P4 6.1 - Volumes of revolution | Edexcel A-Level International Mathematics

Find volumes by rotating a region about the x-axis, using either Cartesian or parametric equations. Understand why the radius is squared, how the limits follow the region, and how to keep a physical volume positive and correctly scaled.

Disc slices about the x-axis

If you double the height of a curve and rotate the region beneath it about the x-axis, does the volume double? Think first about a cylinder: doubling its radius makes each circular end four times as large.

A thin vertical strip between a curve and the x-axis rotates into a thin disc. Its radius is the distance from the curve to the axis, so it is $|y|$ . Its face has area $\pi |y|^2=\pi y^2$ , and its small thickness is $\Delta x$ . Thus its volume is approximately $\pi y^2\Delta x$ . Adding increasingly thin discs gives an integral.

[DIAGRAM: asset_name: A strip becomes a disc; asset_slug: p4-volume-disc-slices; caption: A vertical radius becomes a circular cross-section when the region rotates about the x-axis.; description: Left, the region below y equals x from x equals zero to three with a vertical strip at x equals two. Right, the cone generated by rotating the region, with a circular slice at x equals two labelled radius two and thickness delta x.]
A vertical radius becomes a circular cross-section when the region rotates about the x-axis.

Volume about the x-axis

For the region between $y=f(x)$ , the x-axis and $x=a$ , $x=b$ , with $a<b$ ,

V=\pi\int_a^b y^2\,\mathrm dx.

Use this for a finite solid with disc cross-sections. Not in the formula booklet: learn it.

The integrand is a cross-sectional area. It is not the area under the original curve.

Worked example 1: a constant radius

Question. Rotate the region between $y=2$ , the x-axis, $x=0$ and $x=3$ through one complete revolution about the x-axis. Find the volume.

Plan. Each slice is a disc of radius $2$ ; integrate its constant area along the length.

V=\pi\int_0^3 2^2\,\mathrm dx =4\pi[x]_0^3=12\pi.

Answer. $12\pi$ cubic units.

Check. The solid is a cylinder, so $\pi r^2h=\pi(2)^2(3)=12\pi$ .

Change the radius from a constant to the function $y=x$ . The cross-sectional area now changes with position.

Worked example 2: a changing radius

Question. Rotate the region between $y=x$ , the x-axis and $x=3$ about the x-axis. Find the volume.

Plan. Use radius $x$ on $0\le x\le3$ , so each circular face has area $\pi x^2$ .

V=\pi\int_0^3 x^2\,\mathrm dx =\pi\left[\frac{x^3}{3}\right]_0^3=9\pi.

Answer. $9\pi$ cubic units.

Check. This is a cone with radius $3$ and height $3$ : $\frac13\pi(3)^2(3)=9\pi$ .

The cone check also explains why using one maximum-radius cylinder would overestimate the volume.

Watch out

$\pi\int y\,\mathrm dx$ misses the square of the radius. Write the disc area $\pi y^2$ before choosing an integration method.

Set up the slices before evaluating this volume.

Cartesian volume integrals

Squaring the radius comes before integration, and it can make a curve simpler or more complicated.

For $y=\sqrt{9-x^2}$ , the square is $9-x^2$ . For $y=\mathrm e^x$ , it is $\mathrm e^{2x}$ . For $y=1+x$ , expand the entire bracket: $(1+x)^2=1+2x+x^2$ . Choose an integration method for this squared expression, using the limits of the stated region.

Worked example 3: squaring a sum

Question. Find the volume obtained by rotating the region between $y=2+\mathrm e^x$ , the x-axis, $x=0$ and $x=\ln4$ about the x-axis.

Plan. Expand $y^2$ , then integrate the constant and exponential terms separately.

\begin{aligned} V&=\pi\int_0^{\ln4}(4+4\mathrm e^x+\mathrm e^{2x})\,\mathrm dx\\ &=\pi\left[4x+4\mathrm e^x+\frac12\mathrm e^{2x}\right]_0^{\ln4}\\ &=\pi\left(4\ln4+16+8-4-\frac12\right)\\ &=\pi\left(4\ln4+\frac{39}{2}\right). \end{aligned}

Answer. $\pi(4\ln4+39/2)$ cubic units.

Check. Differentiating the antiderivative gives $4+4\mathrm e^x+\mathrm e^{2x}=(2+\mathrm e^x)^2$ . Also $\mathrm e^{2\ln4}=16$ , rather than $8$ .

Change only the sign before the exponential. The curve now crosses the axis at $x=\ln2$ , but a negative ordinate still gives a positive disc area.

Worked example 4: a curve below the axis

Question. Rotate the region between $y=2-\mathrm e^x$ , the x-axis, $x=0$ and $x=\ln4$ about the x-axis. Find the volume.

Plan. Square $2-\mathrm e^x$ throughout the interval; there is no signed-volume cancellation at $x=\ln2$ .

\begin{aligned} V&=\pi\int_0^{\ln4}(4-4\mathrm e^x+\mathrm e^{2x})\,\mathrm dx\\ &=\pi\left[4x-4\mathrm e^x+\frac12\mathrm e^{2x}\right]_0^{\ln4}\\ &=\pi\left(4\ln4-16+8+4-\frac12\right)\\ &=\pi\left(4\ln4-\frac92\right). \end{aligned}

Answer. $\pi(4\ln4-9/2)$ cubic units.

Check. The answer is positive, approximately $3.28$ cubic units. The antiderivative differentiates to $(2-\mathrm e^x)^2$ , so the middle term has the required negative sign.

Each x-position contributes one disc, whether the original region is above or below the axis.

Watch out

$(2-\mathrm e^x)^2\ne4-\mathrm e^{2x}$ . The middle term is $-4\mathrm e^x$ . A curve crossing the axis does not require subtracting volumes on the two sides.

Apply the same square-first method with a different constant.

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P4 6.1 - Volumes of revolution

Disc slices about the x-axis

Cartesian volume integrals

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