How Do You Find the Volume of a Solid of Revolution About the X-Axis or Y-Axis?

Use the disk method or washer method: square the radius, multiply by $\pi$, and integrate along the axis perpendicular to each cross-section. About the x-axis, integrate with respect to $x$; about the y-axis, integrate with respect to $y$.

The Method

A solid of revolution is formed by rotating a region around an axis. Each cross-section perpendicular to that axis is a circular disk or a washer with a hole.

The variable matters because the radius must be written in terms of the integration variable. If a curve is given as $y=f(x)$ but the solid rotates about the y-axis, rearrange it as $x=g(y)$ and change the limits to y-values. For washers, subtract cross-sectional areas, not radii: $\pi R^2-\pi r^2=\pi(R^2-r^2)$.

The key steps are:

1. Identify the axis of rotation.
2. Choose slices perpendicular to that axis.
3. Express every radius using the variable of integration.
4. Square the radii before subtracting and integrating.

Worked Example

The region bounded by $y=\sqrt{x}$, the x-axis and $x=4$ is rotated about the x-axis. The radius is $$, so

$$V=\pi\int_{0}^{4}(\sqrt{x})^2\,\mathrm{d}x =\pi\int_{0}^{4}x\,\mathrm{d}x =8\pi.$$

Therefore, the exact volume is $8\pi$ cubic units.

For a y-axis example, rotating the region between $x=y^2$, the y-axis, $y=0$, and $y=2$ gives

$$V=\pi\int_{0}^{2}(y^2)^2\,\mathrm{d}y=\pi\int_{0}^{2}y^4\,\mathrm{d}y=\frac{32\pi}{5}.$$

This confirms that the radius is a horizontal distance, $x=y^2$, and the bounds are y-values.

A common misconception is to integrate $f(x)$ rather than $[f(x)]^2$. The cross-sectional area is $\pi r^2$, so the radius must be squared.

Exam Technique

This is AHL 5.12, so it is HL-only content; a short-response Paper 1 style is assumed, although it can appear on any GDC-active HL paper. Write the definite integral before giving the GDC value, include cubic units, and give an exact answer or round the final value to three significant figures unless instructed otherwise.