can emerge when two curves are involved, as we see in the following example.

Example 6.20.

Find the volume of the solid of revolution generated when the finite region

R

that lies between

y = 4 - x^{2}

and

y = x + 2

is revolved about the

x

-axis.

Solution.

First, we must determine where the curves

y = 4 - x^{2}

and

y = x + 2

intersect. Substituting the expression for

y

from the second equation into the first equation, we find that

x + 2 = 4 - x^{2} .

Rearranging, it follows that

x^{2} + x - 2 = 0,

and the solutions to this equation are

x = - 2

and

x = 1 .

The curves therefore cross at

(- 2, 0)

and

(1, 1) .

When we revolve the region

R

about the

x

-axis, we get the three-dimensional solid pictured at left in Figure 6.21.

Figure 6.21. At left, the solid of revolution in Example 6.20. At right, a typical slice with inner radius $r (x)$ and outer radius $R (x) .$

Immediately we see a major difference between the solid in this example and the one in Example 6.18: here, the three-dimensional solid of revolution isn’t “solid” because it has open space in its center along the axis of revolution. If we slice the solid perpendicular to the axis of revolution, we observe that the resulting slice is not a solid disk, but rather a washer, as pictured at right in Figure 6.21. At a given location

x

between

x = - 2

and

x = 1,

the small radius

r (x)

of the inner circle is determined by the curve

y = x + 2,

r (x) = x + 2 .

Similarly, the big radius

R (x)

comes from the function

y = 4 - x^{2},

and thus

R (x) = 4 - x^{2} .

To find the volume of a representative slice, we compute the volume of the outer disk and subtract the volume of the inner disk. Since

π R (x)^{2} Δ x - π r (x)^{2} Δ x = π [R (x)^{2} - r (x)^{2}] Δ x,

it follows that the volume of a typical slice is

V_{slice} = π [(4 - x^{2})^{2} - (x + 2)^{2}] Δ x .

Using a definite integral to sum the volumes of the respective slices across the integral, we find that

V = \int_{- 2}^{1} π [(4 - x^{2})^{2} - (x + 2)^{2}] d x .

Evaluating the integral, we find that the volume of the solid of revolution is

V = \frac{108}{5} π .

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This method for finding the volume of a solid of revolution generated by two curves is often called the washer method.

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The Washer Method.

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y = R (x)

and

y = r (x)

are nonnegative continuous functions on

[a, b]

that satisfy

R (x) \geq r (x)

for all

x

[a, b],

then the volume of the solid of revolution generated by revolving the region between them about the

x

-axis over this interval is given by

V = \int_{a}^{b} π [R (x)^{2} - r (x)^{2}] d x .

Example 6.22.

In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated axis. In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid. It is not necessary to evaluate the integrals you find.

The region $S$ bounded by the $x$ -axis, the curve $y = \sqrt{x},$ and the line $x = 4;$ revolve $S$ about the $x$ -axis.
The region $S$ bounded by the $y$ -axis, the curve $y = \sqrt{x},$ and the line $y = 2;$ revolve $S$ about the $x$ -axis.
The finite region $S$ bounded by the curves $y = \sqrt{x}$ and $y = x^{3};$ revolve $S$ about the $x$ -axis.
The finite region $S$ bounded by the curves $y = 2 x^{2} + 1$ and $y = x^{2} + 4;$ revolve $S$ about the $x$ -axis.
The region $S$ bounded by the $y$ -axis, the curve $y = \sqrt{x},$ and the line $y = 2;$ revolve $S$ about the $y$ -axis. How is this problem different from the one posed in part (b)?

Hint.

Use slices perpendicular to the $x$ -axis.
Note that slices perpendicular to the $x$ -axis will be washers.
Find the points where the two curves intersect and draw a labeled graph.
Think about how the solid looks like a donut; what is the outer radius?
Try using slices perpendicular to the $y$ -axis.

Answer.

$V = \int_{0}^{4} π (\sqrt{x})^{2} d x = \int_{0}^{4} π x d x = 8 π .$
$V = \int_{0}^{4} π (4 - (\sqrt{x})^{2}) d x = \int_{0}^{4} π (4 - x) d x = 8 π .$
$V = \int_{0}^{1} π (x - x^{6}) d x = \frac{5}{14} π .$
$V = \int_{- \sqrt{3}}^{\sqrt{3}} π ((x^{2} + 4)^{2} - (2 x^{2} + 1)^{2}) d x = \frac{136 \sqrt{3}}{5} π .$
$V = \int_{0}^{2} π y^{4} d y = \frac{32}{5} π .$

Solution.

