Unit 8: Applications of Integration

Disc & Washer Methods

Finding the volume of a solid of revolution using the disc and washer methods.

Advanced25 min lesson4 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

Area Between Curves

Volume as an Integral of Cross-Sections

A flat area was built by integrating a height function. Before reading on: could a three-dimensional volume be built the same way — by integrating a cross-sectional area function instead of a height?

Definition — Volume by Cross-Sections

If a solid's cross-sections perpendicular to the x-axis all have a known area A(x), the solid's volume is ∫ₐᵇ A(x) dx — the same Riemann-sum logic as area under a curve, just with each thin slice's volume approximated as A(x)·Δx (a thin prism) instead of a rectangle's area. Revolving a region around an axis is one specific way to generate cross-sections: the disc method applies when the cross-sections are full circles, A(x) = π[R(x)]², and the washer method applies when there's a hole, A(x) = π([R(x)]² − [r(x)]²), for outer radius R and inner radius r.

V = \int_a^b A(x)\,dx

Worked Example — The Disc Method

Find the volume when the region under y = √x from x = 0 to x = 4 is revolved around the x-axis. Each cross-section is a circle of radius R(x) = √x, so A(x) = π(√x)² = πx. V = ∫₀⁴ πx dx = π[x²/2]₀⁴ = π(16/2) = 8π.

Worked Example — The Washer Method

The region between y = x and y = x² (intersecting where x = x², i.e. x = 0 and x = 1) is revolved around the x-axis. On (0, 1), y = x is farther from the axis (outer radius R = x) and y = x² is closer (inner radius r = x²) — check at x = 0.5: the line gives 0.5, the parabola gives 0.25, confirming the line is outer. V = ∫₀¹ π[(x)² − (x²)²] dx = π∫₀¹ (x² − x⁴) dx = π[x³/3 − x⁵/5]₀¹ = π(1/3 − 1/5) = π(5/15 − 3/15) = π(2/15) = 2π/15.

Worked Example — A Non-Circular Cross-Section

Using the same base region as the first example — under y = √x from x = 0 to x = 4 — suppose the solid instead has square cross-sections perpendicular to the x-axis, each with side length equal to the region's height, √x. Then A(x) = (√x)² = x (no π, since a square's area is side², not π·radius²). V = ∫₀⁴ x dx = [x²/2]₀⁴ = 8. Compare to the disc method's 8π for the identical base curve — the cross-section's shape genuinely changes the volume, even with the same boundary curve generating it.

Integral Visualizer

f(x) =

x^2

Integration bounds

a — lower bound0

b — upper bound2

Animate b (the play button above) to watch the shaded area — and the accumulation curve below — grow.

Approximation method

Method

Subdivisions (n)10

Show approximationShow exact area

Trace mode

f(x), with the region between a and b shaded

Show accumulation function A(x)

Accumulation function A(x) = ∫ₐˣ f(t) dt — linked to the graph above; panning or zooming either moves both

What the Fundamental Theorem of Calculus tells us

A(x) = ∫ from a to x of f(t) dt is the accumulation function. At x = b ≈ 2, A(b) ≈ 2.67 — the signed area under f from a ≈ 0 to b.
The Fundamental Theorem of Calculus says A'(x) = f(x): differentiating the accumulation function gives the original function back.
Confirmed numerically here: differentiating the accumulation curve gives A'(b) ≈ 4, matching f(b) ≈ 4.

Numerical approximation comparison

Every method's estimate of ∫ₐᵇ f(x) dx at n = 10 subdivisions, against a high-resolution reference — all should converge toward the same value.
Method	Estimate
Left Riemann sum	2.28
Right Riemann sum	3.08
Midpoint Riemann sum	2.66
Trapezoidal Rule	2.68
Simpson's Rule	2.66667
Exact (high-resolution reference)	2.66667

Analysis of f(x)

y-intercept: (0, 0)
x-intercepts: (0, 0)
Extrema: local min at (0, 0)
Inflection points: none found in view
Vertical asymptotes: none found in view
Horizontal asymptotes: none found
Domain: all real numbers in view
Range (estimated): approximately [0, 100]

Tip

Identify the cross-section's shape before writing any integral — a circle (disc), a ring (washer), or something else entirely (square, triangle, semicircle) — since the area formula A(x) depends entirely on that shape, not just on the boundary curves.

Common Mistakes

Using the disc method's formula, π[R(x)]², for a region that actually has a hole in the middle.
Whenever the revolved region doesn't touch the axis of revolution, the resulting solid has a hole — the washer method's π([R(x)]² − [r(x)]²) is needed, not the disc method, or the volume comes out too large.
Squaring the outer and inner radii together, computing [R(x) − r(x)]² instead of [R(x)]² − [r(x)]².
The washer's area is the outer circle's area minus the inner circle's area — two separate squarings subtracted, not the square of a single subtracted radius, which is an entirely different (and smaller) quantity.

Key Takeaways

Volume from known cross-sections is ∫ₐᵇA(x)dx — the same accumulation logic as area under a curve, generalized to any cross-sectional shape.
The disc method (A = πR²) and washer method (A = π(R² − r²)) are the special cases of this idea where revolving a region produces circular or ring-shaped cross-sections.
The cross-section's actual shape — not just the boundary curves — determines the area formula, as the square-cross-section example shows by contrast with the disc method on the identical base region.

Summary

This closes AP Calculus AB: limits and continuity, the derivative and its rules and applications, and the definite integral and its applications together form a complete first course in calculus — connecting graphical, numerical, algebraic, and contextual reasoning about how quantities change and accumulate. AP Calculus BC builds directly on this foundation, extending these same ideas further.

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