P4.6 Volumes of revolution
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Pure Mathematics 4 (A2), specification section 6.1. Rotation about the \(y\)-axis is not in this unit; the integration techniques themselves are on the P3.7 and P4.5 pages.
What this page covers
7 points
- Where the formula \(V=\pi\int y^2\,dx\) comes from, and how it reproduces the cone and sphere formulae
- The volume formed when a region is rotated through \(2\pi\) radians (360°) about the \(x\)-axis
- Evaluating the integral with trigonometric identities, integration by parts, partial fractions and substitution
- Regions between a curve and a line, or between two curves (outer solid minus inner solid)
- Composite solids built from a curved piece and a cone
- Curves given parametrically: \(V=\pi\int y^2\frac{dx}{dt}\,dt\)
- Using volumes of revolution to model real objects, with units and comments on the model
In one minute
5 points
- Rotating the region under \(y=f(x)\) from \(x=a\) to \(x=b\) through 360° about the \(x\)-axis gives a solid of volume \(V=\pi\int_a^b y^2\,dx\): square first, then integrate.
- Make \(y^2\) integrable: expand brackets, use \(\sin^2x=\frac12(1-\cos2x)\) or \(\sec^2x=1+\tan^2x\), split into partial fractions, or integrate by parts.
- Between two graphs: outer solid minus inner solid, \(V=\pi\int_a^b\left(y_{\text{outer}}^2-y_{\text{inner}}^2\right)dx\). Never square the difference.
- Parametric curves: \(V=\pi\int_{t_1}^{t_2}y^2\frac{dx}{dt}\,dt\), with the limits changed to values of \(t\).
- Exact answers keep \(\pi\) (and \(e\), \(\ln\), surds); decimals only when asked. Volumes are never negative, even for parts of the curve below the axis.
Contents
12 sections
- Where does V = π∫y² dx come from?
- Which integration methods might y² need?
- What if the region lies between two graphs?
- How do you find a volume from parametric equations?
- How are volumes of revolution used to model real objects?
- How do you handle a composite solid?
- Exam tips
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
Where does V = π∫y² dx come from?
Cut the solid into thin slices perpendicular to the \(x\)-axis. The slice at \(x\) is (almost) a disc of radius \(y\) and thickness \(\delta x\), so its volume is about \(\pi y^2\,\delta x\). Adding the discs and letting \(\delta x\to0\) turns the sum into an integral, exactly as the sum of thin rectangles became \(\int y\,dx\) for an area. The radius is the distance from the axis to the curve, so it is \(y\) (or \(|y|\)); squaring removes any sign.
Two checks you can do yourself. Rotating the line \(y=\frac{rx}h\) for \(0\le x\le h\) gives a cone: \(\pi\int_0^h\frac{r^2x^2}{h^2}\,dx=\pi\frac{r^2}{h^2}\cdot\frac{h^3}3=\frac13\pi r^2h\). Rotating the semicircle \(y=\sqrt{r^2-x^2}\) for \(-r\le x\le r\) gives a sphere: \(\pi\int_{-r}^r(r^2-x^2)\,dx=\pi\left(2r^3-\frac23r^3\right)=\frac43\pi r^3\).
The region \(R\) is bounded by the curve \(y=x+\frac1x\), the \(x\)-axis and the lines \(x=1\) and \(x=2\). Find the exact volume of the solid formed when \(R\) is rotated through 360° about the \(x\)-axis.
Square and expand: \(y^2=\left(x+\frac1x\right)^2=x^2+2+x^{-2}\).
Integrate: \(V=\pi\int_1^2\left(x^2+2+x^{-2}\right)dx=\pi\left[\frac{x^3}3+2x-\frac1x\right]_1^2\).
Limits: \(\pi\left[\left(\frac83+4-\frac12\right)-\left(\frac13+2-1\right)\right]=\pi\left(\frac{37}6-\frac43\right)=\) \(\dfrac{29\pi}6\ (\approx15.2)\).
Squaring a sum gives a middle term: \(\left(x+\frac1x\right)^2\) is not \(x^2+\frac1{x^2}\). Here the cross term is 2.
Writing \(\pi\left(\int y\,dx\right)^2\), integrating first and squaring the answer. The squaring happens to \(y\) inside the integral, before anything is integrated.
Which integration methods might y² need?