A typical slice (perpendicular to the $x$ -axis) has radius $r (x) = \sqrt{x},$ and thus the volume of such a slice is

$V_{slice} = π (\sqrt{x})^{2} Δ x .$

It follows by the disk method that the volume of the solid of revolution is

$V = \int_{0}^{4} π (\sqrt{x})^{2} d x = \int_{0}^{4} π x d x = 8 π .$
A typical slice (perpendicular to the $x$ -axis) is a washer with outer radius $R (x) = 2$ and inner radius $r (x) = \sqrt{x} .$ The volume of such a slice is

$V_{slice} = π (R (x)^{2} - r (x)^{2}) Δ x = π (2^{2} - (\sqrt{x})^{2}) Δ x .$

It follows by the washer method that the volume of the solid of revolution is

$V = \int_{0}^{4} π (4 - (\sqrt{x})^{2}) d x = \int_{0}^{4} π (4 - x) d x = 8 π .$
The curves $y = \sqrt{x}$ and $y = x^{3}$ intersect at $(0, 0)$ and $(1, 1);$ revolving the region $S$ about the $x$ -axis and taking slices perpendicular to the $x$ -axis, we see that a typical slice is a washer with outer radius $R (x) = \sqrt{x}$ and inner radius $r (x) = x^{3} .$ Hence the volume of a typical slice is

$V_{slice} = π (R (x)^{2} - r (x)^{2}) Δ x = π ((\sqrt{x})^{2} - (x^{3})^{2}) Δ x = π (x - x^{6}) Δ x .$

By the washer method, the volume of the solid is

$V = \int_{0}^{1} π (x - x^{6}) d x = \frac{5}{14} π .$
The curves $y = 2 x^{2} + 1$ and $y = x^{2} + 4$ intersect at the points $(\pm \sqrt{3}, 7);$ after we revolve the region $S$ about the $x$ -axis, cross-sections perpendicular to the axis of rotation are washers with outer radius $R (x) = x^{2} + 4$ and inner radius $r (x) = 2 x^{2} + 1 .$ By the washer method, we see that

$V = \int_{- \sqrt{3}}^{\sqrt{3}} π ((x^{2} + 4)^{2} - (2 x^{2} + 1)^{2}) d x = \frac{136 \sqrt{3}}{5} π .$
The curve $y = \sqrt{x}$ and the line $y = 2$ intersect at $(4, 2);$ when we revolve the region $S$ about the $y$ -axis, slices perpendicular to the $y$ -axis are disks. Here we see that each disk’s radius is a function of $y$ and the disks have thickness $Δ y,$ so we will need to integrate with respect to $y$ and consider the radius function $r (y) = y^{2}$ (which comes from solving $y = \sqrt{x}$ for $x$ in terms of $y .$ Note how this is different from the problem in part (b) because of the axis we rotate about and how this forces us to integrate in terms of $y .$ A typical slice has volume

$V_{slice} = π (y^{2})^{2} Δ y,$

and thus by the disk method the total volume is

$V = \int_{0}^{2} π y^{4} d y = \frac{32}{5} π .$

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Subsection 6.2.2 Revolving About the $y$ -axis

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When we revolve a given region about the

y

-axis, the representative slices now have thickness

Δ y,

which means that we must integrate with respect to

y .

Example 6.23.

Find the volume of the solid of revolution generated when the region

R

that lies between

y = \sqrt{x}

and

y = x^{4}

is revolved about the

y

-axis.

Solution.

These two curves intersect when

x = 1,

hence at the point

(1, 1) .

When we revolve the region

R

about the

y

-axis, we get the three-dimensional solid pictured at left in Figure 6.24.

Figure 6.24. At left, the solid of revolution in Example 6.23. At right, a typical slice with inner radius $r (y)$ and outer radius $R (y) .$

Note that the slices are cylindrical washers only if taken perpendicular to the

y

-axis. We slice the solid horizontally, starting at

y = 0

and proceeding up to

y = 1 .

The thickness of a representative slice is

Δ y,

so we must express the integrand in terms of

y .

The inner radius is determined by the curve

y = \sqrt{x},

so we solve for

x

and get

x = y^{2} = r (y) .

In the same way, we solve the curve

y = x^{4}

(which governs the outer radius) for

x

in terms of

y,

and hence

x = \sqrt[4]{y} .

Therefore, the volume of a typical slice is

V_{slice} = π [R (y)^{2} - r (y)^{2}] = π [(\sqrt[4]{y})^{2} - (y^{2})^{2}] Δ y .

We use a definite integral to sum the volumes of all the slices from

y = 0

y = 1 .

The total volume is

V = \int_{y = 0}^{y = 1} π [(\sqrt[4]{y})^{2} - (y^{2})^{2}] d y .

It is straightforward to evaluate the integral and find that

V = \frac{7}{15} π .

Example 6.25.

