6.3 Volumes of revolution: cylindrical shells  (Page 3/6)

The method of cylindrical shells for a solid revolved around the x -axis

Define $Q$ as the region bounded on the right by the graph of $g(y)=2\sqrt{y}$ and on the left by the $y\text{-axis}$ for $y\in \left[0,4\right].$ Find the volume of the solid of revolution formed by revolving $Q$ around the x -axis.

First, we need to graph the region $Q$ and the associated solid of revolution, as shown in the following figure.

Label the shaded region $Q.$ Then the volume of the solid is given by

$\begin{array}{cc}\hfill V& ={\displaystyle {\int }_c^d\left(2\pi yg(y)\right)}dy\hfill \\ & ={\displaystyle {\int }_0^4\left(2\pi y\left(2\sqrt{y}\right)\right)}dy=4\pi {\displaystyle {\int }_0^4{y}^{3\text{/}2}}dy\hfill \\ & ={4\pi [\frac{2y^{5\text{/}2}}{5}\left]\right|}_0^4=\frac{256\pi }{5}{\text{units}}^3\text{.}\hfill \end{array}$

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For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other than one of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells. Recall that we found the volume of one of the shells to be given by

This was based on a shell with an outer radius of $x_i$ and an inner radius of $x_{i-1}.$ If, however, we rotate the region around a line other than the $y\text{-axis},$ we have a different outer and inner radius. Suppose, for example, that we rotate the region around the line $x=\text{−}k,$ where $k$ is some positive constant. Then, the outer radius of the shell is $x_i+k$ and the inner radius of the shell is $x_{i-1}+k.$ Substituting these terms into the expression for volume, we see that when a plane region is rotated around the line $x=\text{−}k,$ the volume of a shell is given by

As before, we notice that $\frac{x_i+x_{i-1}}{2}$ is the midpoint of the interval $\left[x_{i-1},x_i\right]$ and can be approximated by $x_i^{*}.$ Then, the approximate volume of the shell is

The remainder of the development proceeds as before, and we see that

We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. In each case, the volume formula must be adjusted accordingly. Specifically, the $x\text{-term}$ in the integral must be replaced with an expression representing the radius of a shell. To see how this works, consider the following example.

A region of revolution revolved around a line

Define $R$ as the region bounded above by the graph of $f(x)=x$ and below by the $x\text{-axis}$ over the interval $\left[1,2\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the line $x=\mathrm{-1}.$

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

Note that the radius of a shell is given by $x+1.$ Then the volume of the solid is given by

$\begin{array}{cc}\hfill V& ={\displaystyle {\int }_1^2\left(2\pi \left(x+1\right)f(x)\right)}dx\hfill \\ & ={\displaystyle {\int }_1^2\left(2\pi \left(x+1\right)x\right)}dx=2\pi {\displaystyle {\int }_1^2\left(x^2+x\right)}dx\hfill \\ & ={2\pi \left[\frac{x^3}{3}+\frac{x^2}{2}\right]|}_1^2=\frac{23\pi }{3}{\text{units}}^3\text{.}\hfill \end{array}$

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