Gilbert Strang; Edwin “Jed” Herman

Learning Objectives

6.2.1 Determine the volume of a solid by integrating a cross-section (the slicing method).
6.2.2 Find the volume of a solid of revolution using the disk method.
6.2.3 Find the volume of a solid of revolution with a cavity using the washer method.

In the preceding section, we used definite integrals to find the area between two curves. In this section, we use definite integrals to find volumes of three-dimensional solids. We consider three approaches—slicing, disks, and washers—for finding these volumes, depending on the characteristics of the solid.

Volume and the Slicing Method

Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: $V = l w h .$ The formulas for the volume of a sphere $(V = \frac{4}{3} π r^{3}),$ a cone $(V = \frac{1}{3} π r^{2} h),$ and a pyramid $(V = \frac{1}{3} A h)$ have also been introduced. Although some of these formulas were derived using geometry alone, all these formulas can be obtained by using integration.

We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such as a soup can or a metal rod, in mathematics the word cylinder has a more general meaning. To discuss cylinders in this more general context, we first need to define some vocabulary.

We define the cross-section of a solid to be the intersection of a plane with the solid. A cylinder is defined as any solid that can be generated by translating a plane region along a line perpendicular to the region, called the axis of the cylinder. Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in Figure 6.11 is an example of a cylinder with a noncircular base. To calculate the volume of a cylinder, then, we simply multiply the area of the cross-section by the height of the cylinder: $V = A \cdot h .$ In the case of a right circular cylinder (soup can), this becomes $V = π r^{2} h .$

This graphic has two figures. The first figure is half of a cylinder, on the flat portion. The cylinder has a line through the center labeled “x”. Vertically cutting through the cylinder, perpendicular to the line is a plane. The second figure is a two dimensional cross section of the cylinder intersecting with the plane. It is a semi-circle.

Figure 6.11 Each cross-section of a particular cylinder is identical to the others.

If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula for its volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solid into pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all be parallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, the solid S shown in Figure 6.12, extending along the $x -axis .$

This figure is a graph of a 3-dimensional solid. It has one edge along the x-axis. The x-axis is part of the 2-dimensional coordinate system with the y-axis labeled. The edge of the solid along the x-axis starts at a point labeled “a” and stops at a point labeled “b”.

Figure 6.12 A solid with a varying cross-section.

We want to divide $S$ into slices perpendicular to the $x -axis .$ As we see later in the chapter, there may be times when we want to slice the solid in some other direction—say, with slices perpendicular to the y-axis. The decision of which way to slice the solid is very important. If we make the wrong choice, the computations can get quite messy. Later in the chapter, we examine some of these situations in detail and look at how to decide which way to slice the solid. For the purposes of this section, however, we use slices perpendicular to the $x -axis .$

Because the cross-sectional area is not constant, we let $A (x)$ represent the area of the cross-section at point $x .$ Now let $P = {x_{0}, x_{1} …, X_{n}}$ be a regular partition of $[a, b],$ and for $i = 1, 2,… n,$ let $S_{i}$ represent the slice of $S$ stretching from $x_{i - 1} to x_{i} .$ The following figure shows the sliced solid with $n = 3 .$

Figure 6.13 The solid

S

has been divided into three slices perpendicular to the

x -axis .

Finally, for $i = 1, 2,… n,$ let $x_{i}^{*}$ be an arbitrary point in $[x_{i - 1}, x_{i}] .$ Then the volume of slice $S_{i}$ can be estimated by $V (S_{i}) \approx A (x_{i}^{*}) Δ x .$ Adding these approximations together, we see the volume of the entire solid $S$ can be approximated by

V (S) \approx \sum_{i = 1}^{n} A (x_{i}^{*}) Δ x .

By now, we can recognize this as a Riemann sum, and our next step is to take the limit as $n \to \infty .$ Then we have

V (S) = lim_{n \to \infty} \sum_{i = 1}^{n} A (x_{i}^{*}) Δ x = \int_{a}^{b} A (x) d x .

The technique we have just described is called the slicing method. To apply it, we use the following strategy.