The region $S$ bounded by the $y$ -axis, the curve $y = \sqrt{x},$ and the line $y = 2;$ revolve $S$ about the $y$ -axis.
The region $S$ bounded by the $x$ -axis, the curve $y = \sqrt{x},$ and the line $x = 4;$ revolve $S$ about the $y$ -axis.
The finite region $S$ in the first quadrant bounded by the curves $y = 2 x$ and $y = x^{3};$ revolve $S$ about the $x$ -axis.
The finite region $S$ in the first quadrant bounded by the curves $y = 2 x$ and $y = x^{3};$ revolve $S$ about the $y$ -axis.
The finite region $S$ bounded by the curves $x = (y - 1)^{2}$ and $y = x - 1;$ revolve $S$ about the $y$ -axis

Hint.

Consider slices perpendicular to the $y$ -axis.
Since we revolve around the $y$ -axis, think about slices of thickness $Δ y .$
Note that $y = x^{3}$ lies below $y = 2 x$ on the interval $[0, \sqrt{2}] .$
The curve $y = x^{3}$ can be equivalently expressed as $x = \sqrt[3]{y} .$
Find the intersection points by writing the second curve as $x = y + 1$ and solving $(y - 1)^{2} = y + 1 .$

Answer.

$V = \int_{0}^{2} π y^{4} Δ d y .$
$V = \int_{0}^{2} π (16 - y^{4}) d y .$
$V = i n t_{0}^{\sqrt{2}} π (4 x^{2} - x^{6}) d x .$
$V = \int_{0}^{2 \sqrt{2}} π (y^{2 / 3} - y^{2} / 4) d y .$
$V = \int_{0}^{3} π ((y + 1)^{2} - (y - 1)^{4}) d y .$

Solution.

Observe that horizontal slices with thickness $Δ y$ are all cylindrical disks, each with radius given by $r (y) = y^{2}$ since they are given by the curve $y = \sqrt{x},$ which can be equivalently expressed as $x = y^{2} .$ The volume of each slice is thus

$V_{slice} = π r (y)^{2} Δ y = π y^{4} Δ y .$

The volume of the solid is thus given by the definite integral

$V = \int_{0}^{2} π y^{4} Δ d y .$
Slicing perpendicular to the axis of rotation, we get slices of thickness $Δ y$ that are shaped like washers. The outer radius, $R (y),$ is given by $R (y) = 4;$ the innner radius is given by the function $y = \sqrt{x},$ but expressed with $x$ as a function of $y,$ so $r (y) = y^{2} .$ It follows that a representative washer slice has volume

$V_{slice} = π (R (y)^{2} - r (y)^{2}) Δ y = π (16 - y^{4}) Δ y .$

Further, since $y = \sqrt{x}$ and $x = 4$ intersect at the point $(4, 2),$ we integrate from $y = 0$ to $y = 2$ and thus find that the total volume is

$V = \int_{0}^{2} π (16 - y^{4}) d y .$
The curves $y = x^{3}$ and $y = 2 x$ intersect where $x^{3} = 2 x,$ so where $x^{3} - 2 x = x (x^{2} - 2) = 0 .$ The two nonnegative values of $x$ that make this equation true are $x = 0$ and $x = \sqrt{2},$ and thus the points $(0, 0)$ and $(\sqrt{2}, 2 \sqrt{2})$ form the boundaries of the region $S$ that is being rotated about the $x$ -axis. Since we rotate about the horizontal axis, we slice perpendicular to that axis with slices of thickness $Δ x .$ A representative slice is in the shape of a washer and has volume

$V_{slice} = π (R (x)^{2} - r (x)^{2}) Δ x = π ((2 x)^{2} - (x^{3})^{2}) Δ x,$

and thus the total volume of the solid is

$V = i n t_{0}^{\sqrt{2}} π (4 x^{2} - x^{6}) d x .$
To use the same region $S$ as in part (c), but now revolved about the $y$ -axis, we shift to horizontal slices of thickness $Δ y,$ and thus need to express the curves as functions of $y :$ the rightmost curve is $x = R (y) = \sqrt[3]{y},$ while the leftmost curve is $x = r (y) = \frac{1}{2} y .$ A representative slice is a washer whose volume is

$V_{slice} = π (R (y)^{2} - r (y)^{2}) Δ y = π ((\sqrt[3]{y})^{2} - (\frac{1}{2} y)^{2}) Δ y .$

It follows that the total volume of the solid is

$V = \int_{0}^{2 \sqrt{2}} π (y^{2 / 3} - y^{2} / 4) d y .$
For this pair of curves, we see they intersect where $x = (y - 1)^{2}$ and $x = (y + 1)$ are both true, and hence where $(y - 1)^{2} = y + 1 .$ Expanding, moving all terms to the left, and combining like terms, we see that $y^{2} - 2 y + 1 = y + 1,$ so $y^{2} - 3 y = 0 .$ It follows that $y (y - 3) = 0$ and therefore $y = 3$ or $y = 0 .$ Plotting the region and revolving about the $y$ -axis, we see that horizontal slices of thickness $Δ y$ are shaped like washers. The outer radius is given by $R (y) = y + 1,$ while the inner radius is $r (y) = (y - 1)^{2} .$ A representative slice thus has volume

$V_{slice} = π (R (y)^{2} - r (y)^{2}) Δ y = π ((y + 1)^{2} - ((y - 1)^{2})^{2}) Δ y .$

The volume of the solid of revolution is therefore

$V = \int_{0}^{3} π ((y + 1)^{2} - (y - 1)^{4}) d y .$

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Subsection 6.2.3 Revolving About Horizontal and Vertical Lines other than the Coordinate Axes

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It is possible to revolve a region around any horizontal or vertical line. Doing so adjusts the radii of the cylinders or washers involved by a constant value. A careful, well-labeled plot of the solid of revolution will usually reveal how the different axis of revolution affects the definite integral.

