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

A parametric curve supplies the radius and the axial position separately. You will convert both volume formulae into the parameter, handle decreasing coordinates without a negative volume, and calculate volumes for a curved body joined to a cylindrical section.

Changing the slice thickness into a parameter

The curve $x=t^2$ , $y=t+1$ , $0\le t\le1$ , bounds a region above the $x$ -axis. Why does $\pi\int_0^1(t+1)^2\,\mathrm{d}t$ fail to give its volume of revolution? The radius is correct, but the parameter does not measure horizontal distance: equal changes in $t$ do not give equal slice thicknesses.

You already know

For a full turn about the $x$ -axis, $V=\pi\int y^2\,\mathrm{d}x$ ; about the $y$ -axis, $V=\pi\int x^2\,\mathrm{d}y$ . Parametric equations give $x$ and $y$ in terms of a common parameter. The chain rule gives $\mathrm{d}x=\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t$ in a change of variable.

Here $\frac{\mathrm{d}x}{\mathrm{d}t}=2t$ , so a small parameter change $\Delta t$ gives approximately $2t\Delta t$ of horizontal thickness. Near $t=0$ the curve advances slowly in $x$ ; near $t=1$ it advances faster. The derivative gives each slice its proper weight.

Parametric volume about the x-axis

For $x=f(t)$ , $y=g(t)$ , with $x$ increasing as $t$ runs from $\alpha$ to $\beta$ , the region between the curve and the $x$ -axis gives

V=\pi\int_{\alpha}^{\beta}[g(t)]^2 f'(t)\,\mathrm{d}t =\pi\int_{\alpha}^{\beta}y^2\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t

The bounds correspond to the left and right boundaries. If the axial coordinate reverses direction, split the curve into suitable sections and avoid counting the same solid twice. Not in the formula booklet: learn it.

You do not need to eliminate the parameter. Convert the radius, the differential and the limits together, so every part of the integral uses the same variable.

Worked example 1: a non-constant horizontal speed

Question. The curve $x=t^2$ , $y=t+1$ , $0\le t\le1$ , bounds a region with the $x$ -axis and the lines $x=0$ , $x=1$ . Find the exact volume when it is rotated through $2\pi$ about the $x$ -axis.

Plan. Use $y=t+1$ and $\mathrm{d}x=2t\,\mathrm{d}t$ , then integrate with the given parameter limits.

\begin{aligned} V&=\pi\int_0^1(t+1)^2(2t)\,\mathrm{d}t&&\text{radius squared times thickness}\\ &=\pi\int_0^1(2t^3+4t^2+2t)\,\mathrm{d}t&&\text{expand}\\ &=\pi\left[\frac{t^4}{2}+\frac{4t^3}{3}+t^2\right]_0^1&&\text{integrate each term}\\ &=\frac{17\pi}{6}&&\text{apply the bounds} \end{aligned}

Why this step? The factor $2t$ measures horizontal thickness. Without it you would add the right circular areas using the wrong slice widths.

Answer. $V=\frac{17\pi}{6}$ cubic units.

Check. Eliminating the parameter gives $y=1+\sqrt{x}$ . Thus $\pi\int_0^1(1+2\sqrt{x}+x)\,\mathrm{d}x=\pi(1+\frac43+\frac12)=\frac{17\pi}{6}$ .

The second example changes only the equation for $x$ , from $t^2$ to $t^3$ . The same radius function now needs a different thickness factor.

Worked example 2: changing the horizontal coordinate

Question. The curve $x=t^3$ , $y=t+1$ , $0\le t\le1$ , is used in the same construction about the $x$ -axis. Find the exact volume.

Plan. Replace $\mathrm{d}x$ by $3t^2\,\mathrm{d}t$ , keeping $y=t+1$ .

\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}t}&=3t^2&&\text{differentiate }x\\ V&=\pi\int_0^1(t+1)^2(3t^2)\,\mathrm{d}t&&\text{parametric disc formula}\\ &=\pi\left[\frac{3t^5}{5}+\frac{3t^4}{2}+t^3\right]_0^1&&\text{expand and integrate}\\ &=\frac{31\pi}{10}&&\text{evaluate exactly} \end{aligned}

Why this step? The derivative belongs to the coordinate along the rotation axis, $x$ . Differentiating $y$ instead would measure vertical changes, which are not the slice thickness here.

