Disc Method: Other Axes — Volume with Disc Method: Revolving Around Other Axes - Calculus 2

SECTION 1

Historical Context & Motivation

The problem of computing volumes of curved solids has occupied mathematicians for millennia, beginning with Archimedes and his pioneering work on the sphere, cylinder, and paraboloid. Archimedes' method of exhaustion—slicing solids into thin cross-sections and summing their areas—anticipated the integral calculus that would not be formalized for nearly two thousand years. The key insight, that a solid of revolution can be decomposed into infinitely many thin discs whose volumes can be summed exactly, lies at the heart of what we now call the disc method. While introductory treatments typically revolve regions around the x-axis or y-axis, real applications in engineering and physics frequently require revolution about arbitrary horizontal or vertical lines, making the generalized disc method an essential tool in a calculus student's repertoire.

c. 250 BCE • Archimedes' Method of Exhaustion

Archimedes computes the volume of a sphere by slicing it into thin circular cross-sections and bounding the sum of their areas, establishing the conceptual ancestor of the disc method.

1635 • Cavalieri's Principle

Bonaventura Cavalieri formalizes the idea that solids with equal cross-sectional areas at every height have equal volumes, providing a rigorous bridge between slicing and volume computation.

1665–1676 • Newton & Leibniz Develop Calculus

The invention of the integral allows the summation of infinitely many infinitesimal disc volumes to be expressed as a single definite integral, transforming geometric approximation into exact computation.

18th–19th c. • Engineering Applications Emerge

As industrial design requires computing volumes of axles, pistons, and vessels whose axes of symmetry do not coincide with coordinate axes, the disc method is generalized to arbitrary axes of revolution.

The central question this lesson addresses is: how do we modify the disc method's radius expression when the axis of revolution is not a coordinate axis, but an arbitrary line such as y = k or x = h? Mastering this generalization is critical because it connects geometric intuition about distance to an axis with the algebraic setup of an integral, a skill that transfers directly to the washer method and shell method for more complex solids.

SECTION 2

Core Principles & Definitions

Before diving into computations, it is essential to internalize the foundational ideas that govern the disc method when applied to non-standard axes. The method itself does not change in spirit—every solid of revolution is still approximated by stacking thin circular discs—but the expression for the radius of each disc must be carefully re-derived in terms of the distance from the curve to the new axis of revolution.

Axis of Revolution

The fixed line about which the region rotates. When this is y = k (horizontal) or x = h (vertical), every point in the region traces a circle whose center lies on this line.

Disc Radius as Distance

The radius of each disc equals the perpendicular distance from the curve to the axis of revolution: R(x) = |f(x) − k| for revolution about y = k, or R(y) = |g(y) − h| for revolution about x = h.

Integration Variable

Integrate with respect to the variable that runs along the axis of revolution. For a horizontal axis y = k, discs are perpendicular to that axis, so integrate with respect to x. For a vertical axis x = h, integrate with respect to y.

Limits of Integration

The bounds come from the region being revolved, not from the axis. Identify where the region starts and ends along the integration variable, possibly by solving for intersection points.

No Hollow Interior

The disc method applies when the region extends all the way to the axis of revolution—every cross-section is a full disc (no hole). If a gap exists between the region and the axis, use the washer method instead.

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KEY TAKEAWAY

Think of the axis of revolution as a skewer through a stack of coins. When you move the skewer away from the origin—say from y = 0 to y = 3—the coins don't change thickness, but each coin's radius is now measured from its edge to the new skewer position. The entire challenge of revolving around other axes reduces to correctly computing that new radius as the signed distance from the curve to the axis.

SECTION 3

Visual Explanation

The following diagram illustrates the key geometric idea behind the disc method when the axis of revolution is the line y = −1, rather than the x-axis. The region under the curve y = √x from x = 0 to x = 4 is shown, and representative disc cross-sections are drawn to emphasize how the radius of each disc is measured as the distance from the curve to the axis y = −1.