Example 6.26.

Find the volume of the solid of revolution generated when the finite region

S

that lies between

y = x^{2}

and

y = x

is revolved about the line

y = - 1 .

Solution.

Graphing the region between the two curves in the first quadrant between their points of intersection (

(0, 0)

and

(1, 1)

) and then revolving the region about the line

y = - 1,

we see the solid shown in Figure 6.27. Each slice of the solid perpendicular to the axis of revolution is a washer, and the radii of each washer are governed by the curves

y = x^{2}

and

y = x .

But we also see that there is one added change: the axis of revolution adds a fixed length to each radius. The inner radius of a typical slice,

r (x),

is given by

r (x) = x^{2} + 1,

while the outer radius is

R (x) = x + 1 .

Figure 6.27. The solid of revolution described in Example 6.26.

Therefore, the volume of a typical slice is

V_{slice} = π [R (x)^{2} - r (x)^{2}] Δ x = π [(x + 1)^{2} - (x^{2} + 1)^{2}] Δ x .

Finally, we integrate to find the total volume, and

V = \int_{0}^{1} π [(x + 1)^{2} - (x^{2} + 1)^{2}] d x = \frac{7}{15} π .

Example 6.28.

S

in the first quadrant bounded by the curves

y = 2 x

and

y = x^{3} .

Revolve $S$ about the line $y = - 2 .$
Revolve $S$ about the line $y = 4 .$
Revolve $S$ about the line $x = - 1 .$
Revolve $S$ about the line $x = 5 .$

Hint.

Note that the two curves intersect in the first quadrant at $(0, 0)$ and $(\sqrt{2}, 2 \sqrt{2}) .$
The outer radius is $R (x) = 4 - x^{3} .$
Solve the equations governing the two curves for $x$ in terms of $y .$
The inner radius is $r (y) = 5 - y^{1 / 3} .$

Answer.

$V = \int_{0}^{\sqrt{2}} π ((2 x + 2)^{2} - (x^{3} + 2)^{2}) d x = \frac{4}{21} (21 + 8 \sqrt{2}) π \approx 19.336 .$
$V = \int_{0}^{\sqrt{2}} π ((4 - x^{3})^{2} - (4 - 2 x)^{2}) d x = (8 - \frac{32 \sqrt{2}}{21}) π \approx 18.3626 .$
$V = \int_{0}^{2 \sqrt{2}} π ((y^{1 / 3} + 1)^{2} - (\frac{1}{2} y + 1)^{2}) d y = \frac{2}{15} (15 + 8 \sqrt{2}) π \approx 11.022 .$
$V = \int_{0}^{2 \sqrt{2}} π ((5 - \frac{1}{2} y)^{2} - (5 - y^{1 / 3})^{2}) d y = \frac{2}{15} (75 - 8 \sqrt{2}) π \approx 26.677 .$

Solution.

The two curves intersect in the first quadrant at $(0, 0)$ and $(\sqrt{2}, 2 \sqrt{2}) .$ Revolving the region $S$ about the line $y = - 2,$ we see that a typical slice is a washer with thickness $Δ x,$ outer radius $R (x) = 2 x + 2,$ and inner radius $r (x) = x^{3} + 2 .$ It follows by the washer method that

$V = \int_{0}^{\sqrt{2}} π ((2 x + 2)^{2} - (x^{3} + 2)^{2}) d x = \frac{4}{21} (21 + 8 \sqrt{2}) π \approx 19.336 .$
In this situation, revolving about $y = 4,$ we see that the outer radius of a typical washer is $R (x) = 4 - x^{3},$ while the inner radius is $r (x) = 4 - 2 x,$ and thus the volume is

$V = \int_{0}^{\sqrt{2}} π ((4 - x^{3})^{2} - (4 - 2 x)^{2}) d x = (8 - \frac{32 \sqrt{2}}{21}) π \approx 18.3626 .$
Because we are revolving about $x = - 1,$ when we slice perpendicular to this we get washers of thickness $Δ y,$ and thus we need to write the two curves as functions of $y$ in order to determine the radii of the slices as functions of $y,$ we note that the line can be expressed as $x = \frac{1}{2} y$ and the cubic curve can be rewritten in the form $x = y^{1 / 3} .$ The outer radius of a typical slice is thus $R (y) = y^{1 / 3} + 1$ and the inner radius is $r (y) = \frac{1}{2} y + 1 .$ It follows that the volume of the solid is