Problem-Solving Strategy

Finding Volumes by the Slicing Method

Examine the solid and determine the shape of a cross-section of the solid. It is often helpful to draw a picture if one is not provided.
Determine a formula for the area of the cross-section.
Integrate the area formula over the appropriate interval to get the volume.

Recall that in this section, we assume the slices are perpendicular to the $x -axis .$ Therefore, the area formula is in terms of x and the limits of integration lie on the $x -axis .$ However, the problem-solving strategy shown here is valid regardless of how we choose to slice the solid.

Example 6.6

Deriving the Formula for the Volume of a Pyramid

We know from geometry that the formula for the volume of a pyramid is $V = \frac{1}{3} A h .$ If the pyramid has a square base, this becomes $V = \frac{1}{3} a^{2} h,$ where $a$ denotes the length of one side of the base. We are going to use the slicing method to derive this formula.

Solution

We want to apply the slicing method to a pyramid with a square base. To set up the integral, consider the pyramid shown in Figure 6.14, oriented along the $x -axis .$

This figure has two graphs. The first graph, labeled “a”, is a pyramid on its side. The x-axis goes through the middle of the pyramid. The point of the top of the pyramid is at the origin of the x y coordinate system. The base of the pyramid is shaded and labeled “a”. Inside of the pyramid is a shaded rectangle labeled “s”. The distance from the y-axis to the base of the pyramid is labeled “h”. the distance the rectangle inside of the pyramid to the y-axis is labeled “x”. The second figure is a cross section of the pyramid with the x and y axes labeled. The cross section is a triangle with one side labeled “a”, perpendicular to the x-axis. The distance a is from the y-axis is h. There is another perpendicular line to the x-axis inside of the triangle. It is labeled “s”. It is x units from the y-axis.

Figure 6.14 (a) A pyramid with a square base is oriented along the x-axis. (b) A two-dimensional view of the pyramid is seen from the side.

We first want to determine the shape of a cross-section of the pyramid. We know the base is a square, so the cross-sections are squares as well (step 1). Now we want to determine a formula for the area of one of these cross-sectional squares. Looking at Figure 6.14(b), and using a proportion, since these are similar triangles, we have

\frac{s}{a} = \frac{x}{h} or s = \frac{a x}{h} .

Therefore, the area of one of the cross-sectional squares is

A (x) = s^{2} = {(\frac{a x}{h})}^{2} (step 2) .

Then we find the volume of the pyramid by integrating from $0 to h$ (step $3) :$

\begin{matrix} V & = \int_{0}^{h} A (x) d x \\ = \int_{0}^{h} {(\frac{a x}{h})}^{2} d x = \frac{a^{2}}{h^{2}} \int_{0}^{h} x^{2} d x \\ = {[\frac{a^{2}}{h^{2}} (\frac{1}{3} x^{3})] |}_{0}^{h} = \frac{1}{3} a^{2} h . \end{matrix}

This is the formula we were looking for.

Checkpoint 6.6

Use the slicing method to derive the formula $V = \frac{1}{3} π r^{2} h$ for the volume of a circular cone.

Solids of Revolution

If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution, as shown in the following figure.

This figure has three graphs. The first graph, labeled “a” is a region in the x y plane. The region is created by a curve above the x-axis and the x-axis. The second graph, labeled “b” is the same region as in “a”, but it shows the region beginning to rotate around the x-axis. The third graph, labeled “c” is the solid formed by rotating the region from “a” completely around the x-axis, forming a solid.

Figure 6.15 (a) This is the region that is revolved around the x-axis. (b) As the region begins to revolve around the axis, it sweeps out a solid of revolution. (c) This is the solid that results when the revolution is complete.

Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid of revolution.

Media

Use an online integral calculator to learn more.

Example 6.7

Using the Slicing Method to find the Volume of a Solid of Revolution

Use the slicing method to find the volume of the solid of revolution bounded by the graphs of $f (x) = x^{2} - 4 x + 5, x = 1, and x = 4,$ and rotated about the $x -axis .$

Solution

Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval $[1, 4]$ as shown in the following figure.

