Calculus I - Volumes of Solids of Revolution/Method of Cylinders

Section 6.4 : Volume With Cylinders

2. Use the method of cylinders to determine the volume of the solid obtained by rotating the region bounded by $\displaystyle y = \frac{1}{x}$, $\displaystyle x = \frac{1}{2}$, $x = 4$and the $x$-axis about the $y$-axis.

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Start Solution

We need to start the problem somewhere so let’s start “simple”.

Knowing what the bounded region looks like will definitely help for most of these types of problems since we need to know how all the curves relate to each other when we go to set up the area formula and we’ll need limits for the integral which the graph will often help with.

Here is a sketch of the bounded region with the axis of rotation shown.

$A sketch of the bounded region with the axis of rotation shown. The curve $y = \frac{1}{x}$ drops from $\left( {\frac{1}{2},2} \right)$ on the left down towards the x-axis on the right and the shaded region is the area under it between the vertical lines $x = \frac{1}{2}$ and $x = 4$ and above the x-axis. A small circular arrow on the y-axis marks it as the axis of rotation.$

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Here is a sketch of the solid of revolution.

$A sketch of the solid of revolution obtained by rotating the region about the y-axis. It looks like a wide flat disk or hat brim with a narrow chimney rising out of the middle, since the tall thin part of the region near $x = \frac{1}{2}$ sweeps out the chimney and the long low tail out to $x = 4$ sweeps out the brim.$

Here are a couple of sketches of a representative cylinder. The image on the left shows a representative cylinder with the front half of the solid cut away and the image on the right shows a representative cylinder with a “wire frame” of the back half of the solid (i.e. the curves representing the edges of the of the back half of the solid).

A sketch of the solid of revolution with a representative cylinder drawn in and the front half of the solid cut away so that the cylinder can be seen. The cylinder is centered on the y-axis and stands inside the solid, with its top edge on the curve that forms the upper surface of the solid.

A sketch of the solid of revolution with a representative cylinder drawn in and only a “wire frame” of the back half of the solid shown, so just the curves that form the edges of the solid are drawn. The cylinder is centered on the y-axis and stands inside the solid.

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We now need to find a formula for the surface area of the cylinder. Because we are using cylinders that are centered on the $y$-axis we know that the area formula will need to be in terms of $x$. Therefore, the equation of the curves will need to be in terms of $x$ (which in this case they already are).

Here is another sketch of a representative cylinder with all of the various quantities we need put into it.

$A sketch of a representative cylinder with all of the quantities needed put in. The cylinder is centered on the y-axis with its right edge at some $x$, and the distance from the y-axis across to that edge is labeled radius = $x$. The distance from the x-axis up to the curve $y = \frac{1}{x}$ is labeled height = $\frac{1}{x}$.$

From the sketch we can see the cylinder is centered on the $y$-axis and the right edge of the cylinder is at some $x$.

The radius of the cylinder is just the distance from the $y$-axis to the right edge of the cylinder (i.e. $x$). The height of the cylinder is the distance from the $x$-axis to the curve defining the edge of the solid (a distance of $\frac{1}{x}$).

So, the radius and width of the cylinder are,

\[{\mbox{Radius}} = x\hspace{0.25in}\hspace{0.25in}{\mbox{Height}} = \frac{1}{x}\]

The area of the cylinder is then,

\[A\left( x \right) = 2\pi \left( {{\mbox{Radius}}} \right)\left( {{\mbox{Height}}} \right) = 2\pi \left( x \right)\left( {\frac{1}{x}} \right) = 2\pi \]

Do not expect all the variables to cancel out in the area formula. It may happen on occasion, as it did here, but it is rare with it does.

Show Step 4

The final step is to then set up the integral for the volume and evaluate it.

From the graph from Step 1 we can see that the “first” cylinder in the solid would occur at $x = \frac{1}{2}$ and the “last” cylinder would occur at $x = 4$. Our limits are then : $\frac{1}{2} \le x \le 4$.

The volume is then,

\[V = \int_{{\frac{1}{2}}}^{4}{{2\pi \,dx}} = 2\left. {\pi \left( x \right)} \right|_{\frac{1}{2}}^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{7\pi }}\]