Pc2, Pc3 - Volumes of revolution and mean values | OCR A-Level Further Mathematics B (MEI)

Pc2 builds volumes of revolution from thin cylinders, for rotation about either coordinate axis. Pc3 uses a definite integral to find a function's mean value, connecting the area under a changing height to an equivalent constant height.

Thin discs about the x-axis

Two rectangles each have area $4$ . One has height $1$ and width $4$ ; the other has height $2$ and width $2$ . If each is rotated about its bottom edge on the $x$ -axis, do they make equal volumes? Their areas agree, but the taller rectangle places more material further from the axis.

You already know

A cylinder of radius $r$ and height $h$ has volume $\pi r^2h$ . A definite integral is a limit of sums of thin strips, with $\int_a^b f(x)\,dx$ accumulating height times width.

For a region between $y=f(x)\ge0$ , the $x$ -axis and $x=a,b$ , take a vertical strip of width $\Delta x$ . On rotation about the $x$ -axis it becomes approximately a thin cylinder, or disc, with radius $f(x)$ . Its volume is approximately $\pi f(x)^2\Delta x$ .

[DIAGRAM: asset_name: Vertical strips become discs; asset_slug: h645_b038_discs; caption: Squaring the radius turns an area strip into a volume slice; description: A coordinate graph shows a region beneath y equals square root x between zero and four, with a thin vertical strip at x equals two. The strip is labelled radius y and thickness delta x. A side panel shows a circular cross-section with radius y and states disc volume approximately pi y squared delta x and the limiting integral pi integral y squared dx.]
Squaring the radius turns an area strip into a volume slice

Adding the cylinder volumes, then letting the maximum strip width tend to zero, gives

V=\lim_{\max\Delta x_j\to0}\sum_j\pi f(x_j)^2\Delta x_j =\pi\int_a^b f(x)^2\,dx

This derivation explains the square and the variable of integration: radius squared comes from the circular cross-section, and $dx$ measures thickness along the rotation axis.

Volume about the x-axis

For a bounded region whose cross-section perpendicular to the $x$ -axis is a disc of radius $|y|$ ,

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

If the cross-section has outer radius $R(x)$ and inner radius $r(x)$ , with $R(x)\ge r(x)\ge0$ ,

V=\pi\int_a^b\bigl(R(x)^2-r(x)^2\bigr)\,dx

This subtracts the inner cylinders from the outer ones.

Not in the formula booklet: learn it.

A hole changes the cross-section from a disc into an annulus. Subtract squared radii, rather than squaring their difference.

Worked example 1: a solid with no hole

Question. The region under $y=\sqrt{x}$ , above the $x$ -axis and between $x=0$ and $x=4$ , is rotated through $2\pi$ radians about the $x$ -axis. Find its volume.

Plan. Use radius $\sqrt{x}$ , square it, and integrate the circular cross-sectional area.

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

Why this step? Integrating $y$ would find the original plane area. The rotated cross-section has area $\pi y^2$ .

Answer. $8\pi$ cubic units.

Check. The solid fits inside a cylinder of radius $2$ and length $4$ , whose volume is $16\pi$ ; $8\pi$ is smaller.

Keep the curve and the interval, but change the region to the area above the curve and below $y=3$ . Rotation now leaves a hole of radius $\sqrt{x}$ .

Worked example 2: a solid with a hole

Question. The region between $y=3$ , $y=\sqrt{x}$ , $x=0$ and $x=4$ is rotated about the $x$ -axis. Find its volume.

Plan. Subtract the inner squared radius $x$ from the outer squared radius $9$ .

\begin{aligned} V&=\pi\int_0^4(3^2-(\sqrt{x})^2)\,dx && \text{annular cross-sections} \\ &=\pi\int_0^4(9-x)\,dx && \text{outer minus inner} \\ &=\pi\left[9x-\frac{x^2}{2}\right]_0^4 \\ &=28\pi && 36-8=28 \end{aligned}

Why this step? The empty middle is itself the solid from Worked example 1, so its volume must be subtracted, not its radius.

Answer. $28\pi$ cubic units.

Check. The outer cylinder has volume $36\pi$ . Subtracting the previous $8\pi$ gives $28\pi$ .

