The Disk Method

The disk method is a technique for finding volumes of solids of revolution. When you revolve a region around an axis, the cross-sections perpendicular to that axis are circular disks. This method uses integration to sum the volumes of these infinitely thin disks.

Consider revolving the region between the graph of [latex]f(x)={(x-1)}^{2}+1[/latex] and the [latex]x[/latex]-axis over the interval [latex][-1,3][/latex] around the [latex]x[/latex]-axis. The process creates a three-dimensional solid, and each cross-section perpendicular to the [latex]x[/latex]-axis is a circular disk.

The disk method builds on the general slicing method for volumes. You already know that volume can be found using:

[latex]V=\int_{a}^{b}A(x)dx[/latex]

The key difference with the disk method is that the cross-sectional area [latex]A(x)[/latex] is always the area of a circle: [latex]A(x)=\pi[r(x)]^{2}[/latex], where [latex]r(x)[/latex] is the radius of the disk at position [latex]x[/latex].

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

Show Solution

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

We have

[latex]\begin{array}{cc}\hfill V& ={\displaystyle\int }_{a}^{b}\pi {\left[f(x)\right]}^{2}dx\hfill \\ & ={\displaystyle\int }_{1}^{4}\pi {\left[\sqrt{x}\right]}^{2}dx=\pi {\displaystyle\int }_{1}^{4}xdx\hfill \\ & ={\frac{\pi }{2}{x}^{2}|}_{1}^{4}=\frac{15\pi }{2}.\hfill \end{array}[/latex]

The volume is [latex]\frac{(15\pi )}{2}[/latex] units³.

Revolving Around the [latex]y[/latex]-axis

You can revolve a region around any horizontal or vertical line. When revolving around the [latex]y[/latex]-axis, the mechanics are similar, but you express the function in terms of [latex]y[/latex] and integrate with respect to [latex]y[/latex].

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

Show Solution

Figure 11 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 [latex]y\text{-axis,}[/latex] the disks are horizontal, rather than vertical.

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

To find the volume, we integrate with respect to [latex]y.[/latex] We obtain

[latex]\begin{array}{cc}\hfill V& ={\displaystyle\int }_{c}^{d}\pi {\left[g(y)\right]}^{2}dy\hfill \\ & ={\displaystyle\int }_{0}^{4}\pi {\left[\sqrt{4-y}\right]}^{2}dy=\pi {\displaystyle\int }_{0}^{4}(4-y)dy\hfill \\ & ={\pi \left[4y-\frac{{y}^{2}}{2}\right]|}_{0}^{4}=8\pi .\hfill \end{array}[/latex]

 

The volume is [latex]8\pi[/latex] units³.