Volumes of Revolution: Cylindrical Shells: Learn It 2

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Volumes of Revolution: Cylindrical Shells: Learn It 2

Cylindrical Shells Method Cont.

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 method of cylindrical shells for solids of revolution around the $x$ -axis

Let $g(y)$ be continuous and nonnegative.

Define $Q$ as the region bounded on the right by the graph of $g(y),$ on the left by the $y\text{-axis},$ below by the line $y=c,$ and above by the line $y=d.$

Then, the volume of the solid of revolution formed by revolving $Q$ around the $x\text{-axis}$ is given by:

V = \int_{c}^{d} (2 π y g (y)) d y

$V={\displaystyle\int }_{c}^{d}(2\pi yg(y))dy$

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.

This figure has two graphs. The first graph is labeled “a” and is the curve g(y)=2squareroot(y). It is an increasing curve in the first quadrant beginning at the origin. Between the y-axis and the curve, there is a shaded region labeled “Q”. The shaded region is bounded above by the line y=4. The second graph is the same curve in “a” and labeled “b”. It also has a solid region that has been formed by rotating the curve in “a” about the x-axis. The solid starts at the y-axis and stops at x=4. — Figure 7. (a) The region $Q$ to the left of the function $g(y)$ over the interval $\left[0,4\right].$ (b) The solid of revolution generated by revolving $Q$ around the $x\text{-axis}.$

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

\begin{array}{cc} V & = \int_{c}^{d} (2 π y g (y)) d y \\ = \int_{0}^{4} (2 π y (2 \sqrt{y})) d y = 4 π \int_{0}^{4} y^{3 / 2} d y \\ = {4 π [\frac{2 y^{5 / 2}}{5}] |}_{0}^{4} = \frac{256 π}{5} {units}^{3} . \end{array}

$\begin{array}{cc}\hfill V& ={\displaystyle\int }_{c}^{d}(2\pi yg(y))dy\hfill \\ & ={\displaystyle\int }_{0}^{4}(2\pi y(2\sqrt{y}))dy=4\pi {\displaystyle\int }_{0}^{4}{y}^{3\text{/}2}dy\hfill \\ & ={4\pi \left[\frac{2{y}^{5\text{/}2}}{5}\right]|}_{0}^{4}=\frac{256\pi }{5}{\text{units}}^{3}\text{.}\hfill \end{array}$

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:

\begin{array}{cc} V_{shell} & = f (x_{i}^{*}) (π x_{i}^{2} - π x_{i - 1}^{2}) \\ = π f (x_{i}^{*}) (x_{i}^{2} - x_{i - 1}^{2}) \\ = π f (x_{i}^{*}) (x_{i} + x_{i - 1}) (x_{i} - x_{i - 1}) \\ = 2 π f (x_{i}^{*}) (\frac{x_{i} + x_{i - 1}}{2}) (x_{i} - x_{i - 1}) . \end{array}

$\begin{array}{cc}\hfill {V}_{\text{shell}}& =f({x}_{i}^{*})(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2})\hfill \\ & =\pi f({x}_{i}^{*})({x}_{i}^{2}-{x}_{i-1}^{2})\hfill \\ & =\pi f({x}_{i}^{*})({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1})\hfill \\ & =2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})({x}_{i}-{x}_{i-1}).\hfill \end{array}$

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:

\begin{array}{cc} V_{shell} & = 2 π f (x_{i}^{*}) (\frac{(x_{i} + k) + (x_{i - 1} + k)}{2}) ((x_{i} + k) - (x_{i - 1} + k)) \\ = 2 π f (x_{i}^{*}) ((\frac{x_{i} + x_{i - 2}}{2}) + k) Δ x . \end{array}

$\begin{array}{cc}\hfill {V}_{\text{shell}}& =2\pi f({x}_{i}^{*})(\frac{({x}_{i}+k)+({x}_{i-1}+k)}{2})(({x}_{i}+k)-({x}_{i-1}+k))\hfill \\ & =2\pi f({x}_{i}^{*})((\frac{{x}_{i}+{x}_{i-2}}{2})+k)\text{Δ}x.\hfill \end{array}$

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:

V_{shell} \approx 2 π (x_{i}^{*} + k) f (x_{i}^{*}) Δ x

${V}_{\text{shell}}\approx 2\pi ({x}_{i}^{*}+k)f({x}_{i}^{*})\text{Δ}x$

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

V = \int_{a}^{b} (2 π (x + k) f (x)) d x

$V={\displaystyle\int }_{a}^{b}(2\pi (x+k)f(x))dx$

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.

