1.8 - Volumes of revolution | Cambridge A-Level Mathematics

Rotating a flat region about an axis creates a solid. You will build its volume from circular cross-sections, choose the correct radius and limits for either coordinate axis, and subtract an inner volume when the solid has a hole.

Discs about the x-axis

When the region under $y=\sqrt{x}$ , from $x=0$ to $x=4$ , makes one complete turn about the $x$ -axis, why does its volume involve $x$ , rather than $\sqrt{x}$ ? A thin vertical strip rotates into a thin circular disc. Its radius is $\sqrt{x}$ , so its cross-sectional area is $\pi(\sqrt{x})^2=\pi x$ .

A disc of small thickness $\Delta x$ has volume approximately $\pi y^2\Delta x$ . Adding these and letting the thickness approach zero gives an integral. The squared radius comes from the area of a circle; integrating $y$ instead would calculate a flat area.

Solid of revolution

A solid of revolution is generated by rotating a plane region through one complete turn about a line. In a disc calculation, the radius is the perpendicular distance from the axis to the region's outer boundary.

Distance is non-negative. For a curve below the $x$ -axis the radius is $|y|$ , and its square is still $y^2$ .

Discs about the x-axis

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

V=\pi\int_a^b [f(x)]^2\,dx.

Each cross-section perpendicular to the axis is a disc. Use $x$ -limits because the thickness is $dx$ .

Not in the formula booklet: learn it.

Square the complete expression for the radius before integrating. If $y=x+1$ , the integrand is $(x+1)^2$ , which can be expanded or integrated as a linear bracket.

Worked example 1: discs above the axis

Question. The region between $y=\sqrt{x}$ , the $x$ -axis and $x=4$ is rotated through $360^\circ$ about the $x$ -axis. Find the exact volume.

Plan. Square the radius $\sqrt{x}$ and integrate from $0$ to $4$ .

\begin{aligned} V&=\pi\int_0^4(\sqrt{x})^2\,dx\\ &=\pi\int_0^4 x\,dx\\ &=\pi\left[\frac{x^2}{2}\right]_0^4=8\pi. \end{aligned}

Answer. $8\pi$ cubic units.

Check. The cross-sectional area $\pi x$ increases linearly from $0$ to $4\pi$ ; its average is $2\pi$ . Multiplying by the length $4$ gives $8\pi$ .

Reflect only the region in the $x$ -axis. Its distances from the axis, and therefore its rotating discs, stay the same.

Worked example 2: discs below the axis

Question. The region between $y=-\sqrt{x}$ , the $x$ -axis and $x=4$ is rotated through $360^\circ$ about the $x$ -axis. Find the exact volume.

Plan. Use radius $\sqrt{x}$ , or square the negative boundary value directly.

V=\pi\int_0^4(-\sqrt{x})^2\,dx =\pi\int_0^4 x\,dx=8\pi.

Answer. $8\pi$ cubic units.

Check. The reflection produces exactly the same solid as example 1 after a complete turn. A negative curve height cannot make a volume negative.

The disc picture connects the generating strip with its circular cross-section.

[DIAGRAM: asset_name: A strip becomes a disc; asset_slug: discs-x-axis; caption: The radius is square root of x, so the circular cross-section has area pi times x; description: A left panel shows the region under y equals square root of x from zero to four, with a vertical generating strip at x equals two and radius square root of two. A right panel shows its circular cross-section perpendicular to the x-axis, labelled radius square root of two and area two pi. The formula V equals pi integral zero to four of x dx equals eight pi is shown below.]
The radius is square root of x, so the circular cross-section has area pi times x

Watch out

$V=\pi\int y^2\,dx$ , not $\pi\int y\,dx$ , for these discs about the $x$ -axis. Write the radius and its square before integrating.

Use a linear boundary whose whole expression needs squaring.

Rotation about the y-axis

For rotation about the $y$ -axis, take horizontal strips. Each disc's radius is a horizontal distance $|x|$ , and its thickness is $dy$ . Rewrite the curve so that $x^2$ is expressed in terms of $y$ , then use lower and upper $y$ -coordinates as the limits.

This often avoids solving explicitly for $x$ . For $y=x^2$ , the needed square is already $x^2=y$ . The relation applies to either branch, but specify the actual region before calculating its rotation.

Discs about the y-axis

If $x=g(y)$ bounds a region with the $y$ -axis for $c\le y\le d$ , where $c<d$ , then

V=\pi\int_c^d[g(y)]^2\,dy.

Use the square of the horizontal radius and $y$ -limits.

Not in the formula booklet: learn it.

The integral variable is chosen by the disc thickness, not by which letter appears first in the original equation.

Worked example 3: expressing the squared radius directly

Question. The region in $x\ge0$ bounded by $y=x^2$ , the $y$ -axis and $y=4$ is rotated through $360^\circ$ about the $y$ -axis. Find the exact volume.

Plan. Use horizontal discs from $y=0$ to $y=4$ , with $x^2=y$ .

V=\pi\int_0^4 x^2\,dy =\pi\int_0^4 y\,dy =\pi\left[\frac{y^2}{2}\right]_0^4=8\pi.

Answer. $8\pi$ cubic units.

Check. At height $y$ , the cross-sectional area is $\pi y$ . Its average over $[0,4]$ is $2\pi$ , so the volume over height $4$ is $8\pi$ .

Change only the curve's coefficient to $2$ , keeping the top height $4$ . Every horizontal radius is reduced by a factor $1/\sqrt2$ .

Worked example 4: changing the radius scale

Question. The region in $x\ge0$ bounded by $y=2x^2$ , the $y$ -axis and $y=4$ is rotated through $360^\circ$ about the $y$ -axis. Find the exact volume.

Plan. Express the squared radius as $x^2=y/2$ before integrating.

V=\pi\int_0^4\frac y2\,dy =\pi\left[\frac{y^2}{4}\right]_0^4=4\pi.

Answer. $4\pi$ cubic units.

Check. Compared with example 3, each squared radius and hence each disc area is halved, while the height range is unchanged. The volume should be halved, agreeing with $4\pi$ .

The horizontal generating strip makes the use of $dy$ visible.

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1.8 - Volumes of revolution

Discs about the x-axis

Rotation about the y-axis

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