If a region crosses the axis, identify the actual distance to the furthest boundary. A curve below the axis still has a positive radius; squaring the signed ordinate handles a disc, but two overlapping discs must not be counted twice.

Watch out

Wrong: annular area $=\pi(R-r)^2$ .

Right: annular area $=\pi(R^2-r^2)$ . The gap's thickness is not the radius of a new disc.

The same cross-sectional method now uses a smaller outer radius.

Horizontal discs about the y-axis

For rotation about the $y$ -axis, turn the slicing direction through a right angle. A horizontal strip of thickness $\Delta y$ makes a disc whose radius is the horizontal distance $|x|$ . Its approximate volume is $\pi x^2\Delta y$ .

Therefore the cylinder sum tends to $\pi\int x^2\,dy$ . If the curve is supplied as $y=f(x)$ , rearrange it to express the relevant horizontal boundary in terms of $y$ , and use $y$ -limits.

Volume about the y-axis

For a region whose perpendicular cross-sections are discs of radius $|x|$ ,

V=\pi\int_c^d x^2\,dy

For annular cross-sections with horizontal radii $R(y)\ge r(y)\ge0$ ,

V=\pi\int_c^d\bigl(R(y)^2-r(y)^2\bigr)\,dy

Both follow as limits of sums of cylinders of thickness $\Delta y$ .

Not in the formula booklet: learn it.

Use the boundary's distance from the new axis. Keeping $y^2\,dx$ after changing the rotation axis describes a different solid.

Worked example 3: rearranging the curve

Question. The region bounded by $y=x^2$ , the $y$ -axis and $y=4$ , with $x\ge0$ , is rotated about the $y$ -axis. Find its volume.

Plan. At each height $y$ , the radius is $x=\sqrt{y}$ ; integrate horizontal disc areas from $y=0$ to $4$ .

\begin{aligned} y=x^2&\ \Longrightarrow\ x^2=y && \text{squared radius} \\ V&=\pi\int_0^4y\,dy && \text{horizontal thickness }dy \\ &=\pi\left[\frac{y^2}{2}\right]_0^4=8\pi \end{aligned}

Why this step? The radius is horizontal now. The curve supplies $x^2$ directly, so there is no need to square $y=x^2$ .

Answer. $8\pi$ cubic units.

Check. The solid has height $4$ and maximum radius $2$ , so its volume is below the enclosing cylinder's $16\pi$ .

Keep the curve and heights, but move to the region between the curve and $x=3$ . This introduces an inner radius without changing the integration variable.

Worked example 4: a hole about the y-axis

Question. The region between $x=3$ , $y=x^2$ with $x\ge0$ , $y=0$ and $y=4$ is rotated about the $y$ -axis. Find its volume.

Plan. Use outer radius $3$ and inner radius $\sqrt{y}$ in a horizontal annulus.

\begin{aligned} V&=\pi\int_0^4(3^2-(\sqrt{y})^2)\,dy && \text{outer minus inner discs} \\ &=\pi\int_0^4(9-y)\,dy \\ &=\pi\left[9y-\frac{y^2}{2}\right]_0^4=28\pi \end{aligned}

Why this step? An inner radius comes from the curve $x=\sqrt y$ , even though the original equation was written with $y$ as the subject.

Answer. $28\pi$ cubic units.

Check. The radius- $3$ , height- $4$ cylinder has volume $36\pi$ . Removing the $8\pi$ solid from the previous example leaves $28\pi$ .

The matching values in the two pairs of examples have a geometric reason: interchanging horizontal and vertical coordinates interchanges the axes and produces congruent solids. The integral changes its letter, while the cross-sectional areas have the same shape.

Exam tip

Before integrating, write the radius, the thickness variable and the limits for that variable. The volume formula must include $\pi$ ; integrating the squared radius alone gives only a volume divided by $\pi$ .

This curve rearranges directly into a horizontal squared radius.

Watch out

Wrong: the annular area is $\pi(3-\sqrt y)^2$ . Squaring the difference of radii does not subtract the hole’s area. The correct cross-section is $\pi(3^2-(\sqrt y)^2)=\pi(9-y)$ , integrated with respect to $y$ .

Identify each disc area separately before taking their difference.

