CP 5.1 - Volumes of revolution from Cartesian equations | Edexcel A-Level Further Mathematics

Rotating a region around an axis builds a solid from circular slices. You will derive the volume formulae for rotation about both coordinate axes, choose the radius and limits correctly, and subtract inner discs when the solid has a hole.

Building a solid from circular slices

Drill a cylindrical hole straight through the centre of a sphere, leaving a ring whose height is $6$ units. Could its volume be the same whatever the original sphere's radius? It sounds unlikely: a larger sphere leaves a wider ring. Slicing the solid will reveal what compensates for that extra width.

Begin with a simpler solid. Rotate the region under $y=x$ , from $x=0$ to $x=2$ , through a full turn about the $x$ -axis. A vertical strip becomes a thin circular disc. Its radius is the strip's height, $x$ , and its thickness is the small horizontal width $\Delta x$ . Near $x=1$ , that disc has cross-sectional area $\pi$ ; near $x=2$ , the area is $4\pi$ .

You already know

A circle of radius $r$ has area $\pi r^2$ , and a cylinder's volume is cross-sectional area multiplied by length. A definite integral adds limiting contributions over an interval; evaluate it by subtracting the antiderivative at the lower limit from its value at the upper limit.

For a general boundary $y=f(x)$ , a slice perpendicular to the $x$ -axis has radius $|f(x)|$ . Its area is $\pi[f(x)]^2$ : squaring handles either side of the axis. A thin disc contributes approximately $\pi[f(x)]^2\Delta x$ . Summing many such contributions, then letting the largest slice thickness tend to zero, gives the integral.

[DIAGRAM: asset_name: A thin strip becomes a circular disc; asset_slug: 9fm0-cp32-discs; caption: Cross-sectional area times a small axial thickness gives a small volume; description: Two equal-scale coordinate panels show the region beneath y equals x from x equals zero to two, with a vertical strip at x equals one, and the corresponding circular slice with radius one. A labelled arrow indicates rotation about the x-axis. The disc has cross-sectional area pi and axial thickness delta x.]
Cross-sectional area times a small axial thickness gives a small volume

Volume about the x-axis

If a region between $y=f(x)$ and the $x$ -axis, from $x=a$ to $x=b$ , is rotated through $2\pi$ about the $x$ -axis, then

V=\pi\int_a^b y^2\,\mathrm{d}x =\pi\int_a^b[f(x)]^2\,\mathrm{d}x

Here $a<b$ , and each cross-section is a disc with radius $|y|$ . Not in the formula booklet: learn it.

This formula measures a three-dimensional volume, so coordinates measured in centimetres give an answer in $\mathrm{cm^3}$ . Integrating $y$ alone would add rectangular strip areas instead of circular cross-sectional areas.

Key idea

Slice perpendicular to the rotation axis. The integrand is a circular area, $\pi\times\text{radius}^2$ , and the differential measures thickness along the axis.

A short error check separates a volume calculation from an area calculation.

Calculating volumes about the x-axis

For rotation about the $x$ -axis, square the curve's $y$ -coordinate before integrating with respect to $x$ .

Write down the region's boundaries first. The limits are horizontal positions, not heights, and the solid includes every circular slice between them. If the equation already gives $y^2$ , you can use it directly without taking a square root first.

Worked example 1: a straight boundary makes a cone

Question. Find the exact volume when the region bounded by $y=x$ , $y=0$ and $x=2$ is rotated through $2\pi$ about the $x$ -axis.

Plan. The radius is $x$ , so integrate $\pi x^2$ from $0$ to $2$ .

\begin{aligned} V&=\pi\int_0^2x^2\,\mathrm{d}x&&\text{square the radius}\\ &=\pi\left[\frac{x^3}{3}\right]_0^2&&\text{power rule}\\ &=\frac{8\pi}{3}&&\text{upper minus lower} \end{aligned}

Why this step? A slice at position $x$ has radius $x$ , not the final cone radius $2$ . Using radius $2$ throughout would calculate the enclosing cylinder.