Squaring \(y\) usually changes the kind of integral you face: a root disappears, a sum gains a cross term, a trigonometric function becomes a square that needs an identity. Recognise the form of \(y^2\) and choose the method.
| Form of \(y^2\) | Method | Example |
|---|---|---|
| powers, roots, a squared bracket | expand, then integrate term by term | \(\left(x+\frac1x\right)^2=x^2+2+x^{-2}\) |
| \(\sin^2\), \(\cos^2\), \(\tan^2\), \(\sin x\cos x\) | double-angle identities or \(\sec^2=1+\tan^2\) | \(\tan^2x=\sec^2x-1\) |
| a product such as \(xe^{2x}\) or \(x\cos x\) | integration by parts | \(\int xe^{2x}\,dx\) |
| a rational function | partial fractions, giving logarithms | \(\frac4{(x+1)(x+3)}\) |
| \(f'(x)\times g(f(x))\) | substitution (or recognition) | \(\sin^2x\cos x\) with \(u=\sin x\) |
The region bounded by \(y=\sin x+\cos x\), the coordinate axes and the line \(x=\frac\pi2\) is rotated through \(2\pi\) radians about the \(x\)-axis. Show that the volume of the solid is \(\frac\pi2(\pi+2)\).
Square and simplify with \(\sin^2x+\cos^2x=1\) and \(2\sin x\cos x=\sin2x\): \(y^2=\sin^2x+2\sin x\cos x+\cos^2x=1+\sin2x\).
Integrate: \(V=\pi\int_0^{\pi/2}(1+\sin2x)\,dx=\pi\left[x-\frac12\cos2x\right]_0^{\pi/2}\).
Limits, with \(\cos\pi=-1\) and \(\cos0=1\): \(\pi\left[\left(\frac\pi2+\frac12\right)-\left(0-\frac12\right)\right]=\pi\left(\frac\pi2+1\right)=\) \(\frac\pi2(\pi+2)\) as required.
The curve \(C\) has equation \(y=\sqrt x\,e^x\), \(x\ge0\). The region bounded by \(C\), the \(x\)-axis and the line \(x=1\) is rotated through 360° about the \(x\)-axis. Find the exact volume of the solid.
Square; the root disappears: \(y^2=xe^{2x}\), so \(V=\pi\int_0^1xe^{2x}\,dx\).
By parts with \(u=x\), \(\frac{dv}{dx}=e^{2x}\), \(v=\frac12e^{2x}\): \(\int xe^{2x}\,dx=\frac12xe^{2x}-\int\frac12e^{2x}\,dx=\frac12xe^{2x}-\frac14e^{2x}\).
Limits: \(\pi\left[\left(\frac12e^2-\frac14e^2\right)-\left(0-\frac14\right)\right]=\) \(\dfrac\pi4\left(e^2+1\right)\approx6.59\).
The lower limit matters: at \(x=0\) the bracket is \(-\frac14\), not 0, because \(e^0=1\).
The region bounded by \(y=\frac2{\sqrt{(x+1)(x+3)}}\), the axes and the line \(x=1\) is rotated through 360° about the \(x\)-axis. Show that the volume of the solid is \(\pi\ln k\), stating the value of \(k\).
Square and split: \(y^2=\frac4{(x+1)(x+3)}=\frac A{x+1}+\frac B{x+3}\); \(4=A(x+3)+B(x+1)\Rightarrow A=2\), \(B=-2\).
Integrate: \(V=\pi\int_0^1\left(\frac2{x+1}-\frac2{x+3}\right)dx=\pi\big[2\ln(x+1)-2\ln(x+3)\big]_0^1\).
Limits, then the laws of logarithms: \(\pi\big[(2\ln2-2\ln4)-(0-2\ln3)\big]=2\pi\ln\frac{2\times3}4=2\pi\ln\frac32=\pi\ln\frac94\), so \(k=\frac94\).
The form "\(\pi\ln k\)" forces the coefficient 2 inside the logarithm: \(2\ln\frac32=\ln\frac94\). As a decimal, \(V\approx2.55\).
Write \(y^2\) on its own line, fully simplified, before the integral sign appears. Most of the difficulty in this topic is the algebra of \(y^2\); once it is in a recognisable form, the integration is routine.
What if the region lies between two graphs?
Rotating a region between an upper graph \(y_1\) and a lower graph \(y_2\) (both above the \(x\)-axis) gives a solid with a hole along the axis. Each slice is a washer: a disc of radius \(y_1\) with a disc of radius \(y_2\) removed. So the volume is (outer solid) − (inner solid): \(\pi\int y_1^2\,dx-\pi\int y_2^2\,dx\).
\(\pi\int(y_1-y_2)^2\,dx\) is wrong: it would be the volume of a solid whose radius is the gap between the graphs, which is a different solid. The washer picture below shows the difference in one slice.