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 $x=-1.$

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

This figure has two graphs. The first graph is labeled “a” and is the line f(x)=x, a diagonal line through the origin. There is a shaded region above the x-axis under the line labeled “R”. This region is bounded to the left by the line x=1 and to the right by the line x=2. There is also the vertical line x=-1 on the graph. The second figure has the same graphs as “a” and is labeled “b”. Also on the graph is a solid formed by rotating the region “R” from the first graph about the line x=-1. — Figure 8. (a) The region $R$ between the graph of $f(x)$ and the $x\text{-axis}$ over the interval $\left[1,2\right].$ (b) The solid of revolution generated by revolving $R$ around the line $x=-1.$

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

\begin{array}{cc} V & = \int_{1}^{2} (2 π (x + 1) f (x)) d x \\ = \int_{1}^{2} (2 π (x + 1) x) d x = 2 π \int_{1}^{2} (x^{2} + x) d x \\ = {2 π [\frac{x^{3}}{3} + \frac{x^{2}}{2}] |}_{1}^{2} = \frac{23 π}{3} {units}^{3} . \end{array}

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

For our final example, let’s look at the volume of a solid of revolution for which the region of revolution is bounded by the graphs of two functions.

Define $R$ as the region bounded above by the graph of the function $f(x)=\sqrt{x}$ and below by the graph of the function $g(x)=\frac{1}{x}$ over the interval $\left[1,4\right].$ Find the volume of the solid of revolution generated 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.

This figure has two graphs. The first graph is labeled “a” and has two curves. The curves are the graphs of f(x)=squareroot(x) and g(x)=1/x. In the first quadrant the curves intersect at (1,1). In between the curves in the first quadrant there is a shaded region labeled “R”, bounded to the right by the line x=4. The second graph is labeled “b” and is the same as the graphs in “a”. Also on this graph is a solid that has been formed by rotating the region “R” from the figure “a” about the y-axis. — Figure 9. (a) The region $R$ between the graph of $f(x)$ and the graph of $g(x)$ over the interval $\left[1,4\right].$ (b) The solid of revolution generated by revolving $R$ around the $y\text{-axis}.$

Note that the axis of revolution is the $y\text{-axis},$ so the radius of a shell is given simply by $x.$ We don’t need to make any adjustments to the $x$ -term of our integrand. The height of a shell, though, is given by $f(x)-g(x),$ so in this case we need to adjust the $f(x)$ term of the integrand.

Then the volume of the solid is given by:

\begin{array}{cc} V & = \int_{1}^{4} (2 π x (f (x) - g (x))) d x \\ = \int_{1}^{4} (2 π x (\sqrt{x} - \frac{1}{x})) d x = 2 π \int_{1}^{4} (x^{3 / 2} - 1) d x \\ = {2 π [\frac{2 x^{5 / 2}}{5} - x] |}_{1}^{4} = \frac{94 π}{5} {units}^{3} . \end{array}

$\begin{array}{cc}\hfill V& ={\displaystyle\int }_{1}^{4}(2\pi x(f(x)-g(x)))dx\hfill \\ & ={\displaystyle\int }_{1}^{4}(2\pi x(\sqrt{x}-\frac{1}{x}))dx=2\pi {\displaystyle\int }_{1}^{4}({x}^{3\text{/}2}-1)dx\hfill \\ & ={2\pi \left[\frac{2{x}^{5\text{/}2}}{5}-x\right]|}_{1}^{4}=\frac{94\pi }{5}{\text{units}}^{3}.\hfill \end{array}$

Volumes of Revolution: Cylindrical Shells: Learn It 2

Cylindrical Shells Method Cont.

the method of cylindrical shells for solids of revolution around the xxx-axis

the method of cylindrical shells for solids of revolution around the $x$ -axis