Disc Method: x- or y-Axis — Volume with Disc Method: Revolving Around the x- or y-Axis - Calculus 2

SECTION 1

Historical Context & Motivation

The problem of computing the volume of curved solids has occupied mathematicians for over two millennia. Ancient Greek geometers sought rigorous methods to determine the volume of spheres, cones, and paraboloids — shapes that defied the straightforward multiplication rules applicable to rectangular prisms. The breakthrough insight, refined across centuries, was that a curved solid could be approximated by stacking infinitely many thin circular slices, each resembling a disc. This idea, which we now formalize as the disc method, stands as one of the most elegant applications of the definite integral: it converts a geometric question about three-dimensional volume into a one-dimensional integration problem.

~250 BCE • Archimedes and the Method of Exhaustion

Archimedes computed the volume of a sphere by inscribing and circumscribing it with known solids, effectively summing thin cross-sectional slices — a precursor to the disc method.

1635 • Cavalieri's Principle

Bonaventura Cavalieri proposed that two solids with equal cross-sectional areas at every height have the same volume, providing a systematic framework for slicing methods.

1665–1676 • Newton and Leibniz Formalize Calculus

The development of integral calculus by Newton and Leibniz gave mathematicians the tools to express the sum of infinitely many disc areas as a definite integral, turning Archimedes' geometric intuition into an algebraic formula.

18th–19th c. • Rigorization and the Riemann Integral

Cauchy and Riemann formalized the definite integral, ensuring that the limiting process underlying the disc method — summing cylindrical volumes as thickness tends to zero — rests on solid mathematical foundations.

The central question the disc method answers is deceptively simple: given a region in the plane bounded by a known curve, what is the volume of the solid obtained by revolving that region around a coordinate axis? While symmetry tells us the resulting solid is rotationally symmetric, the actual volume depends on the specific curve and the axis of revolution. The disc method provides a systematic, general-purpose technique for answering this question using a single definite integral.

SECTION 2

Core Principles & Definitions

The disc method rests on the idea that a solid of revolution — the three-dimensional shape produced by rotating a planar region about an axis — can be decomposed into infinitely many thin circular discs stacked along the axis of rotation. Each disc has a known radius determined by the generating curve and an infinitesimal thickness, so its volume is simply the area of a circle multiplied by the thickness. Integrating these infinitesimal volumes over the appropriate interval yields the total volume of the solid.

Solid of Revolution

A three-dimensional solid formed by rotating a two-dimensional region about a fixed line (the axis of revolution). The resulting solid possesses rotational symmetry about that axis.

Representative Disc

A thin cylindrical slice perpendicular to the axis of revolution. Its radius equals the distance from the axis to the curve, and its thickness is an infinitesimal element dx or dy.

Radius Function

The function R(x) or R(y) that gives the radius of each disc as a function of position along the axis of revolution. This is simply the value of the generating curve at that point.

Axis of Revolution

The fixed line about which the region is rotated. In this lesson, we restrict attention to the x-axis and y-axis; the choice determines whether we integrate with respect to x or y.

Integration Variable

When revolving about the x-axis, discs are stacked along x so we integrate with respect to x. When revolving about the y-axis, discs are stacked along y and we integrate with respect to y.

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KEY TAKEAWAY

Think of the disc method like slicing a loaf of bread perpendicular to its length. Each slice is approximately a flat disc whose area you can compute from its circular face. If you knew the area of every infinitely thin slice and added them all up, you would recover the total volume of the loaf. The disc method does exactly this: it expresses the volume of a solid of revolution as the integral of the cross-sectional area π[R(x)]² (or π[R(y)]²) over the interval that spans the solid.

SECTION 3

Visual Explanation — Revolution About the x-Axis

The diagram below illustrates the geometric foundation of the disc method when a region is revolved about the x-axis. A curve y = f(x) is drawn above the x-axis from x = a to x = b. When this region is rotated about the x-axis, every point on the curve traces out a circle, and the resulting solid looks like a stack of circular discs. A single representative disc at position x has radius R(x) = f(x) and infinitesimal thickness dx.