A mean value is an equivalent constant height

The mean value of a function is the constant height that gives the same signed area over the same interval. If that height is $m$ , the rectangle's signed area is $m(b-a)$ , so it must equal $\int_a^b f(x)\,dx$ .

[DIAGRAM: asset_name: A mean height with equal area; asset_slug: h645_b038_mean; caption: The curve x squared on zero to three has the same area as a rectangle of height three; description: A graph shows y equals x squared from zero to three with its area shaded pale blue. A dashed horizontal line at y equals three marks the mean height, with a pale sage rectangle of width three and height three shown in an adjacent matching panel. Both areas are labelled nine.]
The curve x squared on zero to three has the same area as a rectangle of height three

Mean value of a function

For an integrable function $f$ on $[a,b]$ , with $a<b$ , its mean value is

\overline f=\frac1{b-a}\int_a^b f(x)\,dx

This is a mean of the function's values over the interval, not just the mean of its two endpoint values.

In the formula booklet.

For a continuous function, the mean lies between its minimum and maximum values. Positive and negative signed contributions may cancel: a function can have mean zero without being zero throughout the interval.

Worked example 5: a changing height

Question. Find the mean value of $f(x)=x^2$ on $[0,3]$ .

Plan. Integrate the function, then divide by the interval width $3$ .

\begin{aligned} \overline f&=\frac1{3-0}\int_0^3x^2\,dx && \text{divide accumulated area by width} \\ &=\frac13\left[\frac{x^3}{3}\right]_0^3 \\ &=\frac13(9)=3 \end{aligned}

Why this step? The integral is $9$ , but that is an area. Dividing by width converts it into an average height.

Answer. The mean value is $3$ .

Check. A rectangle of width $3$ and height $3$ has area $9$ , matching the curve's integral. Also $0\le3\le9$ , the function's range here.

Change only the left endpoint to $-3$ . The width doubles, but so does the integral because $x^2$ is even.

Worked example 6: an interval crossing zero

Question. Find the mean value of $f(x)=x^2$ on $[-3,3]$ .

Plan. Use interval width $6$ , evaluate the integral and divide.

\begin{aligned} \overline f&=\frac1{3-(-3)}\int_{-3}^3x^2\,dx && \text{width is }6 \\ &=\frac16\left[\frac{x^3}{3}\right]_{-3}^3 \\ &=\frac16(9-(-9))=3 && \text{subtract lower value} \end{aligned}

Why this step? A symmetric interval does not make an even function's integral zero. Both halves contribute positively.

Answer. The mean value remains $3$ .

Check. The mean of $x$ on this interval is $0$ , while the mean of $x^2$ is $3$ . Averaging first and then squaring gives a different answer.

That distinction explains a useful fact about a sinusoidal signal. Over $[0,2\pi]$ , $\sin x$ has mean $0$ , but $\sin^2x=(1-\cos2x)/2$ has mean $1/2$ . Equal positive and negative swings cancel in the first mean, while squaring preserves both. The identities and the mean-value formula are in the booklet; the conclusion follows by integration.

Watch out

Wrong: mean value of $x^2$ on $[0,3]$ is $(0^2+3^2)/2=9/2$ .

Right: it is $3$ , from the integral divided by the width. Averaging endpoints is exact for a straight-line function, but not for every curve.

The opening rectangles also settle the volume question: their volumes are $\pi(1)^2(4)=4\pi$ and $\pi(2)^2(2)=8\pi$ . Equal plane areas need not produce equal volumes because the disc area depends on radius squared.

Quick Recap

Thin cylinders give $V=\pi\int y^2\,dx$ about the $x$ -axis and $V=\pi\int x^2\,dy$ about the $y$ -axis.
For a hole, subtract the inner squared radius from the outer squared radius.
Choose limits for the variable measuring thickness along the rotation axis.
A function's mean is its signed integral divided by the interval width; it need not equal an endpoint average or the function of an average input.

The two area- $4$ rectangles make volumes $4\pi$ and $8\pi$ : equal areas can give different volumes of revolution.

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Pc2, Pc3 - Volumes of revolution and mean values

Thin discs about the x-axis

Horizontal discs about the y-axis

A mean value is an equivalent constant height

Quick Recap

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