The line \(y=7-x\) meets the curve \(y=\frac6x\), \(x>0\), at the points \(A\) and \(B\). The finite region \(R\) between them is rotated through 360° about the \(x\)-axis. (a) Find the coordinates of \(A\) and \(B\). (b) Find the exact volume of the solid generated.
(a) \(7-x=\frac6x\Rightarrow x^2-7x+6=0\Rightarrow(x-1)(x-6)=0\), so \(A(1,6)\), \(B(6,1)\).
(b) Decide which is outer: at \(x=3\) the line gives 4, the curve gives 2. So the outer radius is \(7-x\) and the inner radius \(\frac6x\).
Outer solid: \(\pi\int_1^6(7-x)^2\,dx=\pi\left[-\frac13(7-x)^3\right]_1^6=\pi\left(-\frac13+72\right)=\frac{215\pi}3\).
Inner solid: \(\pi\int_1^6\frac{36}{x^2}\,dx=\pi\left[-\frac{36}x\right]_1^6=\pi(-6+36)=30\pi\).
Subtract: \(V=\frac{215\pi}3-30\pi=\) \(\dfrac{125\pi}3\approx131\).
The outer solid is a frustum of a cone, so \(\frac{215\pi}3\) can be checked: the line meets the \(x\)-axis at \(x=7\), giving cones of radius 6 (height 6) and radius 1 (height 1): \(\frac13\pi(216-1)=\frac{215\pi}3\).
Using \(\pi\int(y_1-y_2)^2\,dx\). For worked example 5 it gives 35.1 instead of 131. Square each radius separately, then subtract.
How do you find a volume from parametric equations?
When \(x=f(t)\) and \(y=g(t)\), change the variable in \(\pi\int y^2\,dx\): replace \(dx\) by \(\frac{dx}{dt}\,dt\). The limits must become values of \(t\): solve \(f(t)=a\) and \(f(t)=b\) (or use the \(t\)-values given). If \(x\) decreases as \(t\) increases, keep the \(t\)-limits in the order that matches \(x=a\) to \(x=b\); the negative \(\frac{dx}{dt}\) then makes the answer come out positive.
Quick check: the ellipse \(x=a\cos t\), \(y=b\sin t\) rotated about the \(x\)-axis gives \(\frac43\pi ab^2\), which is a sphere when \(a=b\).
The curve \(C\) has parametric equations \(x=\sin t\), \(y=\sin2t\), \(0\le t\le\frac\pi2\). The region bounded by \(C\) and the \(x\)-axis is rotated through \(2\pi\) radians about the \(x\)-axis. (a) Show that the volume is \(4\pi\int_0^{\pi/2}\sin^2t\cos^3t\,dt\). (b) Use the substitution \(u=\sin t\) to find the exact volume.
(a) \(x=0\) at \(t=0\) and \(x=1\) at \(t=\frac\pi2\); \(\frac{dx}{dt}=\cos t\). With \(\sin2t=2\sin t\cos t\): \(V=\pi\int_0^{\pi/2}(2\sin t\cos t)^2\cos t\,dt=4\pi\int_0^{\pi/2}\sin^2t\cos^3t\,dt\) as required.
(b) \(u=\sin t\), \(du=\cos t\,dt\), \(\cos^2t=1-u^2\); limits \(u=0\) to 1. \(4\pi\int_0^1u^2(1-u^2)\,du=4\pi\left[\frac{u^3}3-\frac{u^5}5\right]_0^1=4\pi\cdot\frac2{15}=\) \(\dfrac{8\pi}{15}\approx1.68\).
Check in Cartesian form: \(y^2=4x^2(1-x^2)\), and \(\pi\int_0^14x^2(1-x^2)\,dx=\frac{8\pi}{15}\) as well.
Keeping the \(x\)-limits after changing to \(t\), or leaving out \(\frac{dx}{dt}\). Every part of the integral must be in terms of \(t\): the integrand, the "\(dt\)" and both limits.
How are volumes of revolution used to model real objects?
Many objects (bowls, vases, bottles, lampshades, buoys) have a circular cross-section, so their shape can be described by a profile curve rotated about the axis of symmetry. Put that axis along the \(x\)-axis. Keep the units: lengths in cm give volumes in cm³ (1 litre = 1000 cm³), and mass = density × volume. "Filled to a depth \(d\)" means integrate from the base to \(x=d\), so a volume condition often leads to an equation for \(d\). Be ready to criticise the model: real walls have thickness, bases are flat or have a foot, the profile is only an approximation, and the material may not be uniform.
