Title: Default for 14.3 Volumes of Revolution
Course: HELM Workbooks 2023-2024

Volumes generated by rotating curves about the x-axis

The figure below shows a graph of the function \(y=2x\) for \(x\) between 0 and 3.

[Image: A graph of the function y = 2x, for x between 0 and 3.]

Imagine rotating the line \(y=2x\) by one complete revolution (\(360^{\circ}\) or \(2\pi\) radians) around the \(x\)-axis. The surface so formed is the surface of a cone as shown in the figure below. Such a three-dimensional shape is known as a solid of revolution. We now discuss how to obtain the volumes of such solids of revolution.

[Image: When line y = 2x is rotated around the axis a solid is generated. It's a cone, vertex at origin, axis = x-axis, base at]

Exercise

Find the volume of the cone generated by rotating \(y=[value]\), for \([value]\le x\le [value]\), around the \(x\)-axis. In order to find the volume of this solid we assume that it is composed of lots of thin circular discs all aligned perpendicular to the \(x\)-axis, such as that shown in the diagram. From the diagram below we note that a typical disc has radius \(y\), which in this case equals \([value]\), and thickness \(\delta x\).

[Image: Representation of a circular disk on the cone of revolution.]

The cone is divided into a number of thin circular discs.

The volume of a circular disc is the circular area multiplied by the thickness.

Write down an expression for the volume of this typical disc. You can type \(\delta x\) as deltax.

\(\quad V_{\mbox{disc}}=\)

To find the total volume we must sum the contributions from all discs and find the limit of this sum as the number of discs tends to infinity and \(\delta x\) tends to zero. That is \[ \lim_{\delta x\rightarrow 0} \sum_{x=[value]}^{x=[value]} [value] \delta x\] This is the definition of a definite integral.

Write down the corresponding integral:

\(\displaystyle \int\) |  | \(\mathrm{d}x\)

Find the required volume by performing the integration:

\(\quad V =\)
Feedback: Solution A disc is like a cylinder. So the volume of the disc will be given by the area of the circle 'base', which has radius \([value]\), multiplied by the thickness of the disc, \(\delta x\), so that \[V_{\mbox{disc}} = [value].\] The integral corresponding to the limit given is \[\int_[value]^[value] [value] \,\mathrm{d}x.\] Evaluating that integral, we obtain the volume of the cone: \[V=[value].\]

Exercise

A graph of the function \(y=[value]\) for \(x\) between \([value]\) and \([value]\) is shown in the diagram. The graph is rotated around the \(x\)-axis to produce the solid shown. Find its volume.

[Image: Representation of a disk on the solid of revolution formed by the parabola around the x-axis.]

The solid of revolution is divided into a number of thin circular discs.

As in the previous Task, the solid is considered to be composed of lots of circular discs of radius \(y\), (which in this example is equal to \([value]\)), and thickness \(\delta x\).

Write down the volume of each disc. You can type \(\delta x\) as deltax.

\(\quad V_{\mbox{disc}}=\)

Write down the expression which represents summing the volumes of all such discs:

\(x=\) |
\(\displaystyle \sum\) |
\(x=\) |

Write down the integral which results from taking the limit of the sum as \(\delta x \rightarrow 0\):

\(\displaystyle \int\) |  | \(\mathrm{d}x\)

Perform the integration to find the volume of the solid:

\(\quad V =\)
Feedback: Solution A disc is like a cylinder. So the volume of the disc will be given by the area of the circle 'base', which has radius \([value]\), multiplied by the thickness of the disc, \(\delta x\), so that \[V_{\mbox{disc}} = [value].\] Summing all such disks that exist between \(x=[value]\) and \(x=[value]\) can be written as \[\sum_{x=[value]}^{x=[value]} [value].\] The integral corresponding to the limit given is \[\int_[value]^[value] [value] \,\mathrm{d}x.\] Evaluating that integral, we obtain the volume of the cone: \[V=[value].\]

Exercise

In general, suppose the graph of \(y(x)\) between \(x=a\) and \(x=b\) is rotated about the \(x\)-axis, and the solid so formed is considered to be composed of lots of circular discs of thickness \(\delta x\).

Write down an expression for the radius of a typical disc:

\(\quad r_{\mbox{disc}}=\)

Write down an expression for the volume of a typical disc:

\(\quad V_{\mbox{disc}}=\)

The total volume is found by summing these individual volumes and taking the limit as \(\delta x\) tends to zero: \[ \lim_{\delta x\rightarrow 0}\sum_{x=a}^{x=b}\pi y^2\delta x\]

Write down the definite integral which this sum defines:

\(\displaystyle \int\) |  | \(\mathrm{d}x\)
Feedback: Solution The radious of a typical disc is the \(y\)-coordinate of the function. Hence, \(r_{\mbox{disc}} = [value]\). Given the radius and the thickness of the disc, we can compute its volume as \(V_{\mbox{disc}} = [value]\). In order to find the total volume bounded by rotating the graph about the \(x\)-axis, we take the limit as \(\delta x\) goes to \(0\) of the sum of the volumes of each disc: \[ \lim_{\delta x \to 0} \sum_{x=a}^{x=b} [value] \] This sum defines the definite integral \[ \int_{[value]}^{[value]} {[value]} \,\mathrm{d}x. \]

Key Point

If the graph of \(y(x)\), between \(x = a\) and \(x = b\), is rotated about the \(x\)-axis the volume of the solid formed is \[\int_a^b \pi y^2 \, \mathrm{d}x\]

Exercise

Find the volume of the solid formed when the part of the curve \(y=[value]\) between \(x=[value]\) and \(x=[value]\) is rotated about the \(x\)-axis.