The shaded region under y = f(x) from x = a to x = b is revolved about the x-axis. The cyan rectangle represents a single representative disc at position x with radius R(x) = f(x) and thickness dx. The volume of that disc is π[f(x)]² dx, and integrating from a to b yields the total volume.

Notice that the representative disc is drawn perpendicular to the axis of revolution. This is a hallmark of the disc method: the slicing direction is always orthogonal to the rotation axis. Because each cross-section is a full circle (not an annulus or a shell), the disc method applies when the region extends from the axis of revolution outward to a single bounding curve. If there were a gap between the axis and the region, we would instead use the washer method — a close relative of the disc method that accounts for a hollow core.

SECTION 4

Mathematical Framework

We now derive the disc method formulas rigorously. The key idea is to partition the interval of integration into n subintervals, approximate the solid's volume with n thin cylinders, and then pass to the limit as n → ∞.

Revolution About the x-Axis

Suppose y = f(x) ≥ 0 on [a, b], and the region between the curve and the x-axis is revolved about the x-axis. Partition [a, b] into n subintervals of width Δx = (b − a)/n, and choose a sample point x_k* in the k-th subinterval. The representative disc at x_k* has radius R = f(x_k*) and thickness Δx. Its volume is ΔV_k = π[f(x_k*)]² Δx. Summing and taking the limit gives the definite integral.

DISC METHOD — REVOLUTION ABOUT THE x-AXIS

V = π ∫ₐᵇ [f(x)]² dx

Here f(x) is the radius of each disc, [a, b] is the interval along the x-axis, and the integrand [f(x)]² represents the square of the radius so that π[f(x)]² gives the area of the circular cross-section.

Revolution About the y-Axis

When the axis of revolution is the y-axis, we slice the solid with horizontal discs perpendicular to the y-axis. The radius of each disc is now a function of y — specifically, x = g(y) — and the discs are stacked from y = c to y = d. The derivation follows the same Riemann-sum argument as before, but with the roles of x and y interchanged.

DISC METHOD — REVOLUTION ABOUT THE y-AXIS

V = π ∫꜀ᵈ [g(y)]² dy

Here g(y) is the horizontal distance from the y-axis to the curve (i.e., x expressed as a function of y), and [c, d] is the interval along the y-axis over which the solid extends.

⚠

Choosing the Correct Variable

A common source of error is integrating with respect to the wrong variable. Remember: you always integrate along the axis of revolution. If you revolve around the x-axis, your integration variable is x and your radius function must be in terms of x. If you revolve around the y-axis, you integrate with respect to y and must express the radius as a function of y — which may require solving your equation for x in terms of y.

VOLUME OF A SINGLE DISC ELEMENT

dV = π R² · (thickness) = π [R(t)]² dt, t = x or y

This differential form emphasizes that each infinitesimal disc contributes a volume equal to the area of a circle (πR²) multiplied by an infinitesimal thickness dt, where t is the integration variable matching the axis of revolution.

SECTION 5

Detailed Setup — x-Axis vs. y-Axis Revolution

The choice of axis of revolution fundamentally changes how you set up the integral, even for the same curve. This section provides a side-by-side comparison and a second diagram illustrating revolution about the y-axis, so that you can see exactly how the geometry determines the integration variable and the radius function.

Left panel: revolving y = √x about the x-axis produces vertical disc slices with radius R = f(x) and integration in x. Right panel: revolving the same curve about the y-axis produces horizontal disc slices with radius R = g(y) = y² and integration in y. Note how the shaded regions differ because the region must touch the axis of revolution.

Feature	Revolution About x-Axis	Revolution About y-Axis
Integration variable	x	y
Disc orientation	Vertical cross-section (perpendicular to x-axis)	Horizontal cross-section (perpendicular to y-axis)
Radius function	R(x) = f(x) = y-value of curve	R(y) = g(y) = x-value of curve
Limits of integration	x = a to x = b (endpoints along x-axis)	y = c to y = d (endpoints along y-axis)
Curve rewriting needed?	Only if curve is given as x = h(y)	Must solve y = f(x) for x = g(y)

Comparison of setup details for the two standard axes of revolution

A critical prerequisite for the disc method is that the region being revolved must be bounded on one side by the axis of revolution. If the curve does not touch the axis, or if the region lies between two curves neither of which is the axis, then the cross-sections will be washers (annuli) rather than full discs, and the washer method is required. Always sketch the region and the axis before writing the integral to verify that the disc method is appropriate.