This figure is a graph of the parabola f(x)=x^2-4x+5. The parabola is the top of a shaded region above the x-axis. The region is bounded to the left by a line at x=1 and to the right by a line at x=4.

Figure 6.16 A region used to produce a solid of revolution.

Next, revolve the region around the x-axis, as shown in the following figure.

This figure has two graphs of the parabola f(x)=x^2-4x+5. The parabola is the top of a shaded region above the x-axis. The region is bounded to the left by a line at x=1 and to the right by a line at x=4. The first graph has a shaded solid below the parabola. This solid has been formed by rotating the parabola around the x-axis. The second graph is the same as the first, with the solid being rotated to show the solid.

Figure 6.17 Two views, (a) and (b), of the solid of revolution produced by revolving the region in Figure 6.16 about the

x -axis .

Since the solid was formed by revolving the region around the $x -axis,$ the cross-sections are circles (step 1). The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by $f (x) .$ Use the formula for the area of the circle:

A (x) = π r^{2} = π {[f (x)]}^{2} = π {(x^{2} - 4 x + 5)}^{2} (step 2) .

The volume, then, is (step 3)

\begin{matrix} V & = \int_{a}^{b} A (x) d x \\ = \int_{1}^{4} π {(x^{2} - 4 x + 5)}^{2} d x = π \int_{1}^{4} (x^{4} - 8 x^{3} + 26 x^{2} - 40 x + 25) d x \\ = {π (\frac{x^{5}}{5} - 2 x^{4} + \frac{26 x^{3}}{3} - 20 x^{2} + 25 x) |}_{1}^{4} = \frac{78}{5} π . \end{matrix}

The volume is $78 π / 5 .$

Checkpoint 6.7

Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function $f (x) = 1 / x$ and the $x -axis$ over the interval $[1, 2]$ around the $x -axis .$ See the following figure.

This figure has two graphs. The first graph is the curve f(x)=1/x. It is a decreasing curve, above the x-axis in the first quadrant. The graph has a shaded region under the curve between x=1 and x=2. The second graph is the curve f(x)=1/x in the first quadrant. Also, underneath this graph, there is a solid between x=1 and x=2 that has been formed by rotating the region from the first graph around the x-axis.

The Disk Method

When we use the slicing method with solids of revolution, it is often called the disk method because, for solids of revolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution generated by revolving the region between the graph of the function $f (x) = {(x - 1)}^{2} + 1$ and the $x -axis$ over the interval $[−1, 3]$ around the $x -axis .$ The graph of the function and a representative disk are shown in Figure 6.18(a) and (b). The region of revolution and the resulting solid are shown in Figure 6.18(c) and (d).

This figure has four graphs. The first graph, labeled “a” is a parabola f(x)=(x-1)^2+1. The curve is above the x-axis and intersects the y-axis at y=2. Under the curve in the first quadrant is a vertical rectangle starting at the x-axis and stopping at the curve. The second graph, labeled “b” is the same parabola as in the first graph. The rectangle under the parabola from the first graph has been rotated around the x-axis forming a solid disk. The third graph labeled “c” is the same parabola as the first graph. There is a shaded region bounded above by the parabola, to the left by the line x=-1 and to the right by the line x=3, and below by the x-axis. The fourth graph labeled “d” is the same parabola as the first graph. The region from the third graph has been revolved around the x-axis to form a solid.

Figure 6.18 (a) A thin rectangle for approximating the area under a curve. (b) A representative disk formed by revolving the rectangle about the

x -axis .

(c) The region under the curve is revolved about the

x -axis,

resulting in (d) the solid of revolution.

We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that

V = \int_{a}^{b} A (x) d x .

The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle. This gives the following rule.