Answer. $V=\frac{8\pi}{3}$ cubic units.

Check. A cone of height $2$ and base radius $2$ has volume $\frac13\pi(2^2)(2)=\frac{8\pi}{3}$ .

The second example changes the boundary to $y=\sqrt{x}$ , keeping the same interval. Squaring the radius now removes the root.

Worked example 2: a curved boundary

Question. Find the exact volume when the region between $y=\sqrt{x}$ , the $x$ -axis and $x=2$ is rotated through $2\pi$ about the $x$ -axis.

Plan. Substitute $y^2=x$ directly into the disc formula.

\begin{aligned} V&=\pi\int_0^2(\sqrt{x})^2\,\mathrm{d}x&&\text{radius is }\sqrt{x}\\ &=\pi\int_0^2x\,\mathrm{d}x&&\text{square first}\\ &=\pi\left[\frac{x^2}{2}\right]_0^2=2\pi&&\text{integrate and evaluate} \end{aligned}

Why this step? The formula contains $y^2$ , so integrating $\sqrt{x}$ would calculate the wrong quantity even if the integration were flawless.

Answer. $V=2\pi$ cubic units.

Check. The enclosing cylinder has radius $\sqrt2$ , height $2$ and volume $4\pi$ . The solid occupies half of that volume, consistent with its cross-sectional area $\pi x$ increasing linearly from zero to $2\pi$ .

Keep $\pi$ outside the integral until the final exact value. If a decimal is requested, evaluate that exact expression once, at the end.

Watch out

Wrong: $V=\pi\left(\int_a^by\,\mathrm{d}x\right)^2$ . You must square each slice's radius before adding the slices. Right: $V=\pi\int_a^by^2\,\mathrm{d}x$ .

The next question changes the scale in the curve, while preserving the same disc method.

Calculating volumes about the y-axis

For rotation about the $y$ -axis, use horizontal slices: the radius is the horizontal distance $|x|$ , and the thickness is measured in $y$ .

A strip of height $\Delta y$ becomes a disc of area $\pi x^2$ . Adding its volume contributions and taking the same limiting process as before gives the second required formula.

Volume about the y-axis

If a region between $x=g(y)$ and the $y$ -axis, from $y=c$ to $y=d$ , is rotated through $2\pi$ about the $y$ -axis, then

V=\pi\int_c^d x^2\,\mathrm{d}y =\pi\int_c^d[g(y)]^2\,\mathrm{d}y

Here $c<d$ , and each slice has radius $|x|$ . Not in the formula booklet: learn it.

Notice the two linked changes: use $x^2$ for the radius squared and $\mathrm{d}y$ for the slice thickness. Your limits must now be heights. This can make a difficult-looking curve pleasantly simple.

Worked example 3: a sideways parabola

Question. The region bounded by $x=\sqrt{y}$ , the $y$ -axis and $y=3$ is rotated through $2\pi$ about the $y$ -axis. Find its exact volume.

Plan. Use horizontal discs with $x^2=y$ , between heights $0$ and $3$ .

\begin{aligned} V&=\pi\int_0^3x^2\,\mathrm{d}y&&\text{rotation about }y\\ &=\pi\int_0^3y\,\mathrm{d}y&&x^2=y\\ &=\pi\left[\frac{y^2}{2}\right]_0^3=\frac{9\pi}{2}&&\text{evaluate at heights} \end{aligned}

Why this step? The distance to the $y$ -axis is $x$ , so a vertical-axis rotation does not use $y^2$ .

Answer. $V=\frac{9\pi}{2}$ cubic units.

Check. The enclosing cylinder has radius $\sqrt3$ and height $3$ , giving volume $9\pi$ ; the linearly growing disc areas again give half that value.

The second example changes the radius-squared profile from linear to quadratic. It makes a familiar solid, giving a strong independent check.

Worked example 4: a hemisphere

Question. The region bounded by $x=\sqrt{4-y^2}$ , the non-negative coordinate axes and $0\le y\le2$ is rotated through $2\pi$ about the $y$ -axis. Find its exact volume.