$V = \int_{0}^{2 \sqrt{2}} π ((y^{1 / 3} + 1)^{2} - (\frac{1}{2} y + 1)^{2}) d y = \frac{2}{15} (15 + 8 \sqrt{2}) π \approx 11.022 .$
Here, the outer radius is $R (y) = 5 - \frac{1}{2} y$ and the inner radius is $r (y) = 5 - y^{1 / 3},$ so the total volume of the solid is

$V = \int_{0}^{2 \sqrt{2}} π ((5 - \frac{1}{2} y)^{2} - (5 - y^{1 / 3})^{2}) d y = \frac{2}{15} (75 - 8 \sqrt{2}) π \approx 26.677 .$

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Subsection 6.2.4 Finding the Length of a Curve

🔗

We can also use the definite integral to find the length of a portion of a curve. We use the same fundamental principle: we slice the curve up into small pieces whose lengths we can easily approximate. Specifically, we subdivide the curve into small approximating line segments, as shown at left in Figure 6.29.

🔗
Figure 6.29. At left, a continuous function $y = f (x)$ whose length we seek on the interval $a = x_{0}$ to $b = x_{3} .$ At right, a close up view of a portion of the curve.

🔗

We estimate the length

L_{slice}

of each portion of the curve on a small interval of length

Δ x .

We use the right triangle with legs parallel to the coordinate axes and hypotenuse connecting the endpoints of the slice, as seen at right in Figure 6.29. The length,

h,

of the hypotenuse approximates the length,

L_{slice},

of the curve between the two selected points. Thus,

L_{slice} \approx h = \sqrt{(Δ x)^{2} + (Δ y)^{2}} .

🔗

Next we use algebra to rearrange the expression for the length of the hypotenuse into a form that we can integrate. By removing a factor of

(Δ x)^{2},

we find

\begin{aligned} L_{slice} & \approx \sqrt{(Δ x)^{2} + (Δ y)^{2}} \\ = \sqrt{(Δ x)^{2} (1 + \frac{(Δ y)^{2}}{(Δ x)^{2}})} \\ = \sqrt{1 + \frac{(Δ y)^{2}}{(Δ x)^{2}}} \cdot Δ x . \end{aligned}

🔗

Then, as

n \to \infty

and

Δ x \to 0,

we have that

\frac{Δ y}{Δ x} \to \frac{d y}{d x} = f^{'} (x) .

Thus, we can say that

L_{slice} \approx \sqrt{1 + f^{'} (x)^{2}} Δ x .

🔗

Taking a Riemann sum of all of these slices and letting

n \to \infty,

we arrive at the following fact.

🔗

Arc Length.

🔗

Given a differentiable function

f

on an interval

[a, b],

the total arc length,

L,

along the curve

y = f (x)

from

x = a

x = b

is given by

L = \int_{a}^{b} \sqrt{1 + f^{'} (x)^{2}} d x .

Example 6.30.

Each of the following questions somehow involves the arc length along a curve.

Use the definition and appropriate computational technology to determine the arc length along $y = x^{2}$ from $x = - 1$ to $x = 1 .$
Find the arc length of $y = \sqrt{4 - x^{2}}$ on the interval $- 2 \leq x \leq 2 .$ Find this value in two different ways: (a) by using a definite integral, and (b) by using a familiar property of the curve.
Determine the arc length of $y = x e^{3 x}$ on the interval $[0, 1] .$
Will the integrals that arise calculating arc length typically be ones that we can evaluate exactly using the First FTC, or ones that we need to approximate? Why?
A moving particle is traveling along the curve given by $y = f (x) = 0.1 x^{2} + 1,$ and does so at a constant rate of 7 cm/sec, where both $x$ and $y$ are measured in cm (that is, the curve $y = f (x)$ is the path along which the object actually travels; the curve is not a “position function”). Find the position of the particle when $t = 4$ sec, assuming that when $t = 0,$ the particle’s location is $(0, f (0)) .$

Hint.

Recall that $L = \int_{a}^{b} \sqrt{1 + f^{'} (x)^{2}} d x .$
Remember that $x^{2} + y^{2} = 4$ generates a circle centered at $(0, 0)$ of radius 4.
Apply the arc length formula. Use technology to evaluate the integral.
Most expressions involving square roots are difficult to antidifferentiate.
Here you can determine the arc length and then experiment to find the upper limit of integration, which will help you determine position.

Answer.