Rule: The Disk Method

Let $f (x)$ be continuous and nonnegative. Define $R$ as the region bounded above by the graph of $f (x),$ below by the $x -axis,$ on the left by the line $x = a,$ and on the right by the line $x = b .$ Then, the volume of the solid of revolution formed by revolving $R$ around the $x -axis$ is given by

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

(6.3)

The volume of the solid we have been studying (Figure 6.18) is given by

\begin{matrix} V & = \int_{a}^{b} π {[f (x)]}^{2} d x \\ = \int_{−1}^{3} π {[{(x - 1)}^{2} + 1]}^{2} d x = π \int_{−1}^{3} [{(x - 1)}^{4} + 2 {(x - 1)}^{2} + 1] d x \\ = π {[\frac{1}{5} {(x - 1)}^{5} + \frac{2}{3} {(x - 1)}^{3} + x] |}_{−1}^{3} = π [(\frac{32}{5} + \frac{16}{3} + 3) - (- \frac{32}{5} - \frac{16}{3} - 1)] = \frac{412 π}{15} {units}^{3} . \end{matrix}

Let’s look at some examples.

Example 6.8

Using the Disk Method to Find the Volume of a Solid of Revolution 1

Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of $f (x) = \sqrt{x}$ and the $x -axis$ over the interval $[1, 4]$ around the $x -axis .$

Solution

The graphs of the function and the solid of revolution are shown in the following figure.

This figure has two graphs. The first graph labeled “a” is the curve f(x) = squareroot(x). It is an increasing curve above the x-axis. The curve is in the first quadrant. Under the curve is a region bounded by x=1 and x=4. The bottom of the region is the x-axis. The second graph labeled “b” is the same curve as the first graph. The solid region from the first graph has been rotated around the x-axis to form a solid region.

Figure 6.19 (a) The function

f (x) = \sqrt{x}

over the interval

[1, 4] .

(b) The solid of revolution obtained by revolving the region under the graph of

f (x)

about the

x -axis .

We have

\begin{matrix} V & = \int_{a}^{b} π {[f (x)]}^{2} d x \\ = \int_{1}^{4} π {[\sqrt{x}]}^{2} d x = π \int_{1}^{4} x d x \\ = {\frac{π}{2} x^{2} |}_{1}^{4} = \frac{15 π}{2} . \end{matrix}

The volume is $(15 π) / 2$ units³.

Checkpoint 6.8

Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of $f (x) = \sqrt{4 - x}$ and the $x -axis$ over the interval $[0, 4]$ around the $x -axis .$

So far, our examples have all concerned regions revolved around the $x -axis,$ but we can generate a solid of revolution by revolving a plane region around any horizontal or vertical line. In the next example, we look at a solid of revolution that has been generated by revolving a region around the $y -axis .$ The mechanics of the disk method are nearly the same as when the $x -axis$ is the axis of revolution, but we express the function in terms of $y$ and we integrate with respect to y as well. This is summarized in the following rule.

Rule: The Disk Method for Solids of Revolution around the y-axis

Let $g (y)$ be continuous and nonnegative. Define $Q$ as the region bounded on the right by the graph of $g (y),$ on the left by the $y -axis,$ below by the line $y = c,$ and above by the line $y = d .$ Then, the volume of the solid of revolution formed by revolving $Q$ around the $y -axis$ is given by

V = \int_{c}^{d} π {[g (y)]}^{2} d y .

(6.4)

The next example shows how this rule works in practice.

Example 6.9

Using the Disk Method to Find the Volume of a Solid of Revolution 2

Let $R$ be the region bounded by the graph of $g (y) = \sqrt{4 - y}$ and the $y -axis$ over the $y -axis$ interval $[0, 4] .$ Use the disk method to find the volume of the solid of revolution generated by rotating $R$ around the $y -axis .$

Solution

Figure 6.20 shows the function and a representative disk that can be used to estimate the volume. Notice that since we are revolving the function around the $y -axis,$ the disks are horizontal, rather than vertical.

This figure has two graphs. The first graph labeled “a” is the curve g(y) = squareroot(4-y). It is a decreasing curve starting on the y-axis at y=4. Between the curve and the y-axis is a horizontal rectangle. The rectangle starts at the y-axis and stops at the curve. The second graph labeled “b” is the same curve as the first graph. The rectangle from the first graph has been rotated around the y-axis to form a horizontal disk.

Figure 6.20 (a) Shown is a thin rectangle between the curve of the function

g (y) = \sqrt{4 - y}

and the

y -axis .

(b) The rectangle forms a representative disk after revolution around the

y -axis .

