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Section 6-4 : Volume With Cylinders

In the previous section we started looking at finding volumes of solids of revolution. In that section we took cross sections that were rings or disks, found the cross-sectional area and then used the following formulas to find the volume of the solid.

\[V = \int_{{\,a}}^{{\,b}}{{A\left( x \right)\,dx}}\hspace{0.75in}V = \int_{{\,c}}^{{\,d}}{{A\left( y \right)\,dy}}\]

In the previous section we only used cross sections that were in the shape of a disk or a ring. This however does not always need to be the case. We can use any shape for the cross sections as long as it can be expanded or contracted to completely cover the solid we’re looking at. This is a good thing because as our first example will show us we can’t always use rings/disks.

Example 1 Determine the volume of the solid obtained by rotating the region bounded by \(y = \left( {x - 1} \right){\left( {x - 3} \right)^2}\)and the \(x\)-axis about the \(y\)-axis.
Show Solution

As we did in the previous section, let’s first graph the bounded region and solid. Note that the bounded region here is the shaded portion shown. The curve is extended out a little past this for the purposes of illustrating what the curve looks like.

This is the graph of $y=\left( x-1 \right){{\left( x-3 \right)}^{2}}$ on the domain 1<x<3.  The portion of this graph that we are interested in lies in the 1st quadrant and looks vaguely like a parabola with vertex at approximately (0.8, 1.5), opening downward and touches the x-axis at x=1 and x=3.  This is the bounded region we are after for this example. This is the graph of the solid we get from rotating the graph from above about the y-axis.  It looks like  the top half of a bagel or doughnut whose “bottom” is on the x-axis.

So, we’ve basically got something that’s roughly doughnut shaped. If we were to use rings on this solid here is what a typical ring would look like.

In this graph we cut away the front half of the solid from the previous image and put in a typical ring cross section that is centered on the y-axis.  The ring will be described in more detail in the next image. For this image we do away with the solid completely and put in just the bounding curves from the first image in this example and their mirrored image around the y-axis.   Both the inner and outer radius of the ring are distances from the y-axis to a (different) portion of the bounding curve.

This leads to several problems. First, both the inner and outer radius are defined by the same function. This, in itself, can be dealt with on occasion as we saw in a example in the Area Between Curves section. However, this usually means more work than other methods so it’s often not the best approach.

This leads to the second problem we got here. In order to use rings we would need to put this function into the form \(x = f\left( y \right)\). That is NOT easy in general for a cubic polynomial and in other cases may not even be possible to do. Even when it is possible to do this the resulting equation is often significantly messier than the original which can also cause problems.

The last problem with rings in this case is not so much a problem as it’s just added work. If we were to use rings the limit would be \(y\) limits and this means that we will need to know how high the graph goes. To this point the limits of integration have always been intersection points that were fairly easy to find. However, in this case the highest point is not an intersection point, but instead a relative maximum. We spent several sections in the Applications of Derivatives chapter talking about how to find maximum values of functions. However, finding them can, on occasion, take some work.

So, we’ve seen three problems with rings in this case that will either increase our work load or outright prevent us from using rings.

What we need to do is to find a different way to cut the solid that will give us a cross-sectional area that we can work with. One way to do this is to think of our solid as a lump of cookie dough and instead of cutting it perpendicular to the axis of rotation we could instead center a cylindrical cookie cutter on the axis of rotation and push this down into the solid. Doing this would give the following picture,

In this graph we cut away the front half of the solid from the previous image and put in a typical cylinder cross section that is centered on the y-axis.  The cylinder will be described in more detail in the next image. For this image we do away with the solid completely and put in just the bounding curves from the first image in this example and their mirrored image around the y-axis.   The radius of the cylinder is the distance from the y-axis to the edge of the cylinder and that is just x.  The height of the cylinder is the distance from the bottom of the cylinder (which is one the x-axis) to the bounding curve and this is just the function value.

