What is the formula for the volume of revolution?

About the x-axis: V = π∫ y² dx between the x-limits. About the y-axis: V = π∫ x² dy between the y-limits. About a line y = k, replace y by (y − k).

Is the shell method needed for DSE M2?

No. The curriculum notes state that the shell method is not required; volumes are found with the disc method about a coordinate axis or a line parallel to one.

Why is it hard to picture a solid of revolution?

A flat sketch shows only the generating region. Seeing the solid in 3D and rotating it makes it obvious what each disc looks like and which radius goes into the formula.

Area and Volume of Revolution (DSE M2)

The area between two curves is ∫_a^b (upper − lower) dx. The volume of the solid formed by revolving y = f(x) about the x-axis is π∫_a^b y² dx (disc method); about the y-axis it is π∫_c^d x² dy.

Solve Area and Volume of Revolution questions with AI

The solid from Example 1: the region under y = √x from 0 to 4, revolved about the x-axis. Its volume is 8π. Try it yourself →

Area and Volume Formulas

Area under a curve∫_a^b f(x) dx (f(x) ≥ 0)

Area between two curves∫_a^b [f(x) − g(x)] dx, f ≥ g

Volume about the x-axisπ∫_a^b y² dx

Volume about the y-axisπ∫_c^d x² dy

About the line y = kπ∫_a^b (y − k)² dx

Region between two curves, about the x-axisπ∫_a^b (y₁² − y₂²) dx

Worked Examples — Area and Volume of Revolution

Example 1: Volume of a solid of revolution about the x-axis

The region bounded by y = √x, the x-axis and the line x = 4 is revolved about the x-axis. Find the volume of the solid generated.

V = π∫₀⁴ y² dx = π∫₀⁴ x dx
= π [x²/2]₀⁴
= π × 8 = 8π

Answer: 8π cubic units

Example 2: Area between a line and a parabola

Find the area of the region bounded by y = x² and y = x + 2.

Intersections: x² = x + 2 gives x = −1 or x = 2
The line is above the parabola between them, so area = ∫₋₁² (x + 2 − x²) dx
= [x²/2 + 2x − x³/3]₋₁² = 10/3 − (−7/6)
= 9/2

Answer: 9/2 square units

💡 Exam Traps & Secrets

⚠️ Square first, then subtractFor a region between two curves revolved about the x-axis, the volume is π∫(y₁² − y₂²) dx, not π∫(y₁ − y₂)² dx. The solid is a large disc with a smaller disc removed.

⚠️ About the y-axis, integrate with respect to yRewrite the curve as x in terms of y and use y-limits: π∫ x² dy. The shell method is not in the syllabus, so the disc method is the one examiners expect.

✏️ Self-Test: Quick Interactive Practice

How to study: Click any question below to instantly load it on the coordinate grid and check your steps.

(Disc method)Find the volume generated when the region under y = e^x from x = 0 to x = 1 is revolved about the x-axis.

(About the y-axis)Find the volume when the region bounded by y = x² and y = 4 is revolved about the y-axis.

(Area between curves)Find the area enclosed by y = x³ and y = x in the first quadrant.