SECTION 6

Worked Examples

Example 1: Revolution About the x-Axis

Find the volume of the solid generated by revolving the region bounded by y = x², y = 0, x = 0, and x = 2 about the x-axis.

Volume by Revolution About the x-Axis

1
Step 1 — Sketch and Identify the Region

The region lies above the x-axis (y = 0) and below the parabola y = x², from x = 0 to x = 2. Since the lower boundary is the x-axis itself — the axis of revolution — every cross-section is a full disc (no hollow core).
2
Step 2 — Determine the Radius Function

Each disc is perpendicular to the x-axis. Its radius is the vertical distance from the x-axis to the curve: R(x) = f(x) = x².

R(x) = x²
3
Step 3 — Write the Volume Integral

Using the disc formula V = π ∫ₐᵇ [f(x)]² dx with a = 0 and b = 2, we obtain V = π ∫₀² (x²)² dx = π ∫₀² x⁴ dx.

V = π ∫₀² x⁴ dx
4
Step 4 — Evaluate the Integral

The antiderivative of x⁴ is x⁵/5. Evaluating from 0 to 2: V = π [x⁵/5]₀² = π [(2⁵)/5 − 0] = π (32/5) = 32π/5.

V = 32π/5 ≈ 20.11 cubic units

Example 2: Revolution About the y-Axis

Find the volume of the solid generated by revolving the region bounded by y = x², x = 0, y = 0, and y = 4 about the y-axis.

Volume by Revolution About the y-Axis

1
Step 1 — Sketch and Identify the Region

The region is bounded on the left by the y-axis (x = 0), on the right by the curve x = √y (obtained by solving y = x² for x ≥ 0), below by y = 0, and above by y = 4. Since the left boundary is the y-axis — the axis of revolution — the disc method applies.
2
Step 2 — Determine the Radius Function

Each disc is now perpendicular to the y-axis. Its radius is the horizontal distance from the y-axis to the curve: R(y) = g(y) = √y.

R(y) = √y
3
Step 3 — Write the Volume Integral

Using V = π ∫꜀ᵈ [g(y)]² dy with c = 0 and d = 4: V = π ∫₀⁴ (√y)² dy = π ∫₀⁴ y dy.

V = π ∫₀⁴ y dy
4
Step 4 — Evaluate the Integral

The antiderivative of y is y²/2. Evaluating from 0 to 4: V = π [y²/2]₀⁴ = π [(16)/2 − 0] = 8π.

V = 8π ≈ 25.13 cubic units

💡

Observation

Notice that revolving the "same" parabola about different axes yields different volumes (32π/5 vs. 8π). This is because the region being revolved changes when the axis changes — the boundary conditions redefine which area is swept through space. Always re-examine the bounded region when the axis changes.

SECTION 7

Disc Method — Strengths, Limitations, and Alternatives

The disc method is one of three principal techniques for computing volumes of solids of revolution. Understanding when it is the most efficient choice — and when an alternative is preferable — saves significant setup time on exams and in applications. The table below compares the disc method with the washer method and the shell method, highlighting each technique's natural domain of applicability.

Criterion	Disc Method	Washer Method	Shell Method
Cross-section shape	Full circle (solid disc)	Annulus (ring with hole)	Thin cylindrical shell
When to use	Region touches the axis of revolution on one side	Region bounded by two curves, neither of which is the axis	Slicing parallel to the axis simplifies the integrand
Slicing direction	Perpendicular to axis of revolution	Perpendicular to axis of revolution	Parallel to axis of revolution
Typical integrand	π[R(t)]²	π([R(t)]² − [r(t)]²)	2πr(t) · h(t)
Limitation	Cannot handle hollow solids or regions not touching the axis	Requires two radius functions; algebra can be heavier	May require splitting into cases if the height function changes form

Comparison of the three main methods for computing volumes of revolution

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WHEN TO REACH FOR THE DISC METHOD

The disc method is your go-to tool when the region is bounded on one side by the axis of revolution and on the other by a single curve. Think of it like a potter's wheel: the pot wall (the curve) spins around the axle (the axis), and the interior is completely solid clay. If the pot were hollow — with an inner wall separated from the axis — you'd need the washer method instead. And if it's easier to describe the pot's height rather than its radius at each level, the shell method may prove more efficient.

