Modelling with Volumes of Revolution
Vases, bowls, beads, machine parts — anything with rotational symmetry can be captured by a curve and an axis, turning a real object into an integral.
Volumes of revolution as real-world models
Volumes of revolution are frequently used to model physical objects with rotational symmetry: vases, bowls, wine glasses, bottles, and machine parts. The strategy is always the same: identify the generating curve and axis of rotation from the problem description, set up the correct integral (disc or washer method), and interpret the result in context — including units.
At $x=4$, $y=4$. Rearranging: $x^2=4y$.
$V=\pi\displaystyle\int_0^4 4y\,dy=\pi\left[2y^2\right]_0^4=\pi(32)=32\pi\approx100.5\text{ cm}^3$ (3 s.f.).
$V=\pi\displaystyle\int_0^9(k\sqrt x)^2\,dx=\pi\int_0^9k^2x\,dx=\pi k^2\left[\dfrac{x^2}{2}\right]_0^9=\pi k^2\left(\dfrac{81}{2}\right)$.
Setting equal to $81\pi$: $\dfrac{81\pi k^2}{2}=81\pi \Rightarrow k^2=2 \Rightarrow k=\sqrt2$ (taking the positive root, since $k>0$).
Limits: solve $\sqrt{9-x^2}=1 \Rightarrow 9-x^2=1 \Rightarrow x^2=8 \Rightarrow x=\pm2\sqrt2$.
$V=\pi\displaystyle\int_{-2\sqrt2}^{2\sqrt2}\big[(9-x^2)-1\big]\,dx=\pi\int_{-2\sqrt2}^{2\sqrt2}(8-x^2)\,dx$.
By symmetry: $=2\pi\displaystyle\int_0^{2\sqrt2}(8-x^2)\,dx=2\pi\left[8x-\dfrac{x^3}{3}\right]_0^{2\sqrt2}$.
$8(2\sqrt2)=16\sqrt2$; $(2\sqrt2)^3=16\sqrt2$, so $\dfrac{(2\sqrt2)^3}{3}=\dfrac{16\sqrt2}{3}$.
$=2\pi\left(16\sqrt2-\dfrac{16\sqrt2}{3}\right)=2\pi\cdot\dfrac{32\sqrt2}{3}=\dfrac{64\sqrt2\,\pi}{3}$.
Modelling questions often ask you to criticise the model afterwards (e.g. "state one limitation of this model") — common valid answers include that real containers have walls of non-zero thickness, that the model assumes perfect rotational symmetry, or that the curve may not be an accurate fit to the true cross-sectional profile near the boundaries.
Forgetting to convert between diameter and radius is a frequent error in modelling contexts — if a problem states a container's diameter at some height, remember the curve $y=f(x)$ (or $x=g(y)$) describes the radius, so the given value must be halved before being used in the integral.
Exam-style questions
Question 1 5 marks
A vase is modelled by rotating $y=\ln(x+1)$, for $0\leq x\leq e^2-1$ (measurements in cm), about the $y$-axis. Find the exact capacity of the vase in cm³.
When $x=e^2-1$, $y=\ln(e^2)=2$. Rearranging: $x=e^y-1$, so $x^2=(e^y-1)^2=e^{2y}-2e^y+1$ (M1, A1)
$V=\pi\displaystyle\int_0^2(e^{2y}-2e^y+1)\,dy=\pi\left[\dfrac{e^{2y}}{2}-2e^y+y\right]_0^2$ (M1, A1)
At $y=2$: $\dfrac{e^4}{2}-2e^2+2$. At $y=0$: $\dfrac12-2+0=-\dfrac32$.
$V=\pi\left(\dfrac{e^4}{2}-2e^2+2+\dfrac32\right)=\pi\left(\dfrac{e^4}{2}-2e^2+\dfrac72\right)\text{ cm}^3$ (A1)
Question 2 6 marks
A component is formed by rotating the region under $y=k(4-x)$, for $0\leq x\leq4$, about the $x$-axis, where $k>0$. Given the volume is $\dfrac{256\pi}{3}$ cm³, find $k$.
$V=\pi\displaystyle\int_0^4k^2(4-x)^2\,dx=\pi k^2\left[-\dfrac{(4-x)^3}{3}\right]_0^4=\pi k^2\left(0-\left(-\dfrac{64}{3}\right)\right)=\dfrac{64\pi k^2}{3}$ (M1, A1, A1)
$\dfrac{64\pi k^2}{3}=\dfrac{256\pi}{3} \Rightarrow k^2=4 \Rightarrow k=2$ (taking $k>0$) (M1, A1, A1)
Question 3 7 marks
A ring-shaped bead is modelled by rotating the region between $y=\sqrt{16-x^2}$ and $y=2$ about the $x$-axis, for the valid range of $x$. Find the exact volume of the bead, and suggest one limitation of this model.
Limits: $\sqrt{16-x^2}=2 \Rightarrow 16-x^2=4 \Rightarrow x^2=12 \Rightarrow x=\pm2\sqrt3$ (M1, A1)
$V=\pi\displaystyle\int_{-2\sqrt3}^{2\sqrt3}\big[(16-x^2)-4\big]\,dx=2\pi\int_0^{2\sqrt3}(12-x^2)\,dx$ (M1, A1)
$=2\pi\left[12x-\dfrac{x^3}{3}\right]_0^{2\sqrt3}=2\pi\left(24\sqrt3-\dfrac{24\sqrt3}{3}\right)=2\pi\left(24\sqrt3-8\sqrt3\right)=32\sqrt3\,\pi$ (M1, A1)
Limitation: the model assumes the bead has a perfectly smooth rotationally symmetric surface with no drilled-hole thickness variation, which real manufactured beads may not have exactly. (B1)