The use of definite integrals to find the volume of a solid of revolution generated by rotating a region about the $x$-axis or $y$-axis, using the disc and washer (annulus) methods, and the setting up of the appropriate integral

What this dot point is asking

VCAA wants you to compute the volume of a solid formed when a region is rotated about the $x$-axis or $y$-axis, by setting up and evaluating a definite integral. You need the disc method for a region against the axis and the washer (annulus) method when there is a gap, and you must choose the integration variable to match the axis of rotation.

The disc method

Slice the solid into thin discs perpendicular to the axis of rotation. Rotating the region under $y = f(x)$ about the $x$-axis, a slice at position $x$ is a disc of radius $f(x)$ and thickness $dx$, with volume $\pi [f(x)]^2\,dx$. Summing over the interval gives

The radius is the distance from the axis to the curve, here simply $y = f(x)$. The square is essential: the disc area is $\pi r^2$.

The washer (annulus) method

When the region lies between two curves $y_{\text{outer}}$ and $y_{\text{inner}}$ (both measured from the axis), each slice is a washer: a disc with a hole. Its area is the outer disc minus the inner disc, so

Subtract the squares of the radii, not the radii themselves; $[y_o]^2 - [y_i]^2 \ne (y_o - y_i)^2$.

Rotation about the $y$-axis

To rotate about the $y$-axis, slice horizontally. Each disc has radius $x$ (the distance from the $y$-axis to the curve) and thickness $dy$, so you express $x$ as a function of $y$ and integrate with $y$-limits:

The rule is: match the integration variable to the axis. Rotation about the $x$-axis integrates in $x$ with radius $y$; rotation about the $y$-axis integrates in $y$ with radius $x$.

Examples in context

Example 1. Rotating $y = r$ (a horizontal line) for $0 \le x \le h$ about the $x$-axis gives a cylinder, $V = \int_0^h \pi r^2\,dx = \pi r^2 h$.

Example 2. Rotating the region between $y = x$ and $y = x^2$ on $[0, 1]$ about the $x$-axis uses the washer $\pi(x^2 - x^4)$, since $y = x$ is outer.

Try this

Q1. Write the disc-method volume for rotating $y = f(x)$, $a \le x \le b$, about the $x$-axis. [1 mark]

• Cue. $\int_a^b \pi [f(x)]^2\,dx$.

Q2. Find the volume when $y = \sqrt{x}$, $0 \le x \le 4$, is rotated about the $x$-axis. [3 marks]

• Cue. $\int_0^4 \pi x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi$.

Q3. State the washer integrand for outer radius $R(x)$ and inner radius $r(x)$. [1 mark]

• Cue. $\pi(R^2 - r^2)$.