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

Section 6.4 : Volume With Cylinders

1. Use the method of cylinders to determine the volume of the solid obtained by rotating the region bounded by $x = {\left( {y - 2} \right)^2}$, the $x$-axis and the $y$-axis about the $x$-axis.

Show All Steps Hide All Steps

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 lower half of the parabola $x = {\left( {y - 2} \right)^2}$ runs from $\left( {0,2} \right)$ down to $\left( {4,0} \right)$ and the shaded region is the area under it, bounded above by the curve, on the left by the y-axis and below by the x-axis. A small circular arrow on the x-axis marks it as the axis of rotation.$

Note that we only used the lower half of the parabola here because if we also included the upper half there would be nothing to bound the region above it. Therefore, in order for the $x$-axis and $y$-axis to be bounding curves we have to use the portion below the lower half of the parabola.

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 x-axis. It is a shape rather like a horn, wide and rounded at the left end where it meets the y-axis and tapering to a point at $x = 4$ on the x-axis.$

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 x-axis, lies inside the horn shaped solid near the wide left end, and its edges run from the y-axis out to the curve that forms the edge 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 x-axis and lies inside the solid near the wide left end.

Show Step 3

We now need to find a formula for the surface area of the cylinder. Because we are using cylinders that are centered on the $x$-axis we know that the area formula will need to be in terms of $y$. Therefore, the equation of the curves will need to be in terms of $y$ (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 x-axis with its upper edge at some $y$, and the distance from the x-axis up to that edge is labeled radius = $y$. The distance from the y-axis across to the curve $x = {\left( {y - 2} \right)^2}$ is labeled width = ${\left( {y - 2} \right)^2}$.$

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

The radius of the cylinder is just the distance from the $x$-axis to the upper edge of the cylinder (i.e. $y$). The width of the cylinder is the distance from the $y$-axis to the curve defining the edge of the solid (a distance of ${\left( {y - 2} \right)^2}$).

So, the radius and width of the cylinder are,

\[{\mbox{Radius}} = y\hspace{0.25in}\hspace{0.25in}{\mbox{Width}} = {\left( {y - 2} \right)^2}\]

The area of the cylinder is then,

\[A\left( y \right) = 2\pi \left( {{\mbox{Radius}}} \right)\left( {{\mbox{Width}}} \right) = 2\pi \left( y \right){\left( {y - 2} \right)^2} = 2\pi \left( {4y - 4{y^2} + {y^3}} \right)\]

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 $y = 0$ and the “last” cylinder would occur at $y = 2$. Our limits are then : $0 \le y \le 2$.

The volume is then,

\[V = \int_{0}^{2}{{2\pi \left( {4y - 4{y^2} + {y^3}} \right)\,dy}} = 2\left. {\pi \left( {2{y^2} - \frac{4}{3}{y^3} + \frac{1}{4}{y^4}} \right)} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{8}{3}\pi }}\]