# Volume of Solids of Revolution

> Volume of solids of revolution in JEE Mathematics: disc, washer and shell methods, revolution about off-axis lines and Pappus theorem for the torus.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/volume-solid-revolution
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Volume of Solids of Revolution", PrepElephant, https://prepelephant.com/topics/jee/mathematics/volume-solid-revolution

## Direct answer

Rotating a plane region about a line sweeps a solid whose volume calculus computes by slicing. About the x-axis, the disc method gives V = π ∫ y^2 dx, each slice being a disc of radius y. If the region has a hole (rotation about an axis with the region offset), the washer method subtracts: V = π ∫ (R^2 - r^2) dx with R and r the outer and inner radii. About the y-axis, the shell method wraps cylinders: V = 2π ∫ x y dx, each shell contributing circumference 2πx times height y times thickness dx. Off-axis rotation replaces the radius by the distance from the curve to the axis: revolving y = f(x) about y = c uses π ∫ (f(x) - c)^2 dx. Pappus's theorem compresses special cases to arithmetic: V = (area) × (2π × distance travelled by the centroid), giving the torus volume 2π^2 R r^2 instantly.

## What you must remember

- **Disc method:** about the x-axis, V = π ∫[a to b] y^2 dx; about the y-axis for x = g(y), V = π ∫ x^2 dy.
- **Washer method:** V = π ∫ (R^2 - r^2) dx when the swept region has a hole — outer radius minus inner radius, both squared first.
- **Shell method:** about the y-axis, V = 2π ∫ x y dx; ideal when solving for x is ugly or the height is naturally a function of x.
- **Off-axis rotation:** about the line y = c, radius = |f(x) - c|, so V = π ∫ (f(x) - c)^2 dx; about x = c for shells, V = 2π ∫ |x - c| y dx.
- **Sphere check:** y = √(a^2 - x^2) revolved about the x-axis gives π ∫[-a to a] (a^2 - x^2) dx = 4πa^3/3 — the built-in sanity test for any method.
- **Pappus (Guldinus) theorem:** V = A × 2π d̄, where d̄ is the centroid's distance from the axis; the torus from a circle of radius r centred (R, 0) has V = πr^2 × 2πR = 2π^2 R r^2.
- **Limits follow the variable of integration:** dx for x-axis discs and y-axis shells, dy for y-axis discs — a mismatch of limits and variable is the commonest computational wreck.

## Generating a sphere and a torus

The sphere first, because it calibrates everything: rotate y = √(a^2 - x^2) about the x-axis. V = π ∫[-a to a] (a^2 - x^2) dx = π [a^2 x - x^3/3] from -a to a = π [(a^3 - a^3/3) - (-a^3 + a^3/3)] = π (4a^3/3) = 4πa^3/3, the textbook value confirming the machinery. Now the torus by two routes. By washers about the y-axis, the circle (x - R)^2 + y^2 = r^2 (R > r) gives outer radius R + √(r^2 - (x - R)^2) and inner radius R - √(r^2 - (x - R)^2); squaring and subtracting leaves 4R√(r^2 - (x - R)^2), so V = π ∫ 4R√(r^2 - (x - R)^2) dx over [R - r, R + r], which is 4πR times a half-circle area πr^2/2 — that is, 2π^2 R r^2 after the standard substitution. By Pappus the same answer is one line: area πr^2, centroid at the circle's centre travelling 2πR, so V = 2π^2 R r^2. The two routes' agreement is the theorem's proof in miniature and the reason examiners accept the shortcut when the centroid is known.

## How the exam frames it

JEE Main tests the disc and shell formulae on elementary regions — y = x^2 about the x-axis (πa^5/5 over [0, a]), y = sin x, a rectangle about one side — mostly as numerical-value questions. Advanced prefers off-axis rotation (y = x^2 about y = 4, or the region between two curves about the y-axis), the washer construction with genuinely two curves, and Pappus applied to triangles and semicircles whose centroids are known from coordinate geometry. The trademark slips: using the shell formula with y^2 instead of y (mixing the two templates); revolving about y = c but leaving the radius as f(x) rather than f(x) - c; forgetting that washers subtract squared radii, not radii themselves — π(R - r)^2 is the single most marked error in this chapter. Also note the dimensional check habit: a volume answer must carry length cubed, which instantly exposes a forgotten π or a wrong power. Solids of revolution belong to the application-of-integrals unit in both syllabi.

## Frequently asked questions

### What is the disc method formula for volume of revolution?

About the x-axis, V = π ∫ y^2 dx, each slice of thickness dx being a disc of radius y and volume πy^2 dx.

### When is the washer method needed instead of the disc method?

When the region does not touch the axis, leaving a hole: V = π ∫ (R^2 - r^2) dx with outer radius R and inner radius r.

### How does the shell method compute volumes about the y-axis?

By cylindrical shells, V = 2π ∫ x y dx — circumference 2πx times height y times thickness dx — best when y is easier to express as a function of x.

### How does rotation about a line y = c change the formula?

The radius becomes the distance |f(x) - c|, so V = π ∫ (f(x) - c)^2 dx; failing to shift the radius is the chapter's most frequent error.

### What does Pappus's theorem say about volumes of revolution?

V equals the generating area times the distance its centroid travels: V = 2π d̄ A, giving the torus 2π^2 R r^2 without any integration.