The curve y = √x (cyan) is revolved about the dashed pink line y = −1. Each yellow vertical segment represents the radius R of a disc at a given x-value. Notice that R = √x − (−1) = √x + 1, which is strictly the distance from the curve to the axis, not merely the function value.

In the diagram above, three representative disc radii are drawn at different x-values. At x = 2 (approximately), the curve sits at y = √2 ≈ 1.41, so the disc radius is √2 − (−1) = √2 + 1 ≈ 2.41 units. Compare this to what the radius would be if we revolved about the x-axis: it would simply be √2 ≈ 1.41. The shift of the axis downward by 1 unit increases every radius by 1, producing a larger solid. This observation generalizes: when the axis moves farther from the curve, radii increase and the volume grows; when the axis moves closer, radii decrease and the volume shrinks.

SECTION 4

Mathematical Framework

We now formalize the disc method for arbitrary axes. The fundamental volume formula remains V = π∫[R(variable)]² d(variable), but the expression for R depends on the orientation and position of the axis.

Case 1: Revolution About a Horizontal Line y = k

DISC VOLUME — HORIZONTAL AXIS y = k

V = π ∫ₐᵇ [f(x) − k]² dx

Here f(x) is the curve bounding the region, k is the y-value of the axis of revolution, and [a, b] is the interval over which the region extends along the x-axis. The integrand [f(x) − k]² gives the squared radius of each disc.

When the curve lies above the axis (f(x) > k for all x in [a, b]), the expression f(x) − k is positive and represents the disc radius directly. When the curve lies below the axis (f(x) < k), the expression f(x) − k is negative, but squaring it produces the correct positive area. In either case, the formula works without needing absolute values because of the squaring.

Case 2: Revolution About a Vertical Line x = h

DISC VOLUME — VERTICAL AXIS x = h

V = π ∫꜀ᵈ [g(y) − h]² dy

Here g(y) expresses the curve as x = g(y), h is the x-value of the vertical axis of revolution, and [c, d] is the interval along the y-axis. Each disc is horizontal, with radius |g(y) − h|.

Deriving the Radius Expression

The derivation follows from the definition of distance. Consider a point (x, f(x)) on the curve and the axis of revolution y = k. When this point is rotated about y = k, it traces a circle of radius equal to the vertical distance between the point and the line, namely |f(x) − k|. The area of the resulting disc is A(x) = π[f(x) − k]². Summing these infinitesimal disc volumes over [a, b] yields V = ∫ₐᵇ A(x) dx = π∫ₐᵇ [f(x) − k]² dx. The same logic applies to a vertical axis x = h, replacing the vertical distance with the horizontal distance |g(y) − h|.

GENERAL RADIUS FORMULA

R = |curve value − axis value|

This single principle governs every disc method setup. Whether the axis is x = 5, y = −3, or any other line parallel to a coordinate axis, the radius is always the absolute difference between the function value and the axis constant.

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Common Pitfall

Students frequently forget to adjust the radius when the axis shifts. If revolving y = x² about y = 4 (an axis above the parabola on [0, 2]), the radius is 4 − x², not x². Drawing a representative disc from the curve to the axis and labeling the radius before writing the integral will prevent this error every time.

SECTION 5

Detailed Breakdown: Axis Position Cases

The position of the axis relative to the region determines the sign convention in the radius expression. We can classify the most common configurations into four cases, summarized in the table below and illustrated in the second diagram. Understanding these cases prevents sign errors and ensures a correct integral setup on the first attempt.

Axis	Curve Relative Position	Radius R	Example
y = k (below curve)	f(x) > k	f(x) − k	y = √x about y = −2 → R = √x + 2
y = k (above curve)	f(x) < k	k − f(x)	y = x² about y = 5 → R = 5 − x²
x = h (right of curve)	g(y) < h	h − g(y)	x = y² about x = 4 → R = 4 − y²
x = h (left of curve)	g(y) > h	g(y) − h	x = y + 1 about x = −1 → R = y + 2

Radius expressions for common axis-curve configurations

Four common configurations. In each panel, the pink dashed line is the axis, the cyan curve is the boundary, and the yellow segment is the disc radius. The key idea: always subtract in the direction from the curve to the axis (or vice versa) so the radius is positive.

