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A **solid of revolution **is when a function is rotated about an axis. Using integration, we are able to determine the **volume of solids of revolution**. Consider a portion of the graph of \( f(x) = x^2 \):

** **

Figure 1 Figure 2

The **vertical disk method **is used when rotating a graph around a *horizontal *axis. Figure 2 from above is a good example of when we wish to use this method. Before integrating, it is helpful to draw a diagram; we will use \( f(x) = x^2 \) once again. After drawing the solid of revolution we will proceed to draw a "slice" of it:

We can view the slice as a *disk *that has the *x-axis *going through the middle of it. Notice that the *radius *of the disk is simply \( f(x) \) at that particular \( x \)-value. Taking the formula for the area of a circle, we get that this disk has area \( \pi(f(x))^2 \) and thus has volume \( V_{i} = \pi(f(x))^2 \Delta x \) (where \( \Delta x \) is the width). It follows then that the *volume *of this shape is simply the addition of the volumes of all of the disks*: *\[ V \cong \sum_{i=0}^{n} \pi f(x_{i})^2 \Delta x_{i} \]

Recall our definition of the Definite Integral: \[\int_{a}^{b} f(x)dx = \lim_{n\to\infty} \sum_{i=0}^{n-1} f(x_i)\Delta x \]

If we take the limit of our \( V \) as \( n \rightarrow \infty \) we get that the volume of our solid is \[ V = \int_{a}^{b} \pi(f(x))^2 dx \]

__EXAMPLE__

Draw the graph of \( f(x) = \sqrt{x} \) and determine the volume resulting from rotating \( f(x) \) about the x-axis in the interval of [1,5].

See the video below for the solution:

When dealing with **volumes of solids of revolution**, in many instances the section that we are rotating exists between *two functions*. Consider \( f(x) = x \) and \( g(x) = (x-2)^4 + 1 \)

If we rotate the region between the functions on the interval of [2,3], we see that the *radius *of the disk is \( f(x) - g(x) \).

Rewriting our above formula, we obtain a new formula for the volume of a solid obtained by rotating a region between two functions:

\[ V = \int_{a}^{b} \pi(f(x) - g(x))^2 dx \]

The horizontal disk method is used when rotating a graph around a *vertical *axis. We will use the region between the graphs of \( x = (y-1)^2 \) and \( x = y +1 \) for this example (note that these are functions in terms of \( y \)). Similarly to the vertical disk method, after drawing the solid of revolution we will proceed to draw a "slice" of it:

The **radius **of the "slice"/disk from \( x = y + 1 \) to the y-axis is simply \( R_1 = x \). The radius of the "slice"/disk from \( x = (y-1)^2 \) to the y-axis is \( R_2 = x \). Similarly to the vertical disk method, the radius of the disk formed by rotating the region between the two curves is found by subtracting the distance that is outlined by the grey dashes in the diagram above. We now see that \( R = R_1 - R_2 \).

Recall our formula for volume when dealing with two curves:

\[ V = \int_{a}^{b} \pi(f(x)-g(x))^2 dx \]

We can substitute out \( R \), but we still need to determine our \( a \) and \( b \) values. From the diagram we can see that the points of intersection of the two functions are at \( x = 1, 4 \)

__DO NOT SUBSTITUTE__ __\( x = 1 \)__ __AND__ __\( x = 4 \)__ __AS THE ENDPOINTS!__

Although it may be tempting, since we are dealing with functions in terms of \( y \), our \( a \) and \( b \) values need to be in terms of \( y \) as well. We can use either function to determine what our \( y \)-values are:

\( \Longrightarrow 1 = y + 1 \)

\( \Longrightarrow 0 = y \)

\( \Longrightarrow \) The first point of intersection is \( (1, 0) \)

\( \Longrightarrow 4 = y + 1 \)

\( \Longrightarrow 3 = y \)

\( \Longrightarrow \) The second point of intersection is \( (4, 3) \)

We can now substitute all of our values and proceed to use integration to solve for the volume:

\( \Longrightarrow V = \int_{0}^{3} \pi((y + 1) - (y - 1)^2)^2 dx \)

\( \Longrightarrow \ = \pi(\int_{0}^{3} (y + 1 - y^2 + 2y - 1)^2 dx) \)

\( \Longrightarrow \ = \pi(\int_{0}^{3} (-y^2 + 3y)^2 dx) \)

\( \Longrightarrow \ = \pi(\int_{0}^{3} y^4 -6y^3 + 9y^2 dx) \)

\( \Longrightarrow \ = \pi( \frac{1}{5}y^5 - \frac{3}{2}y^4 + 3y^3 \mid_{0}^{3}) \)

\( \Longrightarrow \ = \pi( \frac{1}{5}(3)^5 - \frac{3}{2}(3)^4 + 3(3)^3 - 0) \)

\( \Longrightarrow \ = \frac{81\pi}{10} \) OR \(8.1\pi \) units cubed

The **cylindrical shell method **is prominently used when rotating a graph about a *vertical *axis, but it can be used when rotating about a *horizontal *axis* *as well. Consider \( f(x) = 2x^2 - x^3 \) on the interval of \( [0, 2] \).

If we rotate the region between \( x = 0 \) and \( x = 2 \) about the \( y-axis \), we obtain a **cylindrical shell **as shown below.

We see that the height of the shell is \( f(x) \) and the radius is \( x \). Using our formula for calculating the volume of a cylinder, we obtain the following general formula for determining the volume of a solid of revolution:

\[ V = \int_{a}^{b} 2\pi x^2f(x) dx \]

See the video below for the solution to this example:

Try changing the function and rotating it on different axis. Use the slider to adjust the angle of rotation.

- Last Updated: Apr 19, 2023 1:48 PM
- URL: https://libraryguides.centennialcollege.ca/c.php?g=717032
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