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### Section 6.3 : Volume With Rings

7. Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by \(x = {y^2} - 4\) and \(x = 6 - 3y\) about the line \(x = 24\).

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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.

To get the intersection points shown on the sketch all we need to do is set the two equations equal and solve (we’ll need these in a bit).

\[\begin{align*}{y^2} - 4 & = 6 - 3y\\ {y^2} + 3y - 10 & = 0\\ \left( {y + 5} \right)\left( {y - 2} \right) & = 0\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,\,\,y = - 5,\,\,\,\,y = 2\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\left( {21, - 5} \right)\,\,\,\& \,\,\,\left( {0,2} \right)\end{align*}\]Note that this can be a difficult thing to do especially if you aren’t a very visual person. However, having a representative ring can be of great help when we go to write down the area formula. Also, getting the representative ring can be difficult without a sketch of the solid of revolution. So, do the best you can at getting these sketches.

Here is a sketch of the solid of revolution.

Here are a couple of sketches of a representative ring. The image on the left shows a representative ring with the front half of the solid cut away and the image on the right shows a representative ring 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).

We now need to find a formula for the area of the ring. Because we are using rings 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 \(y\). 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 ring with all of the various quantities we need put into it.

From the sketch we can see the ring is centered on the line \(x = 24\) and placed at some \(y\).

The inner radius of the ring is then the distance from the axis of rotation to the curve defining the inner edge of the solid. To determine a formula for this first notice that the axis of rotation is a distance of 24 from the \(y\)-axis. Next, the curve defining the inner edge of the solid is a distance of \(x = 6 - 3y\) from the \(y\)-axis. The inner radius is then the difference between these two distances or,

\[{\mbox{Inner Radius}} = 24 - \left( {6 - 3y} \right) = 18 + 3y\]The outer radius is computed in a similar manner but is a little trickier. In this case the curve defining the outer edge of the solid occurs on both the left and right of the \(y\)-axis.

Let’s first look at the case as shown in the sketch above. In this case the value of the function defining the outer edge of the solid is to the left of the \(y\)-axis and so has a negative value. The distance of this point from the \(y\)-axis is then \( - \left( {{y^2} - 4} \right)\) where the minus sign turns the negative function value into a positive value that we need for distance. The outer radius for this case is then the sum of the distance of the axis of rotation to the \(y\)-axis (a distance of 24) and the distance of the curve defining the outer edge to the \(y\)-axis (which we found above).

If the curve defining the outer edge of the solid is to the right of the \(y\)-axis then it will have a positive value and so the distance of points on the curve and the \(y\)-axis is just \({y^2} - 4\). We don’t need the minus sign in this case because the function value is already positive, which we need for distance. The outer radius in this case is then the distance from the axis of rotation to the \(y\)-axis (a distance of 24) minus this new distance.

Nicely enough in either case the outer radius is then,

\[{\mbox{Outer Radius}} = 24 - \left( {{y^2} - 4} \right) = 28 - {y^2}\]Note that in cases like this where the curve defining an edge has both positive and negative values the final equation of the radius (inner or outer depending on the problem) will be the same. You just need to be careful in setting up the case you choose to look at. If you get the first case set up correctly you won’t need to do the second as the formula will be the same.

The area of the ring is then,

\[\begin{align*}A\left( x \right) & = \pi \left[ {{{\left( {{\mbox{Outer Radius}}} \right)}^2} - {{\left( {{\mbox{Inner Radius}}} \right)}^2}} \right]\\ & = \pi \left[ {{{\left( {28 - {y^2}} \right)}^2} - {{\left( {18 + 3y} \right)}^2}} \right] = \pi \left( {460 - 108y - 65{y^2} + {y^4}} \right)\end{align*}\] Show Step 4The final step is to then set up the integral for the volume and evaluate it.

From the intersection points of the two curves we found in Step 1 we can see that the “first” ring in the solid would occur at \(y = - 5\) and the “last” ring would occur at \(y = 2\). Our limits are then : \( - 5 \le y \le 2\).

The volume is then,

\[V = \int_{{ - 5}}^{2}{{\pi \left( {460 - 108y - 65{y^2} + {y^4}} \right)\,dy}} = \left. {\pi \left( {460y - 54{y^2} - \frac{{65}}{3}{y^3} + \frac{1}{5}{y^5}} \right)} \right|_{ - 5}^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{31556}}{{15}}\pi }}\]