Plan. Use $x^2=4-y^2$ , and integrate between the two heights.

\begin{aligned} V&=\pi\int_0^2(4-y^2)\,\mathrm{d}y&&\text{horizontal disc areas}\\ &=\pi\left[4y-\frac{y^3}{3}\right]_0^2&&\text{integrate both terms}\\ &=\pi\left(8-\frac83\right)=\frac{16\pi}{3}&&\text{upper minus lower} \end{aligned}

Why this step? The equation $x^2+y^2=4$ describes a circle of radius $2$ . Rotating its first-quadrant region makes the upper hemisphere, so the limits cover half a sphere.

Answer. $V=\frac{16\pi}{3}$ cubic units.

Check. Half the sphere volume is $\frac12\times\frac43\pi(2^3)=\frac{16\pi}{3}$ .

A radius can decrease as height increases. It is the order of the heights, rather than whether the boundary slopes upwards, that determines the integral's sign.

Watch out

Wrong: use $\pi\int_0^2 y^2\,\mathrm{d}y$ for the hemisphere in Worked example 4. The radius about the $y$ -axis is $x$ , not the height $y$ . Right: use $\pi\int_0^2 x^2\,\mathrm{d}y=\pi\int_0^2(4-y^2)\,\mathrm{d}y$ .

Writing the radius and thickness together also makes the rotation axis clear to the examiner.

Exam tip

Write the substituted volume integral with its differential: $\pi\int x^2\,\mathrm{d}y$ about the $y$ -axis, or $\pi\int y^2\,\mathrm{d}x$ about the $x$ -axis. Omitting $\pi$ or the differential can lose Pearson's formula mark even when later integration is correct.

Try a different hemisphere, using its squared radius directly.

Subtracting the hole from each slice

When the rotating region does not reach the axis, a perpendicular slice is a ring-shaped annulus rather than a full disc.

If the outer and inner radii are $R$ and $r$ , its area is $\pi R^2-\pi r^2$ . The gap $R-r$ is a radial thickness, not a circle's radius: $\pi(R-r)^2$ would describe a different area.

Annular slices

For a full rotation about a coordinate axis, with non-negative outer radius $R(s)$ and inner radius $r(s)$ ,

V=\pi\int_a^b\bigl(R(s)^2-r(s)^2\bigr)\,\mathrm{d}s

Use $s=x$ about the $x$ -axis or $s=y$ about the $y$ -axis. The outer radius must be at least the inner radius throughout each interval; split the region if the boundary ordering changes. Not in the formula booklet: learn it.

Identify the two distances from the axis before subtracting. The same idea also allows you to find a hollow solid's material volume by subtracting its cavity volume from its external volume.

Worked example 5: a fixed outer radius

Question. The region between $y=2$ and $y=\sqrt{x}$ , for $0\le x\le4$ , is rotated through $2\pi$ about the $x$ -axis. Find the exact volume.

Plan. Each slice has outer radius $2$ , inner radius $\sqrt{x}$ , and area $\pi(4-x)$ .

\begin{aligned} V&=\pi\int_0^4\left(2^2-(\sqrt{x})^2\right)\,\mathrm{d}x&&\text{outer disc minus inner}\\ &=\pi\left[4x-\frac{x^2}{2}\right]_0^4&&\text{integrate }4-x\\ &=\pi(16-8)=8\pi&&\text{apply limits} \end{aligned}

Why this step? The hole has radius $\sqrt{x}$ , so its area is $\pi x$ . Subtracting the radii before squaring would not remove that hole.

Answer. $V=8\pi$ cubic units.

Check. The external cylinder has volume $16\pi$ , and the removed solid has volume $\pi\int_0^4x\,\mathrm{d}x=8\pi$ . Their difference is $8\pi$ .

The second example changes the outer radius from $2$ to $3$ . The intersection height determines a new horizontal limit.

