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

Section 6.4 : Volume With Cylinders

7. Use the method of cylinders to determine the volume of the solid obtained by rotating the region bounded by $y = {x^2} - 6x + 9$ and $y = - {x^2} + 6x - 1$ about the line $x = 8$.

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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 downward opening parabola $y = - {x^2} + 6x - 1$ is the upper boundary and the upward opening parabola $y = {x^2} - 6x + 9$ is the lower boundary. The two cross at $\left( {1,4} \right)$ and $\left( {5,4} \right)$, both marked and labeled, and the lens shaped region between them is shaded in. A vertical dashed line at $x = 8$ with a small circular arrow on it marks the axis of rotation.$

To get the intersection points shown above, which we’ll need in a bit, all we need to do is set the two equations equal and solve.

\[\begin{align*}{x^2} - 6x + 9 & = - {x^2} + 6x - 1\\ 2{x^2} - 12x + 10 & = 0\\ 2\left( {x - 1} \right)\left( {x - 5} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in} x = 1,\,\,\,x = 5 \hspace{0.25in} \Rightarrow \hspace{0.25in} \left( {1,4} \right)\,\,\,\,\& \,\,\,\,\left( {5,4} \right)\end{align*}\]

Hint : Give a good attempt at sketching what the solid of revolution looks like and sketch in a representative cylinder.

Note that this can be a difficult thing to do especially if you aren’t a very visual person. However, having a representative cylinder can be of great help when we go to write down the area formula. Also, getting the representative cylinder can be difficult without a sketch of the solid of revolution. So, do the best you can at getting these sketches.

Show Step 2

Here is a sketch of the solid of revolution.

$A sketch of the solid of revolution obtained by rotating the region about the line $x = 8$. It is a thick ring or doughnut shaped solid with a hole down the middle around the axis of rotation.$

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 dashed line $x = 8$ and stands inside the ring shaped 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 dashed line $x = 8$ 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 a vertical axis (i.e. parallel to the $y$-axis) we know that the area formula will need to be in terms of $x$. Therefore, the equations 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 dashed line $x = 8$ with its left edge at some $x$, and the distance from that edge across to the axis of rotation is labeled radius = $8 - x$. The vertical distance from the parabola ${x^2} - 6x + 9$ up to the parabola $ - {x^2} + 6x - 1$ is labeled height = $ - 2{x^2} + 12x - 10$, with the two individual distances also marked on the mirrored curves at the right of the sketch.$

Note that we put all the height “lines” on the mirrored curves and not the actual curves. This was done so we could put them in a place that didn’t interfere with the $y$-axis.

From the sketch we can see the cylinder is centered on the line $x = 8$ and the left edge of the cylinder is at some $x$.

The radius of the cylinder is just the distance from the axis of rotation to the left edge of the cylinder (i.e. $8 - x$).

The upper edge of the cylinder is on the curve defining the upper portion of the solid and is a distance of $ - {x^2} + 6x - 1$ from the $x$-axis. The lower edge of the cylinder is on the curve defining the lower portion of the solid and is a distance of ${x^2} - 6x + 9$ from the x‑axis. The height then is the difference of these two.

So the radius and width of the cylinder are,

\[{\mbox{Radius}} = 8 - x\hspace{0.25in}\hspace{0.25in}{\mbox{Width}} = - {x^2} + 6x - 1 - \left( {{x^2} - 6x + 9} \right) = - 2{x^2} + 12x - 10\]

The area of the cylinder is then,

\[\begin{align*}A\left( x \right) & = 2\pi \left( {{\mbox{Radius}}} \right)\left( {{\mbox{Height}}} \right)\\ & = 2\pi \left( {8 - x} \right)\left( { - 2{x^2} + 12x - 10} \right) = 2\pi \left( { - 80 + 106x - 28{x^2} + 2{x^3}} \right)\end{align*}\]

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 $x = 1$ and the “last” cylinder would occur at $x = 5$. Our limits are then : $1 \le x \le 5$.

The volume is then,

\[V = \int_{1}^{5}{{2\pi \left( { - 80 + 106x - 28{x^2} + 2{x^3}} \right)\,dx}} = 2\left. {\pi \left( { - 80x + 53{x^2} - \frac{{28}}{3}{x^3} + \frac{1}{2}{x^4}} \right)} \right|_1^5 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{640}}{3}\pi }}\]