Doing this gives us a cylinder or shell in the object and we can easily find its surface area. The surface area of this cylinder is,

\[\begin{align*}A\left( x \right) & = 2\pi \left( {{\mbox{radius}}} \right)\left( {{\mbox{height}}} \right)\\ & = 2\pi \left( x \right)\left( {\left( {x - 1} \right){{\left( {x - 3} \right)}^2}} \right)\\ & = 2\pi \left( {{x^4} - 7{x^3} + 15{x^2} - 9x} \right)\end{align*}\]

Notice as well that as we increase the radius of the cylinder we will completely cover the solid and so we can use this in our formula to find the volume of this solid. All we need are limits of integration. The first cylinder will cut into the solid at \(x = 1\) and as we increase \(x\) to \(x = 3\) we will completely cover both sides of the solid since expanding the cylinder in one direction will automatically expand it in the other direction as well.

The volume of this solid is then,

\[\begin{align*}V & = \int_{{\,a}}^{{\,b}}{{A\left( x \right)\,dx}}\\ & = 2\pi \int_{{\,1}}^{{\,3}}{{{x^4} - 7{x^3} + 15{x^2} - 9x\,dx}}\\ & = 2\pi \left. {\left( {\frac{1}{5}{x^5} - \frac{7}{4}{x^4} + 5{x^3} - \frac{9}{2}{x^2}} \right)} \right|_{\,1}^{\,3}\\ & = \frac{{24\pi }}{5}\end{align*}\]

The method used in the last example is called the method of cylinders or method of shells. The formula for the area in all cases will be,

\[A = 2\pi \left( {{\mbox{radius}}} \right)\left( {{\mbox{height}}} \right)\]

There are a couple of important differences between this method and the method of rings/disks that we should note before moving on. First, rotation about a vertical axis will give an area that is a function of \(x\) and rotation about a horizontal axis will give an area that is a function of \(y\). This is exactly opposite of the method of rings/disks.

Second, we don’t take the complete range of \(x\) or \(y\) for the limits of integration as we did in the previous section. Instead we take a range of \(x\) or \(y\) that will cover one side of the solid. As we noted in the first example if we expand out the radius to cover one side we will automatically expand in the other direction as well to cover the other side.

Let’s take a look at another example.

Example 2 Determine the volume of the solid obtained by rotating the region bounded by \(y = \sqrt[3]{x}\), \(x = 8\) and the \(x\)-axis about the \(x\)-axis.
Show Solution

First let’s get a graph of the bounded region and the solid.

This is the graph of $y=\sqrt[3]{x}$ (looks basically like $\sqrt{x}$) on the domain 0<x<8.  This is the bounded region we are after for this example. This is the graph of the solid we get from rotating the graph from above about the x-axis.  It looks like a solid cup shaped object laying on its side.

Okay, we are rotating about a horizontal axis. This means that the area will be a function of \(y\) and so our equation will also need to be written in \(x = f\left( y \right)\) form.

\[y = \sqrt[3]{x}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,x = {y^3}\]

As we did in the ring/disk section let’s take a couple of looks at a typical cylinder. The sketch on the left shows a typical cylinder with the back half of the object also in the sketch to give the right sketch some context. The sketch on the right contains a typical cylinder and only the curves that define the edge of the solid.

In this graph we cut away the front half of the solid from the previous image and put in a typical cylinder cross section that is centered on the x-axis.  The cylinder will be described in more detail in the next image. For this image we do away with the solid completely and put in just the bounding curves from the first image in this example and their mirrored image around the x-axis.   The radius of the cylinder is the distance from the x-axis to the edge of the cylinder and that is just y.  The height of the cylinder is the distance from the left edge of the cylinder (which is one the function) to the right edge which is on the line x=8.