Answer. $V=\frac{31\pi}{10}$ cubic units.

Check. The Cartesian radius is $1+x^{1/3}$ , so the volume divided by $\pi$ is $1+\frac32+\frac35=\frac{31}{10}$ .

There is also a useful invariance: tracing the same boundary at a different parameter speed cannot change its volume. In Worked example 1, put $t=u^2$ . Then $x=u^4$ , $y=u^2+1$ , and

\pi\int_0^1(u^2+1)^2(4u^3)\,\mathrm{d}u =\pi\left[\frac{u^8}{2}+\frac{4u^6}{3}+u^4\right]_0^1 =\frac{17\pi}{6}

The derivative compensates exactly for the change of pace. Changing only $x$ , as in Worked example 2, changes the actual curve; changing parameter consistently leaves the solid unchanged.

Watch out

Wrong: $V=\pi\int y^2\,\mathrm{d}t$ . Right: $V=\pi\int y^2\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t$ about the $x$ -axis. Replace the differential as well as the radius.

The next curve changes the vertical offset; its horizontal thickness factor stays the same.

Rotation about the y-axis and decreasing coordinates

About the $y$ -axis, the radius is $x$ and the slice thickness comes from $y$ , so the required derivative changes to $\frac{\mathrm{d}y}{\mathrm{d}t}$ .

The physical integral runs from the lower height to the higher height. The corresponding parameter values might run backwards. Draw a small endpoint table before deciding their order.

Parametric volume about the y-axis

For a full rotation of a region between the curve and the $y$ -axis,

V=\pi\int_{t_{\mathrm{bottom}}}^{t_{\mathrm{top}}}x(t)^2\frac{\mathrm{d}y}{\mathrm{d}t}\,\mathrm{d}t

Use the parameter value at the lower height first and the value at the upper height second. When $y$ decreases with increasing $t$ , these bounds run from a larger $t$ to a smaller $t$ . For a monotonic section, an equivalent form is $\pi\int_{\alpha}^{\beta}x(t)^2|y'(t)|\,\mathrm{d}t$ , with $\alpha<\beta$ . Not in the formula booklet: learn it.

The absolute-value version is a consequence of correct orientation, not a repair for an unexplained negative answer. If a curve retraces a boundary, integrating with absolute values over the whole trace can still count the same volume twice.

Worked example 3: a decreasing height

Question. The curve $x=2\sin t$ , $y=3\cos t$ , $0\le t\le\frac\pi2$ , bounds a first-quadrant region with the coordinate axes. Find its exact volume of revolution about the $y$ -axis.

Plan. The bottom height $0$ occurs at $t=\frac\pi2$ and the top height $3$ at $t=0$ ; use those oriented bounds with $\frac{\mathrm{d}y}{\mathrm{d}t}=-3\sin t$ .

\begin{aligned} V&=\pi\int_{\pi/2}^{0}(2\sin t)^2(-3\sin t)\,\mathrm{d}t&&\text{bottom to top}\\ &=12\pi\int_0^{\pi/2}\sin^3t\,\mathrm{d}t&&\text{reverse limits and sign}\\ &=12\pi\left[-\cos t+\frac{\cos^3t}{3}\right]_0^{\pi/2}&&\sin^3t=\sin t(1-\cos^2t)\\ &=12\pi\left(0-\left(-1+\frac13\right)\right)=8\pi&&\text{evaluate} \end{aligned}

Why this step? Increasing $t$ moves down the curve. Writing the lower height first therefore means writing the larger parameter value first.

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

Check. Eliminating the parameter gives $x^2=4(1-y^2/9)$ , so $\pi\int_0^3(4-\frac49y^2)\,\mathrm{d}y=8\pi$ .

The second example changes only the height equation to reflect the same profile vertically. Heights now increase with the parameter, so the natural parameter order works directly.

Worked example 4: an increasing height

Question. Find the volume about the $y$ -axis for the region between $x=2\sin t$ , $y=3(1-\cos t)$ , $0\le t\le\frac\pi2$ , the $y$ -axis and the line $y=3$ .

