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Active Calculus

Matthew Boelkins

Section 6.2 Using Definite Integrals to Find Volume

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.
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. We first consider a familiar shape in Preview Activity 6.2.1: a circular cone.

Preview Activity 6.2.1.

Consider a circular cone of radius 3 and height 5, which we view horizontally as pictured below. Our goal in this activity is to use a definite integral to determine the volume of the cone.
(a) Find a formula for the linear function \(y = f(x)\) that is pictured above.
\(f(x)=\)
(b) For the representative slice of thickness \(\triangle x\) that is located horizontally at a location \(x\) (somewhere between \(x = 0\) and \(x = 5\)), what is the radius \(r\) of the representative slice? Note that the radius depends on the value of \(x\text{.}\)
\(r =\)
(c) What is the volume \(V_{\small\text{slice}}(x)\) of the representative slice you found in (b)? (Use D as the value for \(\triangle x\) )
\(V_{\small\text{slice}}(x) =\)
(d) What definite integral \(\int_a^b h(x) \ dx\) will sum the volumes of the thin slices across the full horizontal span of the cone?
\(a =\)
\(b =\)
\(h(x) =\)
What is the exact value of this definite integral?
\(\int_a^b h(x) \ dx =\)
(e) Compare the result of your work in (d) to the volume of the cone that comes from using the formula \(V_{\small\text{cone}} = \frac{1}{3} \pi r^2 h.\)
  • Formula is Larger than the integral
  • Formula is equal to the integral
  • Formula is smaller than the integral

Subsection 6.2.1 The Volume of a Solid of Revolution

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 Preview Activity 6.2.1 is the solid of revolution generated by revolving the portion of the line \(y = 3 - \frac{3}{5}x\) from \(x = 0\) to \(x = 5\) 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.
We first consider solids whose slices are thin cylinders. Recall that the volume of a cylinder is given by \(V = \pi r^2 h\text{.}\)

Example 6.2.1.

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 \((2,0)\text{.}\) When we revolve the region \(R\) about the \(x\)-axis, we get the three-dimensional solid pictured in Figure 6.2.2.
Figure 6.2.2. The solid of revolution in Example 6.2.1.
We slice the solid into vertical slices of thickness \(\Delta x\) between \(x = -2\) and \(x = 2\text{.}\) A representative slice is a cylinder of height \(\Delta x\) and radius \(4-x^2\text{.}\) Hence, the volume of the slice is
\begin{equation*} V_{\text{slice} } = \pi (4-x^2)^2 \Delta x\text{.} \end{equation*}
Using a definite integral to sum the volumes of the representative slices, it follows that
\begin{equation*} V = \int_{-2}^{2} \pi (4-x^2)^2 \, dx\text{.} \end{equation*}
It is straightforward to evaluate the integral and find that the volume is \(V = \frac{512}{15}\pi\text{.}\)
For a solid such as the one in Example 6.2.1, 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.
This method for finding the volume of a solid of revolution is often called the disk method.

The Disk Method.

If \(y = r(x)\) is a nonnegative continuous function on \([a,b]\text{,}\) then the volume of the solid of revolution generated by revolving the curve about the \(x\)-axis over this interval is given by
\begin{equation*} V = \int_a^b \pi r(x)^2 \, dx\text{.} \end{equation*}
A different type of solid can emerge when two curves are involved, as we see in the following example.

Example 6.2.3.

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 \(x + 2 = 4-x^2\text{.}\) Rearranging, it follows that
\begin{equation*} x^2 + x - 2 = 0\text{,} \end{equation*}
and the solutions to this equation are \(x = -2\) and \(x = 1\text{.}\) The curves therefore cross at \((-2,0)\) and \((1,1)\text{.}\)
When we revolve the region \(R\) about the \(x\)-axis, we get the three-dimensional solid pictured at left in Figure 6.2.4.
Figure 6.2.4. At left, the solid of revolution in Example 6.2.3. At right, a typical slice with inner radius \(r(x)\) and outer radius \(R(x)\text{.}\)
Immediately we see a major difference between the solid in this example and the one in Example 6.2.1: 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 6.2.4. At a given location \(x\) between \(x = -2\) and \(x = 1\text{,}\) the small radius \(r(x)\) of the inner circle is determined by the curve \(y = x+2\text{,}\) so \(r(x) = x+2\text{.}\) Similarly, the big radius \(R(x)\) comes from the function \(y = 4-x^2\text{,}\) and thus \(R(x) = 4-x^2\text{.}\)
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
\begin{equation*} \pi R(x)^2 \Delta x - \pi r(x)^2 \Delta x = \pi [ R(x)^2 - r(x)^2] \Delta x\text{,} \end{equation*}
it follows that the volume of a typical slice is
\begin{equation*} V_{\text{slice} } = \pi [ (4-x^2)^2 - (x+2)^2 ] \Delta x\text{.} \end{equation*}
Using a definite integral to sum the volumes of the respective slices across the integral, we find that
\begin{equation*} V = \int_{-2}^1 \pi[ (4-x^2)^2 - (x+2)^2 ] \, dx\text{.} \end{equation*}
Evaluating the integral, we find that the volume of the solid of revolution is \(V = \frac{108}{5}\pi\text{.}\)
This method for finding the volume of a solid of revolution generated by two curves is often called the washer method.