A systematic approach to any disc-method problem about a non-standard axis can be distilled to three steps. First, sketch the region and the axis to determine whether you are in Case A, B, C, or D. Second, write the radius R as a function of the integration variable using the appropriate row of the table. Third, substitute into V = π∫R² d(variable) and evaluate.

SECTION 6

Worked Example

Let us compute the volume of the solid generated by revolving the region bounded by y = x², y = 0, and x = 2 about the line y = 4.

Volume by Disc Method — Revolution About y = 4

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Step 1 — Sketch & Identify the Configuration

The region sits below y = x² (which reaches y = 4 at x = 2) and above y = 0. The axis y = 4 lies at or above the curve for x ∈ [0, 2]. This is Case B (axis above the curve). However, notice that the region does not touch y = 4 everywhere—only at x = 2. For 0 ≤ x < 2, there is a gap between y = x² and y = 4. But the problem asks us to revolve the region bounded by y = x², y = 0, x = 2 about y = 4. The outer boundary from the axis is at y = 0, so the farthest point from y = 4 is y = 0. Since the region extends from y = 0 to y = x², and we must verify whether to use discs or washers. The region is bounded on one side by y = 0 and on the other by y = x², with the axis at y = 4. Since the region does not border the axis y = 4 (there is a gap between y = x² and y = 4), this actually requires washers. Let us instead consider a clean disc-method problem: revolve the region between y = x² and y = 4 (for 0 ≤ x ≤ 2) about y = 4.

Region: between y = x² and y = 4, revolved about y = 4. The upper boundary y = 4 coincides with the axis, so the cross-sections are full discs.
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Step 2 — Determine the Disc Radius

Each disc extends from the axis y = 4 down to the curve y = x². The radius at position x is R(x) = 4 − x². This is positive for all x ∈ [0, 2] since x² ≤ 4 on this interval.

R(x) = 4 − x²
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Step 3 — Set Up the Integral

The discs sweep from x = 0 to x = 2. Each disc has area πR² = π(4 − x²)². By symmetry the region is the same from x = −2 to x = 0, but we are told x ranges from 0 to 2, so we integrate from 0 to 2. If the full parabolic region were revolved, we would integrate from −2 to 2. Here we do 0 to 2.

V = π ∫₀² (4 − x²)² dx
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Step 4 — Expand the Integrand

Expand (4 − x²)² = 16 − 8x² + x⁴.

V = π ∫₀² (16 − 8x² + x⁴) dx
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Step 5 — Evaluate the Integral

Integrate term by term: ∫₀² 16 dx = 32; ∫₀² 8x² dx = 8 × (x³/3)|₀² = 8 × 8/3 = 64/3; ∫₀² x⁴ dx = (x⁵/5)|₀² = 32/5. Therefore V = π(32 − 64/3 + 32/5). Finding a common denominator of 15: V = π(480/15 − 320/15 + 96/15) = π(256/15).

V = 256π/15 ≈ 53.62 cubic units

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Verification Tip

As a sanity check, note that the solid fits inside a cylinder of radius 4 and height 2, whose volume is π(4²)(2) = 32π ≈ 100.5. Our answer 256π/15 ≈ 53.6 is less than 32π, which is consistent since the parabolic boundary curves inward, removing material from the cylinder.

SECTION 7

Disc Method vs. Washer & Shell Methods

The disc method for non-standard axes is one of several techniques for computing volumes of revolution. Choosing the right method depends on the geometry of the region and the axis. The table below compares the disc method with the washer and shell methods, highlighting when each is most natural for problems involving axes other than y = 0 or x = 0.

