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

Section 6.3 : Volume With Rings

1. Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by $y = \sqrt x $, $y = 3$ and the $y$-axis about the $y$-axis.

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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 curve $y = \sqrt x $ rises from the origin and flattens out as it moves right, the horizontal line $y = 3$ runs across the top, and the region between them, to the left of the curve and right of the y-axis, is shaded in. 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 is a funnel or trumpet shaped solid, wide and open across the top at $y = 3$ and narrowing down to a point at the origin.$

Here are a couple of sketches of a representative disk. The image on the left shows a representative disk with the front half of the solid cut away and the image on the right shows a representative disk 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 disk drawn in and the front half of the solid cut away so that the disk can be seen. The disk is centered on the y-axis and sits horizontally inside the funnel shaped solid.

A sketch of the solid of revolution with a representative disk 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 disk is centered on the y-axis and sits horizontally inside the solid.

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We now need to find a formula for the area of the disk. Because we are using disks that are centered on the $y$-axis we know that the area formula will need to be in terms of $y$. This in turn means that we’ll need to rewrite the equation of the boundary curve to get into terms of $y$.

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

$A sketch of a representative disk with all of the quantities needed put in. The disk is centered on the y-axis and placed at some $y$, and the distance from the y-axis out to the curve, which has been rewritten as $x = {y^2}$, is labeled radius = ${y^2}$.$

As we can see from the sketch the disk is centered on the $y$-axis and placed at some $y$. The radius of the disk is the distance from the $y$-axis to the curve defining the edge of the solid. In other words,

\[{\mbox{Radius }} = {y^2}\]

The area of the disk is then,

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

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

For the limits on the integral we can see that the “first” disk in the solid would occur at $y = 0$ and the “last” disk would occur at $y = 3$. Our limits are then : $0 \le y \le 3$.

The volume is then,

\[V = \int_{0}^{3}{{\pi {y^4}\,dy}} = \left. {\frac{1}{5}\pi {y^5}} \right|_0^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{243}}{5}\pi }}\]