Hollow Solids of Revolution | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Subtract the missing inner disc

If the region does not reach the rotation axis, each cross-section has a hole. Its shape is a washer: a larger circular disc with a smaller disc removed. Both radii are distances from the same rotation axis.

At x = 1, R² = 7 and r² = 1. The washer area is π(7 − 1), not π(R − r)².

If the outer radius is $R$ and the inner radius is $r$ , the cross-sectional area is $\pi(R^2-r^2)$ , with $R\ge r\ge0$ . Integrate this area along the axis.

Common mistake

The difference of squared radii is not the square of their difference. For radii 3 and 1, the washer area is

\pi(9-1)=8\pi

, not

\pi(3-1)^2=4\pi

A constant inner radius makes a cylinder

Worked example

Rotate the region between y = √(9 − 2x²), y = 1 and x = 0 about the x-axis.

The line and curve meet where $9-2x^2=1$ . On the right of the y-axis, this gives $x=2$ . The outer radius squared is $9-2x^2$ and the inner radius is 1.

\begin{gathered}V=\pi\int_0^2(8-2x^2)\,dx\\=\pi\left[8x-\frac{2x^3}{3}\right]_0^2\\=\frac{32\pi}{3}\end{gathered}

Equivalently, subtract the inner cylinder, volume $\pi(1)^2(2)=2\pi$ , from the full outer solid. If the inner radius changes, integrate its square too; it is not a cylinder.

Your turn9709/12 May/Jun 2020 Q8(a) [5 marks]

The region bounded by $y=6/x$ , $y=2$ and $x=1$ is rotated about the y-axis. The curve passes through $(1,6)$ and $(3,2)$ . Find the volume.

Show worked answer

Horizontal radii are $R=6/y$ and $r=1$ , with y-limits 2 and 6.

\begin{aligned}V&=\pi\int_2^6\left(\frac{36}{y^2}-1\right)\,dy\\&=\pi\left[-\frac{36}{y}-y\right]_2^6\\&=12\pi-4\pi=8\pi\end{aligned}