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Section 6.4 More Volumes: Shifting the Axis of Revolution

We have introduced some methods for creating and calculating the volume of different 3-dimensional solids of revolution.

Subsection What Changes?

Let’s first consider a volume created using disks or washers.

Activity 6.4.1 What Changes (in the Washer Method) with a New Axis?

Let’s revisit Activity 6.3.14 Volumes by Disks/Washers, and ask some more follow-up questions. First, we’ll tinker with the solid we created: instead of revolving around the \(x\)-axis, let’s revolve the same solid around the horizontal line \(y=-3\text{.}\)

(b)

What changes, if any, do you have to make to the area of the “face” \(k\)th washer?

(c)

What changes, if any, do you have to make to the eventual volume integral for this solid?
Now, let’s consider a volume created using shells.

Activity 6.4.2 What Changes (in the Shell Method) with a New Axis?

Let’s revisit Activity 6.3.23 Volume by Shells, and ask some more follow-up questions about the shell method. Again, we’ll tinker with the solid we created: instead of revolving around the \(x\)-axis, let’s revolve the same solid around the horizontal line \(y=9\text{.}\)

(b)

What changes, if any, do you have to make to the area of the rectangle formed by “unrolling” up \(k\)th cylinder?

(c)

What changes, if any, do you have to make to the eventual volume integral for this solid?
In both of these cases, we can notice that the only changes we make are to the radii: we just need to remeasure the distance from axis of revolution to either the ends of the rectangle (in the washer method) or the side of the rectangle (in the shell method).

Subsection Formalizing These Changes in the Washers and Shells

We can look at yet another interactive graph (similar to how we ended Section 6.2 Area Between Curves and Section 6.3 Volumes of Solids of Revolution). This time, we’ll think about how our axis of revolution as well as our choice of rectangle orientation impacts how we construct the washers or shells.
Notice that in each case, we’re remeasuring the radius! Whether we’re measuring the radii of a washer by thinking about how far the function outputs are away from the axis of revolution, or if we’re measuring the radius of a shell by thinking about how far the input variable is away from the axis of revolution, we need to rethink this and do some subtraction.

Activity 6.4.3 More Shifted Axes.

We’re going to spend some time constructing several different volume integrals in this activity. We’ll consider the same region each time, but make changes to the axis of revolution. For each, we’ll want to think about what kind of method we’re using (disks/washers or shells) and how the different axis of revolution gets implemented into our volume integral formulas.
Let’s consider the region bounded by the curves \(y=\cos(x)+3\) and \(y=\dfrac{x}{2}\) between \(x=0\) and \(x=2\pi\text{.}\)

(a)

Let’s start with revolving this around the \(x\)-axis and thinking about the solid formed. While you set up your volume integral, think carefully about which method you’ll be using (disks/washers or shells) as well as which variable you are integrating with regard to (\(x\) or \(y\)).
Hint.
Note that in this region, we definitely want to use rectangles that stand up vertically. That means that they’ll have a very small width, \(\Delta x\text{,}\) and sit perpendicular to the axis of revolution.

(b)

Now, let’s create a different solid by revolving this region around the \(y\)-axis. Set up a volume integral, and continue to think carefully about which method you’ll be using (disks/washers or shells) as well as which variable you are integrating with regard to (\(x\) or \(y\)).
Hint.
We still will use the same tall rectangle with a small \(\Delta x\) side length, but this time it will be parallel to our axis of revolution.

(c)

We’ll start shifting our axis of revolution now. We’ll revolve the same region around the horizontal line \(y=-1\) to create a solid. Set up an integral expression to calculate the volume.
Hint 1.
Note that we’re still using the same rectangle (perpendicular to this horizontal axis), and so still integrating with regard to \(x\text{,}\) and using the washer method.
Hint 2.
Since in the washer method our function outputs represent the radii, we need to remeasure the distance from our curves to the axis of revolution to find each circle’s radius in the washer formula. For a \(y\)-value on each curve, how do we find the vertical distance down to the line \(y=-1\text{?}\)

(d)

Now, revolve the region around the line \(y=5\) to create a solid of revolution, and write down the integral representing the volume.
Hint.
Note, now, that the \(y\)-value of the axis of revolution is larger than all of the \(y\)-values on the curves, meaning that to measure the distance from the axis of revolution to the curves, we might measure them in the opposite direction. Also, which curve is further away from the axis of revolution, representing the larger/outer radius?