\(\quad V=\)
Feedback: Solution Using the Key Point, we have \(y(x)=[value]\), \(a=[value]\) and \(b=[value]\). Then, \[\begin{aligned} V &= \int_a^b \pi y^2 \,\mathrm{d}x = \int_[value]^[value] \pi([value]) \,\mathrm{d}x \\ &= \pi \left[[value]\right]_[value]^[value] \\ &= \pi \left[[value]-[value]\right] \\ &= [value] \end{aligned}\]

Exercise

The parabola \(y^2=[value]\) for \([value]\le x\le [value]\), is rotated around the \(x\)-axis. Find the volume of the solid formed.

\(\quad V=\)
Feedback: Solution Using the Key Point, we have \(y^2=[value]\), \(a=[value]\) and \(b=[value]\). Then, \[\begin{aligned} V &= \int_a^b \pi y^2 \,\mathrm{d}x = \int_[value]^[value] \pi([value]) \,\mathrm{d}x \\ &= \pi \left[[value]\right]_[value]^[value] \\ &= \pi \left[[value]-[value]\right] \\ &= [value] \end{aligned}\]

Volumes generated by rotating curves about the \(y\)-axis

We can obtain a different solid of revolution by rotating a curve around the \(y\)-axis instead of around the \(x\)-axis. See the figure below.

[Image: Representation of a disk on the solid of revolution formed by a parabola around the y-axis.]

To find the volume of this solid it is divided into a number of circular discs as before, but this time the discs are horizontal. The radius of a typical disc is \(x\) and its thickness is \(\delta y\). The volume of the disc will be \(\pi x^2\delta y\).

The total volume is found by summing these individual volumes and taking the limit as \(\delta y\rightarrow 0\). If the lower and upper limits on \(y\) are \(c\) and \(d\), we obtain for the volume: \[ \lim_{\delta y\rightarrow 0}\sum_{y=c}^{y=d}\pi x^2\delta y\]which is the definite integral \[ \int_c^d \pi x^2\,\mathrm{d}y.\]

Key Point

If the graph of \(y(x)\), between \(y = c\) and \(y = d\), is rotated about the \(y\)-axis the volume of the solid formed is \[\int_c^d \pi x^2 \,\mathrm{d}y.\]

Exercise

Find the volume generated when the graph of \(y=[value]\) between \(x=[value]\) and \(x=[value]\) is rotated around the \(y\)-axis.

Using the latest Key Point write down the required integral:

\(\displaystyle \int\) |  | \(\mathrm{d}y\)

This integral can be written entirely in terms of \(y\), using the fact that \(y=[value]\) to eliminate \(x\). Do this now, and then evaluate the integral:

\(\quad V =\)
Feedback: Solution Using the latest Key Point, the required integral is \[\int_[value]^[value] [value] \,\mathrm{d}y,\] since \(c=[value]\) and \(d=[value]\). We have \(y=[value]\), so the volume becomes \[\begin{align} \int_[value]^[value] [value] \,\mathrm{d}y &= \pi \left[ [value] \right]_[value]^[value] \\ &=\pi \left[ [value]-[value] \right] \\ &= [value] \end{align}\]

Exercise

The curve \(y=[value]\) for \([value]<x<[value]\) is rotated about the \(y\)-axis. Find the volume of the solid formed.

\(\quad V =\)
Feedback: Solution Using the latest Key Point, the required integral is \[\int_[value]^[value] [value] \,\mathrm{d}y,\] since \(c=[value]\) and \(d=[value]\). Since at \(x=[value]\) we have \(y=[value]\) and at \(x=[value]\) we have \(y=[value]\). We have \(y=[value]\), so the volume becomes \[\begin{align} \int_[value]^[value] [value]\,\mathrm{d}y &= \pi \left[ [value] \right]_[value]^[value] \\ &=\pi \left[ [value]-[value] \right] \\ &= [value] \end{align}\]

Exercise

The curve \(y=[value]\) for \([value]<x<[value]\) is rotated about the \(y\)-axis. Find the volume of the solid formed.

\(\quad V =\)
Feedback: Solution Using the latest Key Point, the required integral is \[\int_[value]^[value] [value] \,\mathrm{d}y,\] since \(c=[value]\) and \(d=[value]\).Since at \(x=[value]\) we have \(y=[value]\) and at \(x=[value]\) we have \(y=[value]\) We have \(y=[value]\), so \(x^2=[value]\) and the volume becomes \[\begin{align} \int_[value]^[value] [value] \,\mathrm{d}y &= \pi \left[ [value] \right]_[value]^[value] \\ &=\pi \left[ [value]-[value] \right] \\ &= [value] \end{align}\]