The region to be revolved and the full solid of revolution are depicted in the following figure.

This figure has two graphs. The first graph labeled “a” is the curve g(y) = squareroot(4-y). It is a decreasing curve starting on the y-axis at y=4. The region formed by the x-axis, the y-axis, and the curve is shaded. This region is in the first quadrant. The second graph labeled “b” is the same curve as the first graph. The region from the first graph has been rotated around the y-axis to form a solid.

Figure 6.21 (a) The region to the left of the function

g (y) = \sqrt{4 - y}

over the

y -axis

interval

[0, 4] .

(b) The solid of revolution formed by revolving the region about the

y -axis .

To find the volume, we integrate with respect to $y .$ We obtain

\begin{matrix} V & = \int_{c}^{d} π {[g (y)]}^{2} d y \\ = \int_{0}^{4} π {[\sqrt{4 - y}]}^{2} d y = π \int_{0}^{4} (4 - y) d y \\ = {π [4 y - \frac{y^{2}}{2}] |}_{0}^{4} = 8 π . \end{matrix}

The volume is $8 π$ units³.

Checkpoint 6.9

Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of $g (y) = y$ and the $y -axis$ over the interval $[1, 4]$ around the $y -axis .$

The Washer Method

Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. Sometimes, this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen is when an axis of revolution other than the $x -axis$ or $y -axis$ is selected.

When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers (disks with holes in the center). For example, consider the region bounded above by the graph of the function $f (x) = \sqrt{x}$ and below by the graph of the function $g (x) = 1$ over the interval $[1, 4] .$ When this region is revolved around the $x -axis,$ the result is a solid with a cavity in the middle, and the slices are washers. The graph of the function and a representative washer are shown in Figure 6.22(a) and (b). The region of revolution and the resulting solid are shown in Figure 6.22(c) and (d).

This figure has four graphs. The first graph is labeled “a” and has the two functions f(x)=squareroot(x) and g(x)=1 graphed in the first quadrant. f(x) is an increasing curve starting at the origin and g(x) is a horizontal line at y=1. The curves intersect at the ordered pair (1,1). In between the curves is a shaded rectangle with the bottom on g(x) and the top at f(x). The second graph labeled “b” is the same two curves as the first graph. The shaded rectangle between the curves from the first graph has been rotated around the x-axis to form an open disk or washer. The third graph labeled “a” has the same two curves as the first graph. There is a shaded region between the two curves between where they intersect and a line at x=4. The fourth graph is the same two curves as the first with the region from the third graph rotated around the x-axis forming a solid region with a hollow center. The hollow center is represented on the graph with broken horizontal lines at y=1 and y=-1.

Figure 6.22 (a) A thin rectangle in the region between two curves. (b) A representative washer formed by revolving the rectangle about the

x -axis .

(c) The region between the curves over the given interval. (d) The resulting solid of revolution.

The cross-sectional area, then, is the area of the outer circle less the area of the inner circle. In this case,

A (x) = π {(\sqrt{x})}^{2} - π {(1)}^{2} = π (x - 1) .

Then the volume of the solid is

\begin{matrix} V & = \int_{a}^{b} A (x) d x \\ = \int_{1}^{4} π (x - 1) d x = {π [\frac{x^{2}}{2} - x] |}_{1}^{4} = \frac{9}{2} π {units}^{3} . \end{matrix}

Generalizing this process gives the washer method.

Rule: The Washer Method

Suppose $f (x)$ and $g (x)$ are continuous, nonnegative functions such that $f (x) \geq g (x)$ over $[a, b] .$ Let $R$ denote the region bounded above by the graph of $f (x),$ below by the graph of $g (x),$ on the left by the line $x = a,$ and on the right by the line $x = b .$ Then, the volume of the solid of revolution formed by revolving $R$ around the $x -axis$ is given by

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

(6.5)

Example 6.10

Using the Washer Method

Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of $f (x) = x$ and below by the graph of $g (x) = 1 / x$ over the interval $[1, 4]$ around the $x -axis .$

Solution

The graphs of the functions and the solid of revolution are shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and has the two curves f(x)=x and g(x)=1/x. They are graphed only in the first quadrant. f(x) is a diagonal line starting at the origin and g(x) is a decreasing curve with the y-axis as a vertical asymptote and the x-axis as a horizontal asymptote. The graphs intersect at (1,1). There is a shaded region between the graphs, bounded to the right by a line at x=4. The second graph is the same two curves. There is a solid formed by rotating the shaded region from the first graph around the x-axis.