Worked example 6: finding the meeting point

Question. The region enclosed by $y=3$ , $y=\sqrt{x}$ and the $y$ -axis is rotated through $2\pi$ about the $x$ -axis. Find the exact volume.

Plan. Find where the two boundaries meet, then subtract the inner disc area from the outer disc area.

\begin{aligned} \sqrt{x}&=3\quad\Longrightarrow\quad x=9&&\text{intersection}\\ V&=\pi\int_0^9(9-x)\,\mathrm{d}x&&\text{radii }3,\sqrt{x}\\ &=\pi\left[9x-\frac{x^2}{2}\right]_0^9&&\text{integrate}\\ &=\frac{81\pi}{2}&&\text{evaluate} \end{aligned}

Why this step? The limit is $x=9$ , not $3$ : the point where the curves meet has height $3$ but horizontal position $9$ .

Answer. $V=\frac{81\pi}{2}$ cubic units.

Check. The enclosing cylinder has volume $81\pi$ ; the cavity has volume $\pi\int_0^9x\,\mathrm{d}x=\frac{81\pi}{2}$ , leaving the same volume of material.

Now return to the drilled sphere. Let its radius be $R$ , the hole radius be $r$ , and the remaining height be $h$ . At height $y$ , the sphere has outer radius squared $R^2-y^2$ . At the top of the ring, Pythagoras gives $R^2-r^2=(h/2)^2$ . Consequently each remaining slice has area

\pi\bigl((R^2-y^2)-r^2\bigr)=\pi\left(\frac{h^2}{4}-y^2\right)

This is independent of $R$ . Integrating from $-h/2$ to $h/2$ gives

V=\pi\left[\frac{h^2y}{4}-\frac{y^3}{3}\right]_{-h/2}^{h/2} =\frac{\pi h^3}{6}

So every such ring of height $6$ has volume $36\pi$ , whatever the original sphere's radius. A bigger sphere needs a bigger hole to keep that height, and the slice areas balance exactly. This result is derived here; it is not in the formula booklet.

[DIAGRAM: asset_name: Equal-height drilled spheres have equal remaining slice areas; asset_slug: 9fm0-cp32-napkin-ring; caption: The outer and inner squared radii differ by the same amount at each height; description: Two central cross-sections show a sphere of radius five drilled by a cylinder of radius four and a sphere of radius root thirteen drilled by a cylinder of radius two. Both remaining rings extend from y equals minus three to y equals three. A horizontal slice at y equals one labels outer squared radius minus inner squared radius equals eight in each case. The vertical axis is the rotation and drilling axis.]
The outer and inner squared radii differ by the same amount at each height

Watch out

Wrong: use $\pi(R-r)^2$ for an annulus. For example, outer radius $3$ and inner radius $2$ give $\pi(9-4)=5\pi$ , not $\pi$ . Right: subtract the squared radii.

The outer boundary can be curved too. Track the two squared radii in the final check.

Quick Recap

Circular slices give $V=\pi\int y^2\,\mathrm{d}x$ about the $x$ -axis and $V=\pi\int x^2\,\mathrm{d}y$ about the $y$ -axis.
Square the distance from the axis before integrating, and use bounds for the integration variable.
A hollow slice has area $\pi(R^2-r^2)$ ; subtract squared radii, and split intervals if necessary.
Give cubic units, preserve exact values and show the substituted integral before evaluating it.

The drilled-sphere puzzle has a definite answer: height $6$ means volume $36\pi$ , regardless of the original sphere radius.

Last Accessed

Core Pure Mathematics

Further Pure Mathematics 1

Further Pure Mathematics 2

Further Statistics 1

Further Statistics 2

Further Mechanics 1

Further Mechanics 2

Decision Mathematics 1

Decision Mathematics 2

CP 5.1 - Volumes of revolution from Cartesian equations

Building a solid from circular slices

Calculating volumes about the x-axis

Calculating volumes about the y-axis

Subtracting the hole from each slice

Quick Recap

Learn faster with Nova