The Washer Method.

If \(y = R(x)\) and \(y = r(x)\) are nonnegative continuous functions on \([a,b]\) that satisfy \(R(x) \ge r(x)\) for all \(x\) in \([a,b]\text{,}\) 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
\begin{equation*} V = \int_a^b \pi [R(x)^2 - r(x)^2] \, dx\text{.} \end{equation*}

Activity 6.2.2.

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

Subsection 6.2.2 Revolving about the \(y\)-axis

When we revolve a given region about the \(y\)-axis, the representative slices now have thickness \(\Delta y\text{,}\) which means that we must integrate with respect to \(y\text{.}\)

Example 6.2.5.

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 \(x = 1\text{,}\) hence at the point \((1,1)\text{.}\) When we revolve the region \(R\) about the \(y\)-axis, we get the three-dimensional solid pictured at left in Figure 6.2.6.
Figure 6.2.6. At left, the solid of revolution in Example 6.2.5. At right, a typical slice with inner radius \(r(y)\) and outer radius \(R(y)\text{.}\)
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 \(y = 1\text{.}\) The thickness of a representative slice is \(\Delta y\text{,}\) so we must express the integrand in terms of \(y\text{.}\) The inner radius is determined by the curve \(y = \sqrt{x}\text{,}\) so we solve for \(x\) and get \(x = y^2 = r(y)\text{.}\) In the same way, we solve the curve \(y = x^4\) (which governs the outer radius) for \(x\) in terms of \(y\text{,}\) and hence \(x = \sqrt[4]{y}\text{.}\) Therefore, the volume of a typical slice is
\begin{equation*} V_{\text{slice} } = \pi [R(y)^2 - r(y)^2] = \pi[(\sqrt[4]{y})^2 - (y^2)^2] \Delta y\text{.} \end{equation*}
We use a definite integral to sum the volumes of all the slices from \(y = 0\) to \(y = 1\text{.}\) The total volume is
\begin{equation*} V = \int_{y=0}^{y=1} \pi \left[ (\sqrt[4]{y})^2 - (y^2)^2 \right] \, dy\text{.} \end{equation*}
It is straightforward to evaluate the integral and find that \(V = \frac{7}{15} \pi\text{.}\)

Activity 6.2.3.

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

Subsection 6.2.3 Revolving about horizontal and vertical lines other than the coordinate axes

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 6.2.7.

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 \(y = -1\text{.}\)
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 \(y = -1\text{,}\) we see the solid shown in Figure 6.2.8. 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 \(y = x\text{.}\) 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, \(r(x)\text{,}\) is given by \(r(x) = x^2 + 1\text{,}\) while the outer radius is \(R(x) = x+1\text{.}\)
Figure 6.2.8. The solid of revolution described in Example 6.2.7.
Therefore, the volume of a typical slice is
\begin{equation*} V_{\text{slice} } = \pi[ R(x)^2 - r(x)^2 ] \Delta x = \pi \left[ (x+1)^2 - (x^2 + 1)^2 \right] \Delta x\text{.} \end{equation*}
Finally, we integrate to find the total volume, and
\begin{equation*} V = \int_0^1 \pi \left[ (x+1)^2 - (x^2 + 1)^2 \right] \, dx = \frac{7}{15} \pi\text{.} \end{equation*}

Activity 6.2.4.

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. For each prompt, use the finite region \(S\) in the first quadrant bounded by the curves \(y = 2x\) and \(y = x^3\text{.}\)
  1. Revolve \(S\) about the line \(y = -2\text{.}\)
  2. Revolve \(S\) about the line \(y = 4\text{.}\)
  3. Revolve \(S\) about the line \(x=-1\text{.}\)
  4. Revolve \(S\) about the line \(x = 5\text{.}\)

Subsection 6.2.4 Summary

  • We can 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 by taking slices perpendicular to the axis of revolution which will then be circular disks or washers.
  • If we revolve about a vertical line and slice perpendicular to that line, then our slices are horizontal and of thickness \(\Delta y\text{.}\) This leads us to integrate with respect to \(y\text{,}\) as opposed to with respect to \(x\) when we slice a solid vertically.
  • 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.