Figure 6.23 (a) The region between the graphs of the functions

f (x) = x

and

g (x) = 1 / x

over the interval

[1, 4] .

(b) Revolving the region about the

x -axis

generates a solid of revolution with a cavity in the middle.

We have

\begin{matrix} V & = \int_{a}^{b} π [{(f (x))}^{2} - {(g (x))}^{2}] d x \\ = π \int_{1}^{4} [x^{2} - {(\frac{1}{x})}^{2}] d x = {π [\frac{x^{3}}{3} + \frac{1}{x}] |}_{1}^{4} = \frac{81 π}{4} {units}^{3} . \end{matrix}

Checkpoint 6.10

Find the volume of a solid of revolution formed by revolving the region bounded by the graphs of $f (x) = \sqrt{x}$ and $g (x) = 1 / x$ over the interval $[1, 3]$ around the $x -axis .$

As with the disk method, we can also apply the washer method to solids of revolution that result from revolving a region around the y-axis. In this case, the following rule applies.

Rule: The Washer Method for Solids of Revolution around the y-axis

Suppose $u (y)$ and $v (y)$ are continuous, nonnegative functions such that $v (y) \leq u (y)$ for $y \in [c, d] .$ Let $Q$ denote the region bounded on the right by the graph of $u (y),$ on the left by the graph of $v (y),$ below by the line $y = c,$ and above by the line $y = d .$ Then, the volume of the solid of revolution formed by revolving $Q$ around the $y -axis$ is given by

V = \int_{c}^{d} π [{(u (y))}^{2} - {(v (y))}^{2}] d y .

Rather than looking at an example of the washer method with the $y -axis$ as the axis of revolution, we now consider an example in which the axis of revolution is a line other than one of the two coordinate axes. The same general method applies, but you may have to visualize just how to describe the cross-sectional area of the volume.

Example 6.11

The Washer Method with a Different Axis of Revolution

Find the volume of a solid of revolution formed by revolving the region bounded above by $f (x) = 4 - x$ and below by the $x -axis$ over the interval $[0, 4]$ around the line $y = −2 .$

Solution

The graph of the region and the solid of revolution are shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and has the two curves f(x)=4-x and -2. There is a shaded region making a triangle bounded by the decreasing line f(x), the y-axis and the x-axis. The second graph is the same two curves. There is a solid formed by rotating the shaded region from the first graph around the line y=-2. There is a hollow cylinder inside of the solid represented by the lines y=-2 and y=-4.

Figure 6.24 (a) The region between the graph of the function

f (x) = 4 - x

and the

x -axis

over the interval

[0, 4] .

(b) Revolving the region about the line

y = −2

generates a solid of revolution with a cylindrical hole through its middle.

We can’t apply the volume formula to this problem directly because the axis of revolution is not one of the coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the area of the inner circle. Looking at the graph of the function, we see the radius of the outer circle is given by $f (x) + 2,$ which simplifies to

f (x) + 2 = (4 - x) + 2 = 6 - x .

The radius of the inner circle is $g (x) = 2 .$ Therefore, we have

\begin{matrix} V & = \int_{0}^{4} π [{(6 - x)}^{2} - {(2)}^{2}] d x \\ = π \int_{0}^{4} (x^{2} - 12 x + 32) d x = {π [\frac{x^{3}}{3} - 6 x^{2} + 32 x] |}_{0}^{4} = \frac{160 π}{3} {units}^{3} . \end{matrix}

Checkpoint 6.11

Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of $f (x) = x + 2$ and below by the $x -axis$ over the interval $[0, 3]$ around the line $y = −1 .$

Section 6.2 Exercises

58.

Derive the formula for the volume of a sphere using the slicing method.

