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WASpecialist MathematicsSyllabus dot point

How does integrating the area of circular cross-sections give the volume of a solid of revolution?

Find the volume of a solid generated by rotating a region about the x-axis or y-axis using the disc method

WACE Specialist Unit 4 volumes of revolution: the disc method about the x-axis and y-axis, integrating pi times radius squared, rotating between two curves with the washer idea, and setting up the correct limits, with a worked example.

Generated by Claude Opus 4.76 min answer

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  1. What this dot point is asking
  2. The disc method
  3. Rotating about the y-axis
  4. Rotating the region between two curves
  5. Setting limits

What this dot point is asking

SCSA wants you to set up and evaluate volumes of revolution by the disc method, choosing the axis, the variable of integration and the limits correctly.

The disc method

When a region under y=f(x)y = f(x) is rotated about the xx-axis, each thin slice becomes a disc of radius f(x)f(x) and thickness dxdx, with area π[f(x)]2\pi[f(x)]^2. Summing gives

The radius is the distance from the axis to the curve, which here is just f(x)f(x). The squared radius is essential: forgetting to square is the classic error.

Rotating about the y-axis

Rotating about the yy-axis makes the radius the horizontal distance xx, so you express xx as a function of yy and integrate over yy:

V=π∫cd[x(y)]2 dy.V = \pi\int_c^d [x(y)]^2\,dy.

The limits cc and dd are now yy-values. Match the variable of integration to the axis of rotation.

Rotating the region between two curves

Setting limits

The limits are where the region begins and ends along the axis of rotation, often the intersection points of the bounding curves with each other or with the axis. A sketch confirms them.