Exercises 6.2.5 Exercises

1.

The region bounded by \(y=e^{2 x},y=0,x=-1,x=0\) is rotated around the \(x\)-axis. Find the volume.
volume =

2.

Find the volume of the solid obtained by rotating the region in the first quadrant bounded by \(y=x^{6}\text{,}\) \(y=1\text{,}\) and the \(y\)-axis around the \(y\)-axis.
Volume =

3.

Find the volume of the solid obtained by rotating the region in the first quadrant bounded by \(y=x^{6}\text{,}\) \(y=1\text{,}\) and the \(y\)-axis around the \(x\)-axis.
Volume =

4.

Find the volume of the solid obtained by rotating the region in the first quadrant bounded by \(y=x^{6}\text{,}\) \(y=1\text{,}\) and the \(y\)-axis about the line \(y=-5\text{.}\)
Volume =

5.

Referring to the figure above, find the volume generated by rotating the region \(\mathcal{R}_2\) about the line \(OA\text{.}\)
Volume =

6.

Which of the following integrals represents the volume of the solid obtained by rotating the region bounded by the curves \(y=(x-2)^4\) and \(8x-y=16\) about the line \(x=10?\)
  • \(\displaystyle \displaystyle \pi \int_{2}^{\,4} {\left\{\left[10-\left(\frac{1}{8}y+2\right)\right]^2-\left[10-\left(2+\sqrt[4]{y}\right)\right]^2\right\}}\,dy\)
  • \(\displaystyle \displaystyle \pi \int_{0}^{\,16} {\left\{\left[10-\left(\frac{1}{8}y+2\right)^2\right]-\left[10-\left(2+\sqrt[4]{y}\right)^2\right]\right\}}\,dy\)
  • \(\displaystyle \displaystyle \pi \int_{0}^{\,16} {\left\{\left[10-\left(\frac{1}{8}y+2\right)\right]-\left[10-\left(2+\sqrt[4]{y}\right)\right]\right\}^2}\,dy\)
  • \(\displaystyle \displaystyle \pi \int_{2}^{\,4} {\left\{\left[10-\left(\frac{1}{8}y+2\right)\right]-\left[10-\left(2+\sqrt[4]{y}\right)\right]\right\}^2}\,dy\)
  • \(\displaystyle \displaystyle \pi \int_{0}^{\,16} {\left\{\left[10-\left(\frac{1}{8}y+2\right)\right]^2-\left[10-\left(2+\sqrt[4]{y}\right)\right]^2\right\}}\,dy\)
  • \(\displaystyle \displaystyle \pi \int_{2}^{\,4} {\left\{\left[10-\left(\frac{1}{8}y+2\right)^2\right]-\left[10-\left(2+\sqrt[4]{y}\right)^2\right]\right\}}\,dy\)

7.

Find the volume of the solid obtained by rotating the region bounded by \(y=\displaystyle \frac{1}{9}x^2,\) \(x=3,\) and \(y = 0\) about the \(y\)-axis. Below is a graph of the bounded region.
Volume =
Note: You can click on the graph to enlarge the image.

8.

Find the volume of the solid obtained by rotating the region bounded by \(y= x\) and \(y = \sqrt{x}\) about the line \(x=4.\)
Volume =

9.

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 \(f(x) = 0\text{.}\) Let \(R\) denote the region bounded by this portion of \(f\text{,}\) the \(x\)-axis, and the \(y\)-axis.
  1. Set up a definite integral whose value is the exact arc length of \(f\) that lies along the upper boundary of \(R\text{.}\) Use technology appropriately to evaluate the integral you find.
  2. Set up a definite integral whose value is the exact area of \(R\text{.}\) Use technology appropriately to evaluate the integral you find.
  3. 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.
  4. 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.

10.

Consider the curves given by \(y = \sin(x)\) and \(y = \cos(x)\text{.}\) For each of the following problems, you should include a sketch of the region/solid being considered, as well as a labeled representative slice.
  1. 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?
  2. Set up a definite integral whose value is the exact area of \(R\text{.}\)
  3. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the \(x\)-axis.
  4. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the \(y\)-axis.
  5. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the line \(y = 2\text{.}\)
  6. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the line \(x = -1\text{.}\)

11.

Consider the finite region \(R\) that is bounded by the curves \(y = 1+\frac{1}{2}(x-2)^2\text{,}\) \(y=\frac{1}{2}x^2\text{,}\) and \(x = 0\text{.}\)
  1. Determine a definite integral whose value is the area of the region enclosed by the two curves.
  2. 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 \(y = -1\text{.}\)
  3. 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.
  4. Find an expression involving one or more definite integrals whose value is the perimeter of the region \(R\text{.}\)
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