59.

Use the slicing method to derive the formula for the volume of a cone.

60.

Use the slicing method to derive the formula for the volume of a tetrahedron with side length $a .$

61.

Use the disk method to derive the formula for the volume of a trapezoidal cylinder.

62.

Explain when you would use the disk method versus the washer method. When are they interchangeable?

For the following exercises, draw a typical slice and find the volume using the slicing method for the given volume.

63.

A pyramid with height 6 units and square base of side 2 units, as pictured here.

This figure is a pyramid with base width of 2 and height of 6 units.

64.

A pyramid with height 4 units and a rectangular base with length 2 units and width 3 units, as pictured here.

This figure is a pyramid with base width of 2, length of 3, and height of 4 units.

65.

A tetrahedron with a base side of 4 units, as seen here.

This figure is an equilateral triangle with side length of 4 units.

66.

A pyramid with height 5 units, and an isosceles triangular base with lengths of 6 units and 8 units, as seen here.

This figure is a pyramid with a triangular base. The view is of the base. The sides of the triangle measure 6 units, 8 units, and 8 units. The height of the pyramid is 5 units.

67.

A cone of radius $r$ and height $h$ has a smaller cone of radius $r / 2$ and height $h / 2$ removed from the top, as seen here. The resulting solid is called a frustum.

This figure is a 3-dimensional graph of an upside down cone. The cone is inside of a rectangular prism that represents the xyz coordinate system. the radius of the bottom of the cone is “r” and the radius of the top of the cone is labeled “r/2”.

For the following exercises, draw an outline of the solid and find the volume using the slicing method.

68.

The base is a circle of radius $a .$ The slices perpendicular to the base are squares.

69.

The base is a triangle with vertices $(0, 0), (1, 0),$ and $(0, 1) .$ Slices perpendicular to the x-axis are semicircles.

70.

The base is the region under the parabola $y = 1 - x^{2}$ in the first quadrant. Slices perpendicular to the xy-plane and parallel to the y-axis are squares.

71.

The base is the region under the parabola $y = 1 - x^{2}$ and above the $x -axis .$ Slices perpendicular to the $y -axis$ are squares.

72.

The base is the region enclosed by $y = x^{2}$ and $y = 9 .$ Slices perpendicular to the x-axis are right isosceles triangles. The intersection of one of these slices and the base is the leg of the triangle.

73.

The base is the area between $y = x$ and $y = x^{2} .$ Slices perpendicular to the x-axis are semicircles.

For the following exercises, draw the region bounded by the curves. Then, use the disk method to find the volume when the region is rotated around the x-axis.

74.

$x + y = 8, x = 0, and y = 0$

75.

$y = 2 x^{2}, x = 0, x = 4, and y = 0$

76.

$y = e^{x} + 1, x = 0, x = 1, and y = 0$

77.

$y = x^{4}, x = 0, and y = 1 for x \geq 0$

78.

$y = \sqrt{x}, x = 0, x = 4, and y = 0$

79.

$y = sin x, y = cos x, and x = 0$

80.

$y = \frac{1}{x}, x = 2, and y = 3$

81.

$x^{2} - y^{2} = 9 and x + y = 9, y = 0 and x = 0$

For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the y-axis.

82.

$y = 4 - \frac{1}{2} x, x = 0, and y = 0$

83.

$y = 2 x^{3}, x = 0, x = 1, and y = 0$

84.

$y = 3 x^{2}, x = 0, and y = 3$

85.

$y = \sqrt{4 - x^{2}}, y = 0, and x = 0$

86.

$y = \frac{1}{\sqrt{x + 1}}, x = 0, x = 3, and y = 0$

87.

$x = sec (y) and y = \frac{π}{4}, y = 0 and x = 0$

88.

$y = \frac{1}{x + 1}, x = 0,, x = 2, and y = 0$

89.

$y = 4 - x, y = x, and x = 0$

For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the x-axis.

90.

$y = x + 2, y = x + 6, x = 0, and x = 5$

91.

$y = x^{2} and y = x + 2$

92.

$x^{2} = y^{3} and x^{3} = y^{2}$

93.