The inside of a vase is modelled by rotating the curve \(y=6-2\cos\left(\frac{\pi x}{15}\right)\), \(0\le x\le30\), through 360° about the \(x\)-axis, where \(x\) cm is the distance along the axis from the base. (a) Show that \(y^2=38-24\cos\left(\frac{\pi x}{15}\right)+2\cos\left(\frac{2\pi x}{15}\right)\). (b) Find the capacity of the vase in litres, to 3 significant figures. (c) Show that the vase is exactly half full when the water is 15 cm deep, and give one limitation of the model.
(a) With \(\theta=\frac{\pi x}{15}\) and \(\cos^2\theta=\frac12(1+\cos2\theta)\): \(y^2=36-24\cos\theta+4\cos^2\theta=36-24\cos\theta+2+2\cos2\theta=38-24\cos\left(\frac{\pi x}{15}\right)+2\cos\left(\frac{2\pi x}{15}\right)\).
(b) \(V=\pi\left[38x-\frac{360}\pi\sin\left(\frac{\pi x}{15}\right)+\frac{15}\pi\sin\left(\frac{2\pi x}{15}\right)\right]_0^{30}\). Since \(\sin2\pi=\sin4\pi=0\), only the first term survives: \(V=1140\pi\) cm³ \(\approx3581\) cm³ \(=\) 3.58 litres.
(c) The same integral from 0 to 15, with \(\sin\pi=\sin2\pi=0\), gives \(\pi(38\times15)=570\pi=\frac12\times1140\pi\), so the vase is half full at a depth of 15 cm. This also follows from symmetry: the profile is symmetrical about \(x=15\).
Limitations: the model ignores the thickness of the glass in the walls and the base, assumes the profile is exactly a cosine curve, and assumes the base is a perfectly flat disc of radius 4 cm. Any one, stated in context, is enough.
How do you handle a composite solid?
When the region is bounded by a curve, a line and the \(x\)-axis, split it at the meeting point and find each piece separately. Straight-line pieces give cones (\(\frac13\pi r^2h\)) or frustums, which you can find without integrating, and doing it both ways gives an independent check.
The curve \(y=3\sqrt x\) and the line \(y=10-x\) meet at the point \(P\). The region \(R\) is bounded by the curve, the line and the \(x\)-axis. (a) Show that \(P\) has coordinates \((4,6)\). (b) \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis. Find the exact volume of the solid. (c) The solid models a fishing float, with 1 unit representing 1 cm, made of foam of density 0.08 g cm⁻³. Find the mass of the float.
(a) Put \(u=\sqrt x\) in \(3\sqrt x=10-x\): \(u^2+3u-10=0\Rightarrow(u-2)(u+5)=0\Rightarrow u=2\) (as \(u\ge0\)), so \(x=4\), \(y=6\): \(P(4,6)\).
(b) Split at \(x=4\). Curve part: \(\pi\int_0^49x\,dx=\pi\left[\frac92x^2\right]_0^4=72\pi\). Line part: the line meets the \(x\)-axis at \(x=10\), so this is a cone of radius 6 and height 6: \(\frac13\pi\times6^2\times6=72\pi\) (or \(\pi\int_4^{10}(10-x)^2\,dx=72\pi\)). Add: \(V=144\pi\).
(c) Mass = density × volume \(=0.08\times144\pi=36.19\ldots\approx\) 36.2 g.
Both pieces happen to have the same volume, but you cannot assume that; always work out each part.
Before integrating a straight-line piece, ask whether it is a cone or a frustum. \(\frac13\pi r^2h\) is quicker and gives an independent check on your integral.
Exam tips for P4.6
Start every answer with \(V=\pi\int\dots dx\) and carry the \(\pi\) to the end. Dropping \(\pi\) part-way is an easy way to lose the final answer.
Answers such as \(\frac{125\pi}3\), \(\frac\pi4(e^2+1)\) or \(\pi\ln\frac94\) are exact. Give a decimal only when the question asks for one, and then to the accuracy stated.
A quick sketch shows the limits, whether one graph is outside the other, and whether part of the solid is a cone. Find the meeting points algebraically and mark them.
When the answer is given, the working is the whole point: show \(y^2\), the identity or partial fractions used, the integrated expression and the substituted limits.
