Further Maths/Lessons/Volumes of Revolution/Combining & Modelling Volumes
Core Pure · Volumes of Revolution · Year 1

Combining & Modelling Volumes

Subtract one solid of revolution from another to hollow out a shape, then use the same technique the other way round — to model the capacity of a real container.

13 min read 3 worked examples

Adding and subtracting volumes

When a region is enclosed between two curves and rotated about the $x$-axis, the resulting solid has a cavity running through it — like a napkin ring rather than a solid disc. Each thin slice is now a washer: a disc of radius $y_{\text{outer}}$ with a smaller disc of radius $y_{\text{inner}}$ removed from its centre.

Theorem 5.1 — volume between two curves

If $y_1\geq y_2\geq0$ on $[a,b]$, rotating the region between $y=y_1$ and $y=y_2$ about the $x$-axis gives $$V = \pi\int_a^b \left(y_1^2-y_2^2\right)\mathrm dx$$

Diagram showing two curves, an outer curve and an inner curve, with the region between them shaded, and the solid of revolution shown as a ring-shaped washer cross-section
Fig. 1 — the region between two curves rotated about the x-axis sweeps out a washer-shaped cross-section: an outer disc of radius y₁ with an inner disc of radius y₂ removed.
Example 1 · Volume between two parabolas
The curves $y=4-x^2$ and $y=x^2$ intersect where $x=\pm\sqrt2$. The region enclosed between them, for $-\sqrt2\leq x\leq\sqrt2$, is rotated $360°$ about the $x$-axis. Find the exact volume of the solid formed.

For $-\sqrt2\leq x\leq\sqrt2$, $4-x^2\geq x^2$, so the outer curve is $y_1=4-x^2$ and the inner is $y_2=x^2$.

$$y_1^2-y_2^2 = (4-x^2)^2-(x^2)^2 = 16-8x^2+x^4-x^4 = 16-8x^2$$

Since the integrand is even, the integral is twice its value from $0$ to $\sqrt2$:

$$V = 2\pi\int_0^{\sqrt2}(16-8x^2)\,\mathrm dx = 2\pi\left[16x-\frac{8x^3}{3}\right]_0^{\sqrt2} = 2\pi\left(16\sqrt2-\frac{16\sqrt2}{3}\right)=2\pi\cdot\frac{32\sqrt2}{3}=\frac{64\sqrt2\,\pi}{3}$$

Modelling with volumes of revolution

The same integral, run in reverse, is how a curve is used to model the shape of a real container — a vase, a bowl, a wine glass — and predict its capacity, or find an unknown parameter given a stated capacity.

Example 2 · Modelling a vase
A vase is modelled by rotating the curve $y=\dfrac12\sqrt{x}$ ($0\leq x\leq16$, units in cm) $360°$ about the $y$-axis. Find the capacity of the vase, in cm³, when filled to the brim.

At $x=16$: $y=\dfrac12\sqrt{16}=2$, so the vase has height $2$ cm. Rearranging $y=\dfrac12\sqrt x$ gives $x^2=4y^2$.

$$V = \pi\int_0^2 4y^2\,\mathrm dy = 4\pi\left[\frac{y^3}{3}\right]_0^2 = 4\pi\cdot\frac83 = \frac{32\pi}{3}\ \text{cm}^3$$
Example 3 · Finding an unknown parameter
A bowl is modelled by rotating $y=x^2$ ($0\leq x\leq3$, cm) about the $y$-axis. Find the volume of liquid in the bowl when it is filled to a depth of $h=9$ cm.

Since $y=x^2$, we have $x^2=y$ directly.

$$V = \pi\int_0^9 y\,\mathrm dy = \pi\left[\frac{y^2}{2}\right]_0^9 = \pi\cdot\frac{81}{2} = 40.5\pi\ \text{cm}^3$$
Exam technique — modelling questions

Always check the model's domain matches the physical object: the height of a container corresponds to a range of $y$ (or $x$) values, and "filled to a depth of $h$" means integrating up to $y=h$, not necessarily to the container's full height. Sketching the curve alongside a rough cross-section of the real object is the fastest way to avoid setting up the wrong limits.

Exam-style questions

Question 1 7 marks

The line $y=x+1$ and the curve $y=x^2-1$ intersect at $x=-1$ and $x=2$. The region enclosed between the line and the curve, for $-1\leq x\leq2$, is rotated $360°$ about the $x$-axis. Find the exact volume of the solid formed.
(Hint: use $y_1^2-y_2^2$ with $y_1=x+1$ as the outer curve.)

$(x+1)^2-(x^2-1)^2 = (x^2+2x+1)-(x^4-2x^2+1) = -x^4+3x^2+2x$ (M1, A1)

$V=\pi\displaystyle\int_{-1}^2(-x^4+3x^2+2x)\,dx = \pi\left[-\dfrac{x^5}{5}+x^3+x^2\right]_{-1}^2$ (M1, A1)

At $x=2$: $-\dfrac{32}{5}+8+4=\dfrac{28}{5}$. At $x=-1$: $\dfrac15-1+1=\dfrac15$. Difference: $\dfrac{28}{5}-\dfrac15=\dfrac{27}{5}$ (M1, A1)

$V=\dfrac{27\pi}{5}$ (A1)

Question 2 5 marks

A bowl is modelled by rotating the curve $y=\dfrac{x^2}{3}$ ($0\leq x\leq3$, cm) $360°$ about the $y$-axis. Find the capacity of the bowl, in cm³, when filled to the brim ($x=3$).

At $x=3$: $y=3$, so the bowl has height 3 cm. From $y=\dfrac{x^2}{3}$, $x^2=3y$ (M1)

$V = \pi\displaystyle\int_0^3 3y\,dy = 3\pi\left[\dfrac{y^2}{2}\right]_0^3 = 3\pi\cdot\dfrac92 = \dfrac{27\pi}{2}\ \text{cm}^3$ (M1, A1, A1, A1)

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Volumes of Revolution: x- and y-axis