$L \approx 2.95789 .$
$L = \int_{- 2}^{2} \sqrt{\frac{4}{4 - x^{2}}} d x = 2 π .$
$L = \int_{0}^{1} \sqrt{1 + e^{6 x} (9 x^{2} + 6 x + 1)} d x \approx 20.1773 .$
We will usually have to estimate the value of $\int_{a}^{b} \sqrt{1 + f^{'} (x)^{2}} d x$ using computational technology.
Approximately $(9.011, f (9.011)) = (9.011, 9.1198) .$

Solution.

Using the formula for arc length, we know

$L = \int_{- 1}^{1} \sqrt{1 + (2 x)^{2}} d x .$

Evaluating the integral, we find $L \approx 2.95789 .$
The curve is a semi-circle of radius 2, so its length must be $L = \frac{1}{2} \cdot 2 π r = 2 π .$ This is confirmed by the arc length formula, since $f (x) = \sqrt{4 - x^{2}},$ so $f^{'} (x) = \frac{1}{2} (4 - x^{2})^{- 1 / 2} (- 2 x)$

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

It follows
¹
This integral is actually improper because the integrand is undefined at the endpoints, $x = \pm 2 .$ We learned how to evaluate such integrals in Section 5.10.
that

$L = \int_{- 2}^{2} \sqrt{\frac{4}{4 - x^{2}}} d x = 2 π .$
For $y = x e^{3 x},$ we observe that $y^{'} = 3 x e^{3 x} + e^{3 x} = e^{3 x} (3 x + 1),$ and that

$(y^{'})^{2} = e^{6 x} (9 x^{2} + 6 x + 1) .$

Thus, the arc length of the curve on $[0, 1]$ is

$L = \int_{0}^{1} \sqrt{1 + e^{6 x} (9 x^{2} + 6 x + 1)} d x \approx 20.1773 .$
Because $\sqrt{1 + f^{'} (x)^{2}}$ will rarely simplify nicely and rarely have an elementary antiderivative, we will usually have to estimate the value of $\int_{a}^{b} \sqrt{1 + f^{'} (x)^{2}} d x$ using computational technology.
After 4 seconds traveling at 7 cm/sec, the particle has moved 28 cm. So, we have to find the value of $b$ for which the arc length $L$ along $[0, b]$ is 28. Thus, we know that

$28 = L = \int_{0}^{b} \sqrt{1 + (0.2 x)^{2}} d x .$

Rewriting the integral, we see

$28 = \int_{0}^{b} \sqrt{1 + 0.4 x^{2}} d x .$

Experimenting with different values of $b$ in a computational engine like Wolfram|Alpha, we find that for $b \approx 9.011,$ $\int_{0}^{9.011} \sqrt{1 + 0.4 x^{2}} d x \approx 27.9992,$ and thus the position of the particle when $t = 4$ is approximately $(9.011, f (9.011)) = (9.011, 9.1198) .$

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The same technique can be applied to parametric curves. For now consider a curve

(x (t), y (t))

travelling in just two dimensions (but see Example 6.33 for curves in 3D). As before, the length of a small segment will be

\sqrt{(Δ x)^{2} + (Δ y)^{2}},

but here we will factor out

Δ t

to get:

L_{s l i c e} \approx \sqrt{{(\frac{Δ x}{Δ t})}^{2} + {(\frac{Δ y}{Δ t})}^{2}} \cdot Δ t

🔗

Now, taking the limit as

Δ t \to 0,

\frac{Δ x}{Δ t}

and

\frac{Δ y}{Δ t}

turn into

\frac{d x}{d t}

and

\frac{d y}{d t},

so the length of the segment is

L = \int_{t = a}^{t = b} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t .

🔗

We summarize our result here:

🔗

Arc Length for a Parametric Curve.

🔗

Given two differentiable functions

x (t), y (t)

on an interval

a \leq t \leq b,

the total arc length

L

along the parametric curve

(x (t), y (t))

from

t = a

t = b

is given by

L = \int_{t = a}^{t = b} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t .

Example 6.31.

Recall that the unit circle is traced out by the parametric equations $x (t) = \cos (t)$ and $y (t) = \sin (t)$ on the interval $0 \leq t \leq 2 π .$ Set up and evaluate the arc length integral to find the circumference of the unit circle.
An ellipse (a stretched circle) can be traced out by the parametric equations $x (t) = 2 \cos (t)$ and $y (t) = \sin (t) .$ Set up the arc length integral to find the perimeter of this ellipse. You do not have to evaluate the integral.

Hint.

Use the formula for the arc length for a parametric curve to set up the integral. Recall also the Pythagorean Identity for trig functions: $\sin^{2} (t) + \cos^{2} (t) = 1 .$
Use the formula for the arc length for a parametric curve to set up the integral.

Answer.

$L = \int_{0}^{2 π} 1 d t = 2 π$
$L = \int_{0}^{2 π} \sqrt{(- 2 \sin (t)^{2}) + (\cos (t))^{2}} d t$

Solution.