SECTION 8

Connection to Advanced Methods and Theory

The disc method is the simplest member of a family of volume-computation techniques that share a common intellectual core: express volume as the integral of a known cross-sectional area. In more advanced settings, the axis of revolution may not be a coordinate axis, the cross-sections may not be circles, or the solid may exist in higher dimensions. Understanding how the disc method generalizes prepares you for these extensions.

Disc Method (This Lesson)	Advanced Extension
Axis of revolution is x- or y-axis	Washer/disc about arbitrary lines y = k or x = h; adjust radius by subtracting axis offset
Cross-sections are always circles	Cross-sectional area method: A(x) can be squares, equilateral triangles, semicircles, etc.
Region bounded by a single curve and the axis	Washer method handles regions between two curves; shell method handles parallel slicing
Integration in one variable (x or y)	Parametric and polar volume integrals; multivariable calculus extends to triple integrals and change-of-variables (Jacobians)
Volume of solids of revolution	Surface area of revolution (arc-length formulation), moments of inertia, center of mass of solids

How the disc method connects to more advanced volume and solid-geometry techniques

One particularly important generalization is the Pappus theorem (also called the theorem of Pappus–Guldin), which states that the volume of a solid of revolution equals the area of the generating region multiplied by the distance traveled by its centroid. While the disc method requires integrating [R(t)]², Pappus' theorem reduces certain problems to a simple multiplication — provided you know the centroid. This underscores a recurring theme in mathematics: what appears to be a single tool (the disc integral) is actually one viewpoint on a richer structure connecting integration, geometry, and the properties of centroids.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the disc method produces a full circular cross-section rather than an annular (ring-shaped) cross-section. Under what geometric condition on the region and the axis of revolution does the disc method apply, and what happens to the formula if that condition is violated?

Show answer

The disc method yields a full circular cross-section because one boundary of the region is the axis of revolution itself. When you rotate a vertical (or horizontal) strip whose inner edge lies on the axis, the entire strip sweeps out a solid disc with no hollow core. If the region does not touch the axis — for instance, if the region lies between two curves y = f(x) and y = g(x) with 0 < g(x) < f(x) — then each strip sweeps out an annular region (washer) with outer radius R = f(x) and inner radius r = g(x). The formula becomes V = π ∫ₐᵇ ([f(x)]² − [g(x)]²) dx, which is the washer method. The disc method is simply the special case where the inner radius is zero.

Question 2: BASIC CALCULATION

Find the volume of the solid generated by revolving the region bounded by y = 3x, y = 0, and x = 2 about the x-axis.

Show answer

The region lies above the x-axis and below the line y = 3x from x = 0 to x = 2. Since the lower boundary is the x-axis (the axis of revolution), the disc method applies with R(x) = 3x. The volume is V = π ∫₀² (3x)² dx = π ∫₀² 9x² dx = π [9x³/3]₀² = π [3x³]₀² = π(3 · 8 − 0) = 24π. Thus V = 24π ≈ 75.40 cubic units. This solid is a cone with base radius 6 and height 2, and indeed the cone formula V = (1/3)πr²h = (1/3)π(36)(2) = 24π confirms our answer.

Question 3: INTERMEDIATE

Find the volume of the solid obtained by revolving the region bounded by x = y² and x = 4 about the y-axis. (Hint: determine the appropriate y-limits and identify the correct radius function.)

Show answer

The parabola x = y² and the vertical line x = 4 intersect where y² = 4, giving y = −2 and y = 2. When revolving about the y-axis, each horizontal disc at height y has its outer edge at x = 4 and its inner edge at x = y². However, we must recognize this as a washer problem with outer radius R = 4 and inner radius r = y². Alternatively, we can use the disc method on just the region between x = 0 and x = y² revolved about the y-axis, which gives V₁ = π ∫₋₂² (y²)² dy = π ∫₋₂² y⁴ dy = π [y⁵/5]₋₂² = π(32/5 − (−32/5)) = 64π/5. The full cylinder of radius 4 and height 4 has volume π(16)(4) = 64π. The volume of the solid between x = y² and x = 4 revolved about the y-axis is 64π − 64π/5 = 256π/5 ≈ 160.85 cubic units. This problem illustrates the importance of carefully identifying which region is being revolved and whether the disc or washer formulation is needed.

