Find volumes of revolution generated by rotating a region about the x-axis or y-axis, including the volume between two curves

What this dot point is asking

SEAB wants you to find the volume of a solid of revolution formed by rotating a plane region about the $x$-axis or the $y$-axis, set up the correct disc integral, and handle the volume of a region between two curves.

The answer

The disc method about the x-axis

Rotating the region under $y = \mathrm{f}(x)$ between $x = a$ and $x = b$ a full turn about the $x$-axis sweeps out thin discs of radius $y$ and thickness $dx$. Summing $\pi y^2\,dx$ gives

Rotation about the y-axis

Rotating about the $y$-axis instead, the discs have radius $x$ and thickness $dy$, so

with the limits in $y$ and $x^2$ expressed in terms of $y$.

Volume between two curves

When the region lies between two curves $y = \mathrm{f}(x)$ (outer) and $y = \mathrm{g}(x)$ (inner), rotating about the $x$-axis gives a washer (a disc with a hole):

Subtract the squares (not the square of the difference).

Setting it up

1. Sketch the region and the axis of rotation.
2. Decide whether to integrate in $x$ (about the $x$-axis) or $y$ (about the $y$-axis).
3. Square the relevant radius and integrate between the correct limits.

Finding the limits from the geometry

The trickiest setup step is often the limits, which come from where the region starts and ends, not from numbers handed to you. For rotation about the $x$-axis, the limits are the $x$-values bounding the region; for a region between two curves, find them by solving the curves' intersection. To rotate the region between $y = x$ and $y = x^2$, set $x = x^2$ to get $x = 0$ and $x = 1$, which become the integration limits. When rotating about the $y$-axis, the limits are $y$-values instead, so read or compute the region's lowest and highest $y$. Deriving the limits from intersections and the region's extent is the planning step that an H2 answer is marked on.

Choosing the axis and variable consistently

A frequent source of error is mixing the axis of rotation with the variable of integration. The rule is simple but must be applied consistently: rotation about the $x$-axis uses radius $y$ and integrates with respect to $x$ ($dx$, $x$-limits); rotation about the $y$-axis uses radius $x$ and integrates with respect to $y$ ($dy$, $y$-limits). So before integrating, rewrite the curve to express the squared radius in the correct variable, for instance turning $y = x^2$ into $x^2 = y$ when rotating about the $y$-axis. Keeping the radius, the differential, and the limits all in the same variable is what makes the disc integral come out right.

Examples in context

Example 1. Volume of a sphere. Rotating the semicircle $y = \sqrt{r^2 - x^2}$ about the $x$-axis from $-r$ to $r$ gives $\pi\int_{-r}^{r}(r^2 - x^2)\,dx = \dfrac{4}{3}\pi r^3$, deriving the sphere volume formula from integration.

Example 2. Designing a vase. A vase profile rotated about its central axis has a volume given by $\pi\int x^2\,dy$, which a manufacturer uses to compute material and capacity directly from the profile curve.

Try this

Q1. The region under $y = x$ from $0$ to $2$ is rotated about the $x$-axis. Find the volume. [3 marks]

• Cue. $\pi\int_0^2 x^2\,dx = \pi\left[\frac{x^3}{3}\right]_0^2 = \frac{8\pi}{3}$.

Q2. State the formula for the volume when rotating about the $y$-axis. [1 mark]

• Cue. $V = \pi\int x^2\,dy$.

Q3. When rotating a region between two curves about the $x$-axis, what is integrated? [1 mark]

• Cue. $\pi(y_{\text{outer}}^2 - y_{\text{inner}}^2)$.