(e)

Let’s change things up. Revolve the region around the vertical line \(x=-1\) to create a new solid. Set up an integral representing the volume of that solid.
Hint 1.
Note that the same rectangle that we used before is standing parallel to our axis of revolution. We’re going to change methodology, and use the shell method!
Hint 2.
Normally we use the input variable (\(x\) in this case) to measure the radius from the rectangles at different \(x\)-value to the axis of revolution, the \(y\)-axis. Now, though, we’re not looking at the distance from \(x\)-values to \(x=0\text{.}\) We’re looking to find the radius, the distance from \(x\)-values in this region to \(x=-1\text{.}\)

(f)

We’ll do one more solid. Let’s revolve this region around the line \(x=7\text{.}\) Set up an integral representing the volume.
Hint.
Note that this time, the axis of revolution’s \(x\)-value is larger than all of the \(x\)-values in our region. So when we subtract to measure the radius, we need to subtract from \(x=7\) down to the varying \(x\)-values in the region.

Practice Problems Practice Problems

1.

Consider the integral formula for computing volumes of a solid of revolution using the disk/washer method. What part of this integral formula represents the radius/radii of any circle(s)? Why is the radius represented using the function output from the curve(s) defining the region?

2.

Consider the integral formula for computing volumes of a solid of revolution using the shell method. What part of this integral formula represents the radius/radii of any circle(s)? Why is the radius not represented using the function output from the curve(s) defining the region?

3.

For each of the solids described below, set up an integral expression using disks/washers representing the volume of the solid.

(a)

The region bounded by the curve \(y=1-\sqrt{x}\) in the first quadrant, revolved around \(x=2\text{.}\)
Solution.
\begin{align*} V \amp = \pi \int_{y=0}^{y=0} (2)^2 - (2-(1-y^2))^2\;dy \end{align*}

(b)

The region bounded by the curve \(y=1-\sqrt{x}\) in the first quadrant, revolved around \(x=-1\text{.}\)
Solution.
\begin{align*} V \amp = \pi \int_{y=0}^{y=1} (1-y^2-(-1))^2 - (1)^2\;dy \end{align*}

(c)

The region bounded by the curve \(y=1-\sqrt{x}\) in the first quadrant, revolved around \(y=-2\text{.}\)
Solution.
\begin{align*} V \amp = \pi \int_{x=0}^{x=1} (1-\sqrt{x} - (-2))^2 - (-2)^2\;dx \end{align*}

(d)

The region bounded by the curve \(y=1-\sqrt{x}\) in the first quadrant, revolved around \(y=3\text{.}\)
Solution.
\begin{equation*} V = \int_{x=0}^{x=1} (3)^2 - (3-(1-\sqrt{x}))^2\;dx \end{equation*}

4.

For each of the solids described below, set up an integral expression using shells representing the volume of the solid.

(a)

The region bounded by the curves \(y=\sqrt{x}\) and \(y=x\) in the first quadrant, revolved around the line \(x=2\text{.}\)
Solution.
\begin{align*} V \amp = 2\pi \int_{x=0}^{x=1} (2-x)(\sqrt{x}-x)\;dx \end{align*}

(b)

The region bounded by the curves \(y=\sqrt{x}\) and \(y=x\) in the first quadrant, revolved around the line \(x=-1\text{.}\)
Solution.
\begin{align*} V \amp = 2\pi \int_{x=0}^{x=1} (x-(-1))(\sqrt{x}-x)\;dx \end{align*}

(c)

The region bounded by the curves \(y=\sqrt{x}\) and \(y=x\) in the first quadrant, revolved around the line \(y=-2\text{.}\)
Solution.
\begin{align*} V \amp = 2\pi \int_{y=0}^{y=1}(y+2)(y-y^2)\;dy \end{align*}

(d)

The region bounded by the curves \(y=\sqrt{x}\) and \(y=x\) in the first quadrant, revolved around the line \(y=3\text{.}\)
Solution.
\begin{equation*} V = 2\pi \int_{y=0}^{y=1} (3-y)(y-y^2)\;dy \end{equation*}
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