Question 4: APPLIED

A decorative vase has a profile described by the curve y = √(x) + 1 for 0 ≤ x ≤ 9, where x is measured in inches from the bottom of the vase and y is the radius of the vase at height x. The vase is formed by revolving this profile about the x-axis. Find the interior volume of the vase in cubic inches.

Show answer

Here the radius of the vase at height x is given directly as R(x) = √x + 1, and the axis of revolution is the x-axis (the vertical centerline of the vase in this coordinate system). Applying the disc method: V = π ∫₀⁹ (√x + 1)² dx = π ∫₀⁹ (x + 2√x + 1) dx. We integrate term by term: ∫₀⁹ x dx = [x²/2]₀⁹ = 81/2; ∫₀⁹ 2√x dx = 2 · [

2x^{3/2}

/3]₀⁹ = (4/3)(27) = 36; ∫₀⁹ 1 dx = 9. Summing: V = π(81/2 + 36 + 9) = π(81/2 + 45) = π(81/2 + 90/2) = π(171/2) = 171π/2 ≈ 268.61 cubic inches. This is approximately 268.6 in³, or about 1.16 gallons — a reasonable size for a decorative vase.

Question 5: CRITICAL THINKING

Let f be a non-negative, continuous function on [0, a]. The solid S₁ is obtained by revolving the region under y = f(x) about the x-axis, and the solid S₂ is obtained by revolving the region to the left of x = f⁻¹(y) (assuming f is one-to-one) about the y-axis from y = 0 to y = f(a). Prove that Vol(S₁) + Vol(S₂) = πa²f(a), and give a geometric interpretation of this identity.

Show answer

Let b = f(a). Then Vol(S₁) = π ∫₀ᵃ [f(x)]² dx and Vol(S₂) = π ∫₀ᵇ [f⁻¹(y)]² dy. For S₂, substitute y = f(x), so dy = f′(x) dx and f⁻¹(y) = x. When y = 0 we get x = 0 (assuming f(0) = 0 for simplicity; the general case adjusts boundary terms), and when y = b we get x = a. Thus Vol(S₂) = π ∫₀ᵃ x² f′(x) dx. Now consider the cylinder of radius a and height b = f(a), which has volume πa²b. We can write πa²b = π ∫₀ᵃ d/dx [x² f(x)] dx − π ∫₀ᵃ x² f′(x) dx + ... More directly, integrate by parts or use the identity: π ∫₀ᵃ [f(x)]² dx + π ∫₀ᵇ [f⁻¹(y)]² dy = πa²b. This follows because the two integrals, viewed in the (x, y)-plane, partition the rectangle [0, a] × [0, b] — the first integral sums horizontal slices from the y-axis to the curve, and the second sums vertical slices from the x-axis to the curve. Their sum accounts for the entire rectangle, giving πa²f(a). Geometrically, the two solids of revolution together fill a cylinder of radius a and height f(a), minus the overlap, but the identity shows they exactly complement each other within that cylinder.

SUMMARY

Lesson Summary

Summary

The disc method computes the volume of a solid of revolution by summing the volumes of infinitely many thin circular discs stacked along the axis of revolution. When revolving about the x-axis, the formula is V = π ∫ₐᵇ [f(x)]² dx, where f(x) is the radius of each disc. When revolving about the y-axis, express the curve as x = g(y) and integrate: V = π ∫꜀ᵈ [g(y)]² dy.

The method requires that the region touches the axis of revolution on one boundary; if there is a gap, the washer method (which subtracts an inner radius) must be used instead. Always integrate along the axis of revolution, always express the radius as a function of the integration variable, and always sketch the region before writing the integral. Mastery of the disc method provides the foundation for the washer method, the shell method, and the general cross-sectional area method for volumes of known cross-section.