Plan. Heights increase from $0$ to $3$ , so integrate from $t=0$ to $t=\frac\pi2$ , using $\frac{\mathrm{d}y}{\mathrm{d}t}=3\sin t$ .

\begin{aligned} V&=\pi\int_0^{\pi/2}(2\sin t)^2(3\sin t)\,\mathrm{d}t&&\text{increasing heights}\\ &=12\pi\int_0^{\pi/2}\sin^3t\,\mathrm{d}t&&\text{simplify}\\ &=12\pi\left[-\cos t+\frac{\cos^3t}{3}\right]_0^{\pi/2}&&\text{same elementary integral}\\ &=8\pi&&\text{apply the bounds} \end{aligned}

Why this step? The new height is $3$ minus the old height, so its derivative has the opposite sign. The endpoint order changes with it.

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

Check. Reflecting the solid from Worked example 3 in the horizontal plane $y=\frac32$ changes its orientation but preserves every slice area and hence its volume.

Endpoint arrows make the orientation easy to see before any integration.

[DIAGRAM: asset_name: Parameter direction and physical height; asset_slug: 9fm0-cp33-oriented-heights; caption: Increasing the parameter can move either down or up the solid; description: Two equal-scale panels show x equals two sin t with y equals three cos t on the left and y equals three times one minus cos t on the right. The left curve goes from (0, 3) at t equals zero to (2, 0) at t equals pi over two; the right goes from (0, 0) to (2, 3). Arrows along each curve show increasing t, with shaded regions between the curve and the vertical rotation axis.]
Increasing the parameter can move either down or up the solid

Watch out

Wrong: integrate from $0$ to $\frac\pi2$ with $y'=-3\sin t$ , then present a negative volume. Right: use bottom-to-top parameter bounds, or explicitly integrate the positive slice thickness on a monotonic interval.

The following profile has different dimensions. Keep the endpoint reasoning rather than copying a memorised sign.

Joining a curved body to a cylinder

A model may describe only part of a solid parametrically; calculate each non-overlapping component and combine their volumes.

Consider a solid ornament with a curved lower body and a cylindrical neck. The body is formed by revolving the region between its profile and the $y$ -axis. A separate cylinder sits above it, meeting along one circular face. Adding their volumes counts that shared face as no volume, so nothing is duplicated.

For a profile involving $\sin2t$ , expand the squared radius and use $\sin2t=2\sin t\cos t$ . The factors $\sin t$ or $\cos t$ from the thickness derivative often make the resulting integrals manageable. These are the same substitution and trigonometric methods you know from A-level Mathematics.

Worked example 5: a body and neck model

Question. An ornament's body is modelled by revolving $x=1+\sin2t$ , $y=3\cos t$ , $0\le t\le\frac\pi2$ , and the region between this profile and the $y$ -axis, through $2\pi$ about that axis. A cylinder of radius $1$ and height $2$ is attached above the body. All lengths are in centimetres. Find the exact total volume.

Plan. Use positive vertical thickness $3\sin t\,\mathrm{d}t$ from bottom to top, expand the body's squared radius, then add the cylinder.

\begin{aligned} V_{\mathrm{body}}&=3\pi\int_0^{\pi/2}(1+\sin2t)^2\sin t\,\mathrm{d}t&&\text{reverse the descending trace}\\ &=3\pi\int_0^{\pi/2}\left(\sin t+4\sin^2t\cos t+4\sin^3t\cos^2t\right)\,\mathrm{d}t&&\text{expand and use double angle} \end{aligned}

The three elementary integrals are

\begin{aligned} \int_0^{\pi/2}\sin t\,\mathrm{d}t&=1&&[-\cos t]_0^{\pi/2}\\ \int_0^{\pi/2}\sin^2t\cos t\,\mathrm{d}t&=\frac13&&\left[\frac{\sin^3t}{3}\right]_0^{\pi/2}\\ \int_0^{\pi/2}\sin^3t\cos^2t\,\mathrm{d}t&=\frac{2}{15}&&\left[-\frac{\cos^3t}{3}+\frac{\cos^5t}{5}\right]_0^{\pi/2} \end{aligned}

The last antiderivative follows by writing $\sin^3t=\sin t(1-\cos^2t)$ . Hence

V_{\mathrm{body}}=3\pi\left(1+\frac43+\frac8{15}\right)=\frac{43\pi}{5},\qquad V_{\mathrm{neck}}=\pi(1^2)(2)=2\pi

Why this step? The derivative supplies a $\sin t$ factor. Using the double-angle identity exposes exactly the products whose antiderivatives are short powers of sine or cosine.

Answer. $V_{\mathrm{total}}=\frac{53\pi}{5}\,\mathrm{cm^3}$ .