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

94.

[T] $y = cos x, y = e^{− x}, x = 0, and x = 1.2927$

95.

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

96.

$y = sin x, y = 5 sin x, x = 0 and x = π$

97.

$y = \sqrt{1 + x^{2}} and y = \sqrt{4 - x^{2}}$

For the following exercises, draw the region bounded by the curves. Then, use the washer method to find the volume when the region is revolved around the y-axis.

98.

$y = \sqrt{x}, x = 4, and y = 0$

99.

$y = x + 2, y = 2 x - 1, and x = 0$

100.

$y = \sqrt[3]{x} and y = x^{3}$

101.

$x = e^{2 y}, x = y^{2}, y = 0, and y = ln (2)$

102.

$x = \sqrt{9 - y^{2}}, x = e^{− y}, y = 0, and y = 3$

103.

Yogurt containers can be shaped like frustums. Rotate the line $y = \frac{1}{m} x$ around the y-axis to find the volume between $y = a and y = b .$

This figure has two parts. The first part is a solid cone. The base of the cone is wider than the top. It is shown in a 3-dimensional box. Underneath the cone is an image of a yogurt container with the same shape as the figure.

104.

Rotate the ellipse $(x^{2} / a^{2}) + (y^{2} / b^{2}) = 1$ around the x-axis to approximate the volume of a football, as seen here.

This figure has an oval that is approximately equal to the image of a football.

105.

Rotate the ellipse $(x^{2} / a^{2}) + (y^{2} / b^{2}) = 1$ around the y-axis to approximate the volume of a football.

106.

A better approximation of the volume of a football is given by the solid that comes from rotating $y = sin x$ around the x-axis from $x = 0$ to $x = π .$ What is the volume of this football approximation, as seen here?

This figure has a 3-dimensional oval shape. It is inside of a box parallel to the x axis on the bottom front edge of the box. The y-axis is vertical to the solid.

107.

What is the volume of the Bundt cake that comes from rotating $y = sin x$ around the y-axis from $x = 0$ to $x = π ?$

This figure is a graph of a 3-dimensional solid. It is round, bigger towards the bottom. It has a hole in the center that progressively gets smaller towards the bottom. Next to the graph is an image of a bundt cake, resembling the solid.

For the following exercises, find the volume of the solid described.

108.

The base is the region between $y = x$ and $y = x^{2} .$ Slices perpendicular to the x-axis are semicircles.

109.

The base is the region enclosed by the generic ellipse $(x^{2} / a^{2}) + (y^{2} / b^{2}) = 1 .$ Slices perpendicular to the x-axis are semicircles.

110.

Bore a hole of radius $a$ down the axis of a right cone of height $b$ and radius $b$ through the base of the cone as seen here.

This figure is an upside down cone. It has a radius of the top as “b”, center at “a”, and height as “b”.

111.

Find the volume common to two spheres of radius $r$ with centers that are $2 h$ apart, as shown here.

This figure has two circles that intersect. Both circles have radius “r”. There is a line segment from one center to the other. In the middle of the intersection of the circles is point “h”. It is on the line segment.

112.

Find the volume of a spherical cap of height $h$ and radius $r$ where $h < r,$ as seen here.

This figure a portion of a sphere. This spherical cap has radius “r” and height “h”.

113.

Find the volume of a sphere of radius $R$ with a cap of height $h$ removed from the top, as seen here.

This figure is a sphere with a top portion removed. The radius of the sphere is “R”. The distance from the center to where the top portion is removed is “R-h”.

6.2 Determining Volumes by Slicing

Volume and the Slicing Method

Finding Volumes by the Slicing Method

Deriving the Formula for the Volume of a Pyramid

Solution

Solids of Revolution

Using the Slicing Method to find the Volume of a Solid of Revolution

Solution

The Disk Method

Using the Disk Method to Find the Volume of a Solid of Revolution 1

Solution

Using the Disk Method to Find the Volume of a Solid of Revolution 2

Solution

The Washer Method

Using the Washer Method

Solution

The Washer Method with a Different Axis of Revolution

Solution

Section 6.2 Exercises