6.3 Volumes of revolution: cylindrical shells  (Page 2/6)

Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate ( [link] ).

In reality, the outer radius of the shell is greater than the inner radius, and hence the back edge of the plate would be slightly longer than the front edge of the plate. However, we can approximate the flattened shell by a flat plate of height $f(x_i^{*}),$ width $2\pi x_i^{*},$ and thickness $\text{Δ}x$ ( [link] ). The volume of the shell, then, is approximately the volume of the flat plate. Multiplying the height, width, and depth of the plate, we get

which is the same formula we had before.

To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain

Here we have another Riemann sum, this time for the function $2\pi xf(x).$ Taking the limit as $n\to \infty$ gives us

This leads to the following rule for the method of cylindrical shells.

Now let’s consider an example.

The method of cylindrical shells 1

Define $R$ as the region bounded above by the graph of $f(x)=1\text{/}x$ and below by the $x\text{-axis}$ over the interval $\left[1,3\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

First we must graph the region $R$ and the associated solid of revolution, as shown in the following figure.

Then the volume of the solid is given by

$\begin{array}{cc}\hfill V& ={\displaystyle {\int }_a^b\left(2\pi xf(x)\right)}dx\hfill \\ & ={\displaystyle {\int }_1^3\left(2\pi x\left(\frac{1}{x}\right)\right)dx}\hfill \\ & ={\displaystyle {\int }_1^{3}2}\pi dx={2\pi x|}_1^3=4\pi {\text{units}}^3\text{.}\hfill \end{array}$

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The method of cylindrical shells 2

Define R as the region bounded above by the graph of $f(x)=2x-x^2$ and below by the $x\text{-axis}$ over the interval $\left[0,2\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

First graph the region $R$ and the associated solid of revolution, as shown in the following figure.

Then the volume of the solid is given by

$\begin{array}{cc}\hfill V& ={\displaystyle {\int }_a^b\left(2\pi xf(x)\right)}dx\hfill \\ & ={\displaystyle {\int }_0^2\left(2\pi x\left(2x-x^2\right)\right)}dx=2\pi {\displaystyle {\int }_0^2\left(2x^2-x^3\right)}dx\hfill \\ & ={2\pi \left[\frac{2x^3}{3}-\frac{x^4}{4}\right]|}_0^2=\frac{8\pi }{3}{\text{units}}^3\text{.}\hfill \end{array}$

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As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution, revolved around the $x\text{-axis},$ when we want to integrate with respect to $y.$ The analogous rule for this type of solid is given here.