Check. The body's radius lies between $1$ and $2$ , so its volume lies between $3\pi$ and $12\pi$ . The computed $\frac{43\pi}{5}$ lies within those bounds. The cylinder begins at height $3$ , where the body radius is $1$ , so their circular faces fit.

The next example changes only the amplitude of the body's bulge, from $1$ to $2$ . The neck, height and endpoint radii stay the same.

Worked example 6: a wider bulge

Question. An ornament has the same neck as in Worked example 5, but its body profile is $x=1+2\sin2t$ , $y=3\cos t$ , $0\le t\le\frac\pi2$ . Find the exact total volume, in $\mathrm{cm^3}$ .

Plan. Square the new radius before expanding; reuse the same elementary antiderivatives and add the neck volume.

\begin{aligned} V_{\mathrm{body}}&=3\pi\int_0^{\pi/2}(1+2\sin2t)^2\sin t\,\mathrm{d}t&&\text{body slices}\\ &=3\pi\int_0^{\pi/2}\left(\sin t+8\sin^2t\cos t+16\sin^3t\cos^2t\right)\,\mathrm{d}t&&\text{expand carefully}\\ &=3\pi\left[-\cos t+\frac83\sin^3t-\frac{16}{3}\cos^3t+\frac{16}{5}\cos^5t\right]_0^{\pi/2}&&\text{integrate each term}\\ &=3\pi\left(1+\frac83+\frac{32}{15}\right)=\frac{87\pi}{5}&&\text{apply both bounds} \end{aligned}

Why this step? Doubling the bulge does not double the volume: the radius is squared, producing both a doubled cross-term and a quadrupled squared term.

Answer. $V_{\mathrm{total}}=\frac{87\pi}{5}+2\pi=\frac{97\pi}{5}\,\mathrm{cm^3}$ .

Check. This is larger than $\frac{53\pi}{5}$ , as every interior body slice is wider. The body radius is at most $3$ , so its volume is below the enclosing cylinder's $27\pi$ .

The model assumes circular cross-sections perpendicular to the axis and the stated smooth profile. If observations show an asymmetric body or different measured dimensions, those are relevant limitations. A hollow object's external volume would also differ from its material volume unless the cavity were subtracted.

[DIAGRAM: asset_name: Curved body and cylindrical neck; asset_slug: 9fm0-cp33-body-neck; caption: Calculate the curved body and cylindrical neck separately, then add them; description: A central vertical cross-section shows the profile x equals one plus sin two t, y equals three cos t, and its reflection in the y-axis. The body occupies heights zero to three, with radius one at both endpoints and a wider middle bulge. A cylinder of radius one extends from height three to five. Horizontal dotted lines distinguish the join at height three and the top at five.]
Calculate the curved body and cylindrical neck separately, then add them

Watch out

Wrong: expand $(1+2\sin2t)^2$ as $1+4\sin^22t$ . This omits the cross-term, so every interior slice has the wrong area. Right: $(1+2\sin2t)^2=1+4\sin2t+4\sin^22t$ .

Keep that expansion visible before integrating, so each contribution to the body volume can be followed.

Exam tip

When Pearson requests algebraic integration, show how the squared radius becomes an integrable expression and show the substitution of limits. A calculator's volume output does not demonstrate those method steps.

The opening integral missed $2t$ , the horizontal-thickness factor. Correcting it gives $\frac{17\pi}{6}$ ; any consistent reparameterisation gives that same physical volume.

Quick Recap

About the $x$ -axis use $\pi\int y^2x'(t)\,\mathrm{d}t$ ; about the $y$ -axis use $\pi\int x^2y'(t)\,\mathrm{d}t$ , with correctly oriented bounds.
Convert the radius, differential and limits together. The derivative measures the physical slice thickness.
Match parameter bounds to increasing axial position; split non-monotonic or repeated traces before calculating geometric volume.
Expand the whole squared radius, choose useful trigonometric identities and combine non-overlapping component volumes.

The opening curve needs $\pi\int_0^1(t+1)^2(2t)\,\mathrm{d}t=\frac{17\pi}{6}$ : the derivative makes the answer independent of parameter speed.

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CP 5.1 - Volumes of revolution from parametric equations

Changing the slice thickness into a parameter

Rotation about the y-axis and decreasing coordinates

Joining a curved body to a cylinder

Quick Recap

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