Since $\frac{d x}{d t} = - \sin (t)$ and $\frac{d y}{d t} = \cos (t),$ then

$L = \int_{0}^{2 π} \sqrt{(- \sin (t)^{2}) + (\cos (t))^{2}} d t = \int_{0}^{2 π} 1 d t = 2 π$
Since $\frac{d x}{d t} = - 2 \sin (t)$ and $\frac{d y}{d t} = \cos (t),$ then

$L = \int_{0}^{2 π} \sqrt{(- 2 \sin (t)^{2}) + (\cos (t))^{2}} d t = \int_{0}^{2 π} \sqrt{4 \sin^{2} (t) + \cos^{2} (t)} d t$

We do not need to evaluate this integral, and in fact it is an example of an “elliptical integral of the second kind”, which are a famous class of integrals without elementary antiderivatives.

Example 6.32.

Alex drives to work, and their route is along a path given by

x (t) = t^{3},

y (t) = t^{4} + t

for

0 \leq t \leq 1 .

How long is their route to work? (You may wish to use a calculator or computer to numerically evaluate the integral.)

Hint.

Set up the integral using formula for arc length for a parametric curve. Then use a computer to evaluate the integral.

Answer.

L \approx 2.266

Solution.

Since

\frac{d x}{d t} = 3 t^{2}

and

\frac{d y}{d t} = 4 t^{3} + 1,

then

L = \int_{0}^{1} \sqrt{(3 t^{2})^{2} + (4 t^{3} + 1)^{2}} d t \approx 2.266

Example 6.33.

Suppose that one has a parametric curve travelling through 3 dimensions given by

f (t) = (x (t), y (t), z (t)) .

How would you change the “Arc Length for a Parametric Curve” formula to include the third dimension?

Hint.

Treat the

z

-coordinate the same way the

x

- and

y

-coordinates are treated.

Answer.

L = \int_{t = a}^{t = b} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} + {(\frac{d z}{d t})}^{2}} d t

Solution.

The length of a 3D line segment is given by

\sqrt{(Δ x)^{2} + (Δ y)^{2} + (Δ z)^{2}},

so factoring out a factor of

Δ t

and taking the limit as

Δ t \to 0,

we get

L = \int_{t = a}^{t = b} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} + {(\frac{d z}{d t})}^{2}} d t

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Subsection 6.2.5 Summary

🔗

A definite integral may be used to represent the volume of a solid. The idea is to slice the solid into thin shapes whose volume is easy to calculate. The volume of each slice is found by taking the area of the cross-section and multiplying it by the width. Summing up the volumes of the slices and taking the limit as the width $\to 0$ results in an integral representing the volume.
A three-dimensional solid of revolution results from revolving a two-dimensional region about a particular axis or line called the axis of revolution. We can use a definite integral to find the volume of such a solid by taking slices perpendicular to the axis of revolution. The slices will be circular disks or washers.
If we revolve about a horizontal line and slice perpendicular to that line, then our slices are vertical and of thickness $Δ x .$ This leads us to integrate with respect to $x .$
If we revolve about a vertical line and slice perpendicular to that line, then our slices are horizontal and of thickness $Δ y .$ This leads us to integrate with respect to $y .$
If we revolve about a line other than the $x$ - or $y$ -axis, we need to carefully account for the shift that occurs in the radius of a typical slice. Normally, this shift involves taking a sum or difference of the function along with the constant connected to the equation for the horizontal or vertical line; a well-labeled diagram is usually the best way to decide the new expression for the radius.
The arc length, $L,$ along the curve $y = f (x)$ from $x = a$ to $x = b$ is given by

$L = \int_{a}^{b} \sqrt{1 + f^{'} (x)^{2}} d x .$

The definite integral may also be used to find arc length of parametric curves.

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Exercises 6.2.6 Exercises

🔗

1. Solid of revolution from one function about the $x$ -axis.

🔗

The region bounded by

y = e^{2 x}, y = 0, x = - 2, x = 0

is rotated around the

x

-axis. Find the volume.

🔗

volume =

🔗

2. Solid of revolution from one function about the $y$ -axis.

🔗

Find the volume of the solid obtained by rotating the region in the first quadrant bounded by

y = x^{6},

y = 1,

and the

y

-axis around the

y

-axis.

🔗

Volume =

🔗

3. Solid of revolution from two functions about the $x$ -axis.

🔗

Find the volume of the solid obtained by rotating the region in the first quadrant bounded by

y = x^{6},

y = 1,

and the

y

-axis around the

x

-axis.

🔗

Volume =

🔗

4. Solid of revolution from two functions about a horizontal line.

🔗

Find the volume of the solid obtained by rotating the region in the first quadrant bounded by

y = x^{6},

y = 1,

and the

y

-axis about the line

y = - 5 .