In this case the width of the cylinder is not the function value as it was in the previous example. In this case the function value is the distance between the edge of the cylinder and the \(y\)-axis. The distance from the edge out to the line is \(x = 8\) and so the width is then \(8 - {y^3}\). The cross-sectional area in this case is,

\[\begin{align*}A\left( y \right) & = 2\pi \left( {{\mbox{radius}}} \right)\left( {{\mbox{width}}} \right)\\ & = 2\pi \left( y \right)\left( {8 - {y^3}} \right)\\ & = 2\pi \left( {8y - {y^4}} \right)\end{align*}\]

The first cylinder will cut into the solid at \(y = 0\) and the final cylinder will cut in at \(y = 2\) and so these are our limits of integration.

The volume of this solid is,

\[\begin{align*}V & = \int_{{\,c}}^{{\,d}}{{A\left( y \right)\,dy}}\\ & = 2\pi \int_{{\,0}}^{{\,2}}{{8y - {y^4}\,dy}}\\ & = 2\pi \left. {\left( {4{y^2} - \frac{1}{5}{y^5}} \right)} \right|_{\,0}^{\,2}\\ & = \frac{{96\pi }}{5}\end{align*}\]

The remaining examples in this section will have axis of rotation about axis other than the \(x\) and \(y\)-axis. As with the method of rings/disks we will need to be a little careful with these.

Example 3 Determine the volume of the solid obtained by rotating the region bounded by \(y = 2\sqrt {x - 1} \) and \(y = x - 1\) about the line \(x = 6\).
Show Solution

Here’s a graph of the bounded region and solid.

This is the graphs of $y=x-1$ and $y=2\sqrt{x-1}$ on the domain 1<x<5.  The square root is always larger than the line and only intersects the line at the endpoints of the domain.  This is the bounded region we are after for this example. This is the graph of the solid we get from rotating the graph from above about the line x=6.  It looks like an upside down cone whose walls are in the shape of the bounding region.

Here are our sketches of a typical cylinder. Again, the sketch on the left is here to provide some context for the sketch on the right.

In this graph we cut away the front half of the solid from the previous image and put in a typical cylinder cross section that is centered on the line x=6.  The cylinder will be described in more detail in the next image. For this image we do away with the solid completely and put in just the bounding curves from the first image in this example and their mirrored image around the line x=6.   The radius of the cylinder is the distance from the line x=6 to the edge of the cylinder and that is 6-x.  The height of the cylinder is the distance from the bottom of the cylinder (which is one the line) to the top of the cylinder (which is on the square root).

Okay, there is a lot going on in the sketch to the left. First notice that the radius is not just an \(x\) or \(y\) as it was in the previous two cases. In this case \(x\) is the distance from the y‑axis to the edge of the cylinder and we need the distance from the axis of rotation to the edge of the cylinder. That means that the radius of this cylinder is \(6 - x\).

Secondly, the height of the cylinder is the difference of the two functions in this case.

The cross-sectional area is then,

\[\begin{align*}A\left( x \right) & = 2\pi \left( {{\mbox{radius}}} \right)\left( {{\mbox{height}}} \right)\\ & = 2\pi \left( {6 - x} \right)\left( {2\sqrt {x - 1} - x + 1} \right)\\ & = 2\pi \left( {{x^2} - 7x + 6 + 12\sqrt {x - 1} - 2x\sqrt {x - 1} } \right)\end{align*}\]

Now the first cylinder will cut into the solid at \(x = 1\) and the final cylinder will cut into the solid at \(x = 5\) so there are our limits.

Here is the volume.

\[\begin{align*}V & = \int_{{\,a}}^{{\,b}}{{A\left( x \right)\,dx}}\\ & = 2\pi \int_{{\,1}}^{{\,5}}{{{x^2} - 7x + 6 + 12\sqrt {x - 1} - 2x\sqrt {x - 1} \,dx}}\\ & = 2\pi \left. {\left( {\frac{1}{3}{x^3} - \frac{7}{2}{x^2} + 6x + 8{{\left( {x - 1} \right)}^{\frac{3}{2}}} - \frac{4}{3}{{\left( {x - 1} \right)}^{\frac{3}{2}}} - \frac{4}{5}{{\left( {x - 1} \right)}^{\frac{5}{2}}}} \right)} \right|_{\,1}^{\,5}\\ & = 2\pi \left( {\frac{{136}}{{15}}} \right)\\ & = \frac{{272\pi }}{{15}}\end{align*}\]