Find the \(t\)-values at the ends of the region before you write the integral, and write them on the integral sign. A quick table of \((t,x,y)\) values also confirms that \(y\ge0\) over the region.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(V=\pi\left(\int y\,dx\right)^2\). | Each disc has volume \(\pi y^2\,\delta x\); the square belongs to each radius. | Square \(y\) first, then integrate. |
| \((a+b)^2=a^2+b^2\) when squaring \(y\). | The cross term \(2ab\) is missing. | Expand fully: \((a+b)^2=a^2+2ab+b^2\). |
| Between two graphs, \(V=\pi\int(y_1-y_2)^2\,dx\). | That is a solid of radius equal to the gap, not the hollow solid. | \(\pi\int(y_1^2-y_2^2)\,dx\): outer minus inner. |
| A region below the \(x\)-axis gives a negative volume. | \(y^2\ge0\), so \(\pi\int y^2\,dx\) is never negative. | No splitting at the roots is needed for volumes. |
| In the parametric formula, the limits stay as \(x\)-values. | After the change of variable the integral is in \(t\). | Convert the limits to \(t\)-values. |
| A rounded decimal counts as an exact answer. | Rounding loses exactness. | Keep \(\pi\) (and \(e\), \(\ln\), surds) unless a decimal is asked for. |
How this topic connects
Areas (P2) and P4.2 Parametric equations
The same slicing idea: thin rectangles for area, thin discs for volume. Parametric volumes use the same change of variable as the parametric area \(\int y\frac{dx}{dt}\,dt\).
P3.7 and P4.5 Integration techniques
Trigonometric identities, integration by parts, partial fractions and substitution are what turn \(\pi\int y^2\,dx\) into a number; this topic is where they are combined. The cone and sphere formulae from P1 mensuration come out of the disc formula.
Density and modelling
Mass = density × volume links a solid of revolution to a real object, the same density calculation used in Mechanics and in IAL Physics Unit 1 (materials).
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(V=\pi\int_a^by^2\,dx\) | volume of revolution about the \(x\)-axis | No — learn it |
| \(V=\pi\int_a^b\left(y_{\text{outer}}^2-y_{\text{inner}}^2\right)dx\) | region between two graphs (outer minus inner) | No — derive it |
| \(V=\pi\int_{t_1}^{t_2}y^2\frac{dx}{dt}\,dt\) | curve given parametrically | No — derive it |
| \(\cos^2x=\frac12(1+\cos2x)\), \(\sin^2x=\frac12(1-\cos2x)\) | integrating squared trigonometric terms | No — learn it |
| \(\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx\) | products such as \(xe^{2x}\) | Yes |
| cone \(\frac13\pi r^2h\), sphere \(\frac43\pi r^3\) | straight-line pieces of composite solids; checks | No — learn it |
Check your understanding
1. The region under \(y=2x\) for \(0\le x\le3\) is rotated about the \(x\)-axis. Find the volume, and check it with a standard formula.
\(\pi\int_0^34x^2\,dx=\pi\left[\frac43x^3\right]_0^3=36\pi\). It is a cone of radius 6 and height 3: \(\frac13\pi\times36\times3=36\pi\).
2. Why is \(\pi\int(y_1-y_2)^2\,dx\) not the volume of the solid between two graphs?
Each slice is a washer with area \(\pi y_1^2-\pi y_2^2\), not a disc of radius \(y_1-y_2\). With radii 4 and 2: \(12\pi\) against \(4\pi\).
3. Find the exact volume when the region under \(y=e^{-x}\) for \(0\le x\le\ln2\) is rotated about the \(x\)-axis.
\(\pi\int_0^{\ln2}e^{-2x}\,dx=\pi\left[-\frac12e^{-2x}\right]_0^{\ln2}=\pi\left(-\frac18+\frac12\right)=\frac{3\pi}8\).
4. A curve has \(x=t^3\). A region runs from \(x=1\) to \(x=8\). What are the limits of the parametric integral, and what replaces \(dx\)?
\(t=1\) to \(t=2\), and \(dx\) becomes \(3t^2\,dt\).
5. The curve \(y=x^2-4\) is below the \(x\)-axis for \(-2<x<2\). Is the volume of the solid formed by rotating the region between the curve and the axis negative?
No. \(\pi\int_{-2}^2(x^2-4)^2\,dx=\frac{512\pi}{15}\). Squaring makes every disc positive, so no splitting is needed.
Specification checklist
- (P4 6.1) I can explain where \(V=\pi\int y^2\,dx\) comes from and use it for a region rotated through 360° about the \(x\)-axis.
- (P4 6.1) I can simplify \(y^2\) and evaluate the volume integral using identities, partial fractions, substitution or integration by parts, giving exact answers.
- (P4 6.1) I can find the volume generated by a region between a curve and a line, or between two curves, as outer solid minus inner solid.
- (P4 6.1) I can find volumes of composite solids, using cones and other standard shapes where a boundary is straight.
- (P4 6.1) I can find a volume of revolution for a curve given parametrically, using \(\pi\int y^2\frac{dx}{dt}\,dt\) with limits in \(t\).
- (P4 6.1) I can use a volume of revolution to model a real object, working in its units and commenting on the limitations of the model.