Feature	Disc Method	Washer Method	Shell Method
Cross-section shape	Solid disc (no hole)	Annulus (disc with hole)	Thin cylindrical shell
When to use (other axes)	Region boundary touches the axis	Gap between region and axis	Integration perpendicular to axis is simpler
Radius formula	R = \|curve − axis\|	R_outer − R_inner, both measured from axis	r = \|variable − axis\|, height = f − g
Integration direction	Along the axis	Along the axis	Perpendicular to the axis
Limitation	Cannot handle hollow solids	Requires two radius expressions	Sometimes harder to set up bounds

Method comparison for volumes of revolution about non-standard axes

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CHOOSING YOUR METHOD

Imagine you are building the solid from clay. If the region lies flush against the axis of revolution—like wrapping clay directly around a rod—every cross-section is a solid disc, and the disc method is the tool to reach for. If the clay wraps around the rod but leaves an air gap in the center (like a donut shape), you need the washer method. And if expressing the radius is awkward in one variable but straightforward in the other, consider flipping to the shell method.

SECTION 8

Connection to Advanced Theory

The disc method for other axes is a gateway to several more advanced topics in calculus and applied mathematics. Understanding how changing the axis affects the integrand prepares you for the Pappus' theorem, which states that the volume of a solid of revolution equals the area of the cross-section times the distance traveled by its centroid: V = 2πr̄A, where r̄ is the distance from the centroid to the axis. This theorem provides an elegant shortcut when the centroid is known, and its proof relies on exactly the kind of radius-from-axis reasoning developed in this lesson.

This Lesson	Advanced Extension
Disc method about y = k or x = h	Pappus' theorem: V = 2πr̄A, bypassing integration entirely when centroid is known
Single-function boundary (disc)	Multi-function boundaries require the washer method: V = π∫(R_outer² − R_inner²) dx
Revolution about lines parallel to axes	Revolution about oblique lines (e.g., y = x) using parametric or change-of-variable techniques
Volumes of known cross-sections	Generalizes to solids with non-circular cross-sections (squares, equilateral triangles, semicircles)

From disc method to advanced volume techniques

In multivariable calculus and physics, the ideas behind the disc method resurface in the study of moments of inertia, where one integrates r²dm over a solid body. The 'r' in that formula is precisely the distance from a mass element to the axis of rotation—the same geometric quantity that defines a disc radius. Mastering the art of expressing distance-to-axis in your volume integrals thus pays dividends well beyond Calculus 2.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Suppose you revolve the region under y = f(x) (above the x-axis) about the line y = −3 instead of y = 0. Without computing any integral, explain qualitatively how the radius of each disc changes and whether the resulting volume is larger or smaller than the volume obtained by revolving about the x-axis.

Show answer

When the axis shifts from y = 0 to y = −3, the distance from any point (x, f(x)) on the curve to the axis increases from f(x) to f(x) + 3. Since each disc radius is larger, πR² is larger at every x, and the integral accumulates more volume at every point. Therefore the volume is strictly larger than the volume obtained by revolving about y = 0. Intuitively, moving the axis farther from the region "swings" each point through a larger circle, sweeping out more space.

Question 2: BASIC CALCULATION

Find the volume of the solid obtained by revolving the region bounded by y = x, y = 0, and x = 3 about the line y = −1. Use the disc method.

Show answer

The region is a triangle with vertices (0,0), (3,0), and (3,3). However, the region is bounded below by y = 0 and above by y = x, and the axis y = −1 lies below both. The boundary y = 0 does not coincide with y = −1, so the region does not touch the axis—this actually requires the washer method with R_outer = x + 1 and R_inner = 0 + 1 = 1. Thus V = π∫₀³ [(x + 1)² − 1²] dx = π∫₀³ (x² + 2x) dx = π[x³/3 + x²]₀³ = π(9 + 9) = 18π. The volume is 18π cubic units.

Question 3: INTERMEDIATE

Find the volume of the solid generated by revolving the region bounded by x = √y, x = 0, and y = 9 about the line x = −2. Set up and evaluate the appropriate integral.