The integration of the last term is a little tricky so let’s do that here. It will use the substitution,

\[u = x - 1\hspace{0.25in}du = dx\hspace{0.25in}x = u + 1\] \[\begin{align*}\int{{2x\sqrt {x - 1} \,dx}} & = 2\int{{\left( {u + 1} \right){u^{\frac{1}{2}}}\,du}}\\ & = 2\int{{{u^{\frac{3}{2}}} + {u^{\frac{1}{2}}}\,du}}\\ & = 2\left( {\frac{2}{5}{u^{\frac{5}{2}}} + \frac{2}{3}{u^{\frac{3}{2}}}} \right) + c\\ & = \frac{4}{5}{\left( {x - 1} \right)^{\frac{5}{2}}} + \frac{4}{3}{\left( {x - 1} \right)^{\frac{3}{2}}} + c\end{align*}\]

We saw one of these kinds of substitutions back in the substitution section.

Example 4 Determine the volume of the solid obtained by rotating the region bounded by \(x = {\left( {y - 2} \right)^2}\) and \(y = x\) about the line \(y = - 1\).
Show Solution

We should first get the intersection points there.

\[\begin{align*}y & = {\left( {y - 2} \right)^2}\\ y & = {y^2} - 4y + 4\\ 0 & = {y^2} - 5y + 4\\ 0 & = \left( {y - 4} \right)\left( {y - 1} \right)\end{align*}\]

So, the two curves will intersect at \(y = 1\) and \(y = 4\). Here is a sketch of the bounded region and the solid.

This is the graphs of $y=x$ and $x={{\left( y-2 \right)}^{2}}$ on the domain 1<y<4.  The parabola is always to the left of the line and only intersects the line at the endpoints of the domain.  This is the bounded region we are after for this example. This is the graph of the solid we get from rotating the graph from above about the line y=-1.  It looks like a cup shaped object laying on its side.  The interior is hollow and both ends are open.

Here are our sketches of a typical cylinder. The sketch on the left is here to provide some context for the sketch on the right.

In this graph we cut away the front half of the solid from the previous image and put in a typical cylinder cross section that is centered on the line y=-1.  The cylinder will be described in more detail in the next image. For this image we do away with the solid completely and put in just the bounding curves from the first image in this example and their mirrored image around the line y=-1.   The radius of the cylinder is the distance from the line y=-1 to the edge of the cylinder and that is y+1.  The height of the cylinder is the distance from the left edge of the cylinder (which is one the parabola) to the right edge of the cylinder (which is on the line).

Here’s the cross-sectional area for this cylinder.

\[\begin{align*}A\left( y \right) & = 2\pi \left( {{\mbox{radius}}} \right)\left( {{\mbox{width}}} \right)\\ & = 2\pi \left( {y + 1} \right)\left( {y - {{\left( {y - 2} \right)}^2}} \right)\\ & = 2\pi \left( { - {y^3} + 4{y^2} + y - 4} \right)\end{align*}\]

The first cylinder will cut into the solid at \(y = 1\) and the final cylinder will cut in at \(y = 4\). The volume is then,

\[\begin{align*}V & = \int_{{\,c}}^{{\,d}}{{A\left( y \right)\,dy}}\\ & = 2\pi \int_{{\,1}}^{{\,4}}{{ - {y^3} + 4{y^2} + y - 4\,dy}}\\ & = 2\pi \left. {\left( { - \frac{1}{4}{y^4} + \frac{4}{3}{y^3} + \frac{1}{2}{y^2} - 4y} \right)} \right|_{\,1}^{\,4}\\ & = \frac{{63\pi }}{2}\end{align*}\]