🔗

Volume =

🔗

5. Solid of revolution from two functions about a different horizontal line.

🔗

Find the volume of the solid obtained by rotating the region bounded by the curves

y = x^{4}, y = 1

🔗

about the line

y = 5

🔗

Answer:

🔗

6. Solid of revolution from two functions about a vertical line.

🔗

Find the volume of the solid obtained by rotating the region bounded by the given curves about the line

x = - 8

🔗

y = x^{2}, x = y^{2}

🔗

Answer:

🔗

7. Arc length and area of a region, and volume of its solid of revolution.

🔗

Consider the curve

f (x) = 3 \cos (\frac{x^{3}}{4})

and the portion of its graph that lies in the first quadrant between the

y

-axis and the first positive value of

x

for which

f (x) = 0 .

Let

R

denote the region bounded by this portion of

f,

the

x

-axis, and the

y

-axis.

Set up a definite integral whose value is the exact arc length of $f$ that lies along the upper boundary of $R .$ Use technology appropriately to evaluate the integral you find.
Set up a definite integral whose value is the exact area of $R .$ Use technology appropriately to evaluate the integral you find.
Suppose that the region $R$ is revolved around the $x$ -axis. Set up a definite integral whose value is the exact volume of the solid of revolution that is generated. Use technology appropriately to evaluate the integral you find.
Suppose instead that $R$ is revolved around the $y$ -axis. If possible, set up an integral expression whose value is the exact volume of the solid of revolution and evaluate the integral using appropriate technology. If not possible, explain why.

🔗

8. Solid of revolution from a two functions about multiple horizontal and vertical lines.

🔗

Consider the curves given by

y = \sin (x)

and

y = \cos (x) .

For each of the following problems, you should include a sketch of the region/solid being considered, as well as a labeled representative slice.

Sketch the region $R$ bounded by the $y$ -axis and the curves $y = \sin (x)$ and $y = \cos (x)$ up to the first positive value of $x$ at which they intersect. What is the exact intersection point of the curves?
Set up a definite integral whose value is the exact area of $R .$
Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving $R$ about the $x$ -axis.
Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving $R$ about the $y$ -axis.
Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving $R$ about the line $y = 2 .$
Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving $R$ about the line $x = - 1 .$

🔗

9. Area and perimeter of a region and volume of a solid of revolution around multiple lines.

🔗

Consider the finite region

R

that is bounded by the curves

y = 1 + \frac{1}{2} (x - 2)^{2},

y = \frac{1}{2} x^{2},

and

x = 0 .

Determine a definite integral whose value is the area of the region enclosed by the two curves.
Find an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region $R$ about the line $y = - 1 .$
Determine an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region $R$ about the $y$ -axis.
Find an expression involving one or more definite integrals whose value is the perimeter of the region $R .$

🔗

10. Arc length of a curve.

🔗

Find the arc length of the graph of the function

f (x) = 2 \sqrt{x^{3}}

from

x = 4

x = 6 .

🔗

arc length =

🔗

11. Length of a parametric curve.

🔗

The parametric function given by

x (t) = \sin (2 t),

y (t) = \sin (t)

traces out a figure-8 pattern for

0 \leq t \leq 2 π .

Find the length of this curve. (You may wish to use a calculator or computer to numerically evaluate the integral.)

Hint.

Set up the integral using the “Arc Length for a Parametric Curve” formula, then use a computer to evaluate the integral.

Prev Top Next

Section 6.2 Using Definite Integrals to Find Volume by Rotation and Arc Length

Motivating Questions

Example 6.16.

Subsection 6.2.1 The Volume of a Solid of Revolution

Example 6.18.

The Disk Method.

Example 6.20.

The Washer Method.

Example 6.22.

Subsection 6.2.2 Revolving About the y-axis

Example 6.23.

Example 6.25.

Subsection 6.2.3 Revolving About Horizontal and Vertical Lines other than the Coordinate Axes

Example 6.26.

Example 6.28.

Subsection 6.2.4 Finding the Length of a Curve

Arc Length.

Example 6.30.

Arc Length for a Parametric Curve.

Example 6.31.

Example 6.32.

Example 6.33.

Subsection 6.2.5 Summary

Exercises 6.2.6 Exercises

1. Solid of revolution from one function about the x-axis.

2. Solid of revolution from one function about the y-axis.

3. Solid of revolution from two functions about the x-axis.

4. Solid of revolution from two functions about a horizontal line.

5. Solid of revolution from two functions about a different horizontal line.

6. Solid of revolution from two functions about a vertical line.

7. Arc length and area of a region, and volume of its solid of revolution.

8. Solid of revolution from a two functions about multiple horizontal and vertical lines.

9. Area and perimeter of a region and volume of a solid of revolution around multiple lines.

10. Arc length of a curve.

11. Length of a parametric curve.

Subsection 6.2.2 Revolving About the $y$ -axis

1. Solid of revolution from one function about the $x$ -axis.

2. Solid of revolution from one function about the $y$ -axis.

3. Solid of revolution from two functions about the $x$ -axis.