Show answer

The curve x = √y with x = 0 and y = 9 forms a region in the first quadrant. The axis x = −2 is vertical and lies to the left of the region. Since the region's left boundary x = 0 does not coincide with x = −2, we need washers: R_outer = √y − (−2) = √y + 2, R_inner = 0 − (−2) = 2. Thus V = π∫₀⁹ [(√y + 2)² − 4] dy = π∫₀⁹ [y + 4√y + 4 − 4] dy = π∫₀⁹ (y + 4√y) dy. Evaluating: ∫₀⁹ y dy = 81/2, ∫₀⁹

4y^{1/2}

dy = 4 × (2/3)

y^{3/2}

|₀⁹ = (8/3)(27) = 72. Therefore V = π(81/2 + 72) = π(81/2 + 144/2) = 225π/2 cubic units.

Question 4: APPLIED

A bowl is formed by revolving the curve y = x⁴ on [0, 1] about the line y = 1 (so the opening faces downward). Find the volume of the interior of the bowl (the region between y = x⁴ and y = 1, revolved about y = 1).

Show answer

The region is bounded above by y = 1 (which is also the axis) and below by y = x⁴ for x ∈ [−1, 1]. Since y = 1 coincides with the axis, the cross-section at each x is a full disc with radius R(x) = 1 − x⁴ (axis minus curve, since the axis is above the curve). By symmetry about x = 0, V = 2π∫₀¹ (1 − x⁴)² dx = 2π∫₀¹ (1 − 2x⁴ + x⁸) dx. Evaluating: ∫₀¹ 1 dx = 1, ∫₀¹ 2x⁴ dx = 2/5, ∫₀¹ x⁸ dx = 1/9. So V = 2π(1 − 2/5 + 1/9) = 2π(45/45 − 18/45 + 5/45) = 2π(32/45) = 64π/45 ≈ 4.47 cubic units.

Question 5: CRITICAL THINKING

Let R be the region bounded by y = √x and y = 0 for x ∈ [0, 4]. For a given constant k < 0, the volume of revolution of R about y = k is V(k). Show that V(k) is a quadratic function of k and find the value of k that minimizes V(k). Interpret your result geometrically.

Show answer

Since y = 0 does not coincide with y = k (because k < 0), we need washers: R_outer = √x − k (curve to axis) and R_inner = 0 − k = −k (x-axis to axis). V(k) = π∫₀⁴ [(√x − k)² − k²] dx = π∫₀⁴ [x − 2k√x + k² − k²] dx = π∫₀⁴ (x − 2k√x) dx. Evaluating: ∫₀⁴ x dx = 8, ∫₀⁴ 2k√x dx = 2k·(2/3)

x^{3/2}

|₀⁴ = 2k·(16/3) = 32k/3. So V(k) = π(8 − 32k/3). This is linear in k, not quadratic. V(k) increases as k becomes more negative (k → −∞), confirming that moving the axis farther below increases volume. Since V(k) is strictly decreasing as k → 0⁻, the minimum occurs as k → 0⁻, giving V → 8π. Geometrically, bringing the axis as close as possible to the region minimizes the swept volume. The linearity arises because the k² terms cancel in the washer formula, leaving only the cross-term −2k√x.

SUMMARY

Lesson Summary

Summary

The disc method computes volumes of revolution by stacking infinitely many thin circular discs. When the axis of revolution is a horizontal line y = k or a vertical line x = h instead of a coordinate axis, the method remains the same—only the radius expression changes. The radius is always the perpendicular distance from the curve to the axis: R = |f(x) − k| for a horizontal axis, or R = |g(y) − h| for a vertical axis. The volume integral is then V = π∫R² d(variable), with limits determined by the extent of the region along the integration variable.

To set up these integrals correctly, always sketch the region and the axis, draw a representative disc, and label the radius before writing any algebra. If the region does not touch the axis (creating a hollow core), switch to the washer method. The geometric reasoning developed here—measuring distance from a curve to an axis—extends to Pappus' theorem, moments of inertia, and numerous applications in engineering and physics.