Further Maths/Lessons/Volumes of Revolution (Year 2)/Modelling
Core Pure · Volumes of Revolution · Year 2

Modelling with Volumes of Revolution

Put the whole toolkit to work modelling real rotationally-symmetric objects — vases, bottles, bowls, and hollow containers with a wall thickness.

8 min read 2 worked examples

Modelling with volumes of revolution

Method — real-world solids

Volumes of revolution model real objects with rotational symmetry — vases, bottles, bowls, containers. In modelling questions, the curve (Cartesian or parametric) represents the object's outline in cross-section, and you're often asked to find a volume of liquid, material used, or capacity, sometimes requiring the difference of two volumes (a "washer" style calculation) to model a hollow object with a wall thickness.

Example 1 · A modelling problem
A garden ornament is modelled by rotating the region bounded by $y=\sqrt{x}$, the $x$-axis and the lines $x=0$ and $x=4$ (all in cm) through $2\pi$ radians about the $x$-axis. The ornament is hollow, with a wall of negligible thickness, and is filled with water to a depth corresponding to $x=1$. Find the exact volume of water, in $\text{cm}^3$.

The water fills the solid of revolution between $x=0$ and $x=1$: $V=\pi\displaystyle\int_0^1y^2\,dx=\pi\int_0^1x\,dx=\pi\left[\frac{x^2}2\right]_0^1=\frac\pi2\text{ cm}^3$.

Example 2 · A washer-style modelling problem
A bead is modelled as a solid cylinder of radius $4$ cm and length $2$ cm, with a hole drilled through its axis. In cross-section, the hole's boundary is $y=x^2$ (in cm) for $0\leq x\leq2$. Find the exact volume of material in the bead, in $\text{cm}^3$.

The bead's material is the region between the outer radius $R=4$ and the inner (hole) radius $r=x^2$, rotated about the $x$-axis: $V=\pi\displaystyle\int_0^2\left(4^2-(x^2)^2\right)dx=\pi\int_0^2(16-x^4)\,dx$.

$=\pi\left[16x-\dfrac{x^5}5\right]_0^2=\pi\left(32-\dfrac{32}5\right)=\pi\times\dfrac{128}5=\dfrac{128\pi}5\text{ cm}^3$.

Exam-style questions

Question 1 6 marks

A vase is modelled by rotating the region bounded by $y=\dfrac13x^2+2$, the $x$-axis, and the lines $x=0$ and $x=3$ (all in cm) through $2\pi$ radians about the $x$-axis. Find the exact volume of material needed to make the vase (assume it is solid), in $\text{cm}^3$.

$y^2=\left(\dfrac13x^2+2\right)^2=\dfrac19x^4+\dfrac43x^2+4$. (M1, A1)

$V=\pi\displaystyle\int_0^3\left(\dfrac19x^4+\dfrac43x^2+4\right)dx=\pi\left[\dfrac1{45}x^5+\dfrac49x^3+4x\right]_0^3$. (M1, A1)

$=\pi\left(\dfrac{243}{45}+12+12\right)=\pi\left(\dfrac{27}5+24\right)=\dfrac{147\pi}5$. (A1, A1)

Question 2 7 marks

A napkin ring is modelled as a solid cylinder of radius $5$ cm and length $2$ cm, with a hole drilled through its axis. In cross-section, the hole's boundary is $y=x^2+1$ (in cm) for $0\leq x\leq2$. Find the exact volume of material in the napkin ring, in $\text{cm}^3$.

$V=\pi\displaystyle\int_0^2\left(5^2-(x^2+1)^2\right)dx=\pi\int_0^2\left(25-(x^4+2x^2+1)\right)dx$. (M1, A1)

$=\pi\displaystyle\int_0^2(24-2x^2-x^4)\,dx=\pi\left[24x-\dfrac23x^3-\dfrac{x^5}5\right]_0^2$. (M1, A1)

$=\pi\left(48-\dfrac{16}3-\dfrac{32}5\right)=\pi\times\dfrac{720-80-96}{15}=\dfrac{544\pi}{15}$. (A1, A1, A1)

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