Disc Method: Other Axes — Volume with Disc Method: Revolving Around Other Axes - Calculus 1

SECTION 1

Historical Context & Motivation

For thousands of years, mathematicians struggled to find the volumes of curved, rounded objects. Ancient Greek scholars like Archimedes discovered clever ways to compute the volume of a sphere by imagining it as a stack of thin circular slices. This idea of breaking a solid into infinitely many flat pieces is the heart of what we now call the disc method.

When you first learn the disc method, you typically revolve a region around the x-axis or the y-axis. But many real-world solids — like a donut, a vase on a pedestal, or a decorative column — are formed by spinning a shape around a line that is shifted away from the coordinate axes. Engineers and architects encounter these shapes constantly, so extending the disc method to other axes of revolution is an essential skill.

~250 BC • Archimedes' Method of Exhaustion

Archimedes approximated volumes of spheres and paraboloids by summing thin circular slices, laying the conceptual groundwork for the disc method.

1615 • Kepler's Volume of Wine Barrels

Johannes Kepler used the idea of infinitesimally thin discs to compute volumes of barrels, advancing the practical application of integration before calculus was formalized.

1665–1676 • Newton & Leibniz Formalize Calculus

Isaac Newton and Gottfried Wilhelm Leibniz independently developed the fundamental theorem of calculus, giving mathematicians precise integral formulas for volumes of revolution.

Modern Era • Engineering & CAD Applications

Today, the disc method with arbitrary axes of rotation is used in computer-aided design (CAD) and 3D printing to model axially symmetric parts like turbine blades, pipes, and pottery.

The central question this lesson addresses is: How do we adapt the disc method when the axis of revolution is a horizontal line y = k or a vertical line x = k, rather than the x-axis or y-axis? The answer lies in carefully redefining the radius of each disc.

SECTION 2

Core Principles & Definitions

Before tackling non-standard axes, let's solidify the foundations. The disc method works by slicing a solid of revolution into thin circular discs perpendicular to the axis of rotation. Each disc has a small thickness (dx or dy) and a radius determined by the distance from the curve to the axis. You then integrate the area of these discs over the interval to get the total volume.

Axis of Revolution

The straight line around which the region is rotated. It can be any horizontal line y = k or vertical line x = k, not just the coordinate axes.

Radius of the Disc

The perpendicular distance from the curve to the axis of revolution. When the axis shifts, the radius expression changes: R = f(x) − k or R = k − f(x), depending on orientation.

Volume Element (dV)

Each thin disc contributes a volume of π × R² × (thickness). The thickness is dx for horizontal integration or dy for vertical integration.

Integration Direction

For a horizontal axis of revolution, slice perpendicular to that axis (use dx). For a vertical axis of revolution, slice perpendicular to it (use dy). Always integrate along the axis.

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KEY TAKEAWAY

Think of the axis of revolution like the skewer through a kebab. When the skewer moves, every piece of food is now a different distance from the skewer. In the disc method, the radius is the distance from the curve to the skewer (the axis), so when you move the skewer, you must recalculate that distance. The formula structure stays the same — you just adjust the radius expression.

SECTION 3

Visual Explanation

The diagram below illustrates the key difference between revolving a curve around the x-axis (y = 0) versus revolving it around a different horizontal line, such as y = −1. Notice how the radius of each disc changes when the axis shifts downward.

Left panel: revolving around the x-axis (y = 0), the radius is simply f(x). Right panel: revolving around y = −1, the radius becomes f(x) − (−1) = f(x) + 1. The yellow dashed segment represents the radius of the disc at a sample x-value.

In the left panel, the pink dashed line sits right on the x-axis, so the radius of each disc is just f(x). In the right panel, the axis has moved down to y = −1. Now every point on the curve is farther from the axis than before, so the radius is f(x) + 1. This bigger radius produces a larger solid, which makes intuitive sense — you're sweeping the curve in a wider circle.

SECTION 4

Mathematical Framework

Let's formalize the disc method for non-standard axes. The key idea is always the same: identify the radius as the distance from the curve to the axis of revolution, then integrate π × R² over the appropriate interval.

Case 1: Revolving Around a Horizontal Line y = k

HORIZONTAL AXIS y = k

V = π ∫ₐᵇ [f(x) − k]² dx

Here, f(x) is the curve being revolved, k is the y-value of the axis of revolution, and [a, b] is the interval of integration along x. The radius of each disc is R(x) = |f(x) − k|.

Case 2: Revolving Around a Vertical Line x = k

VERTICAL AXIS x = k

V = π ∫_c^d [g(y) − k]² dy

Here, g(y) is the curve expressed as a function of y, k is the x-value of the vertical axis, and [c, d] is the interval along y. The radius is R(y) = |g(y) − k|.

Determining the Correct Radius Sign

Because the radius is a distance, it must be non-negative. If the curve lies above the axis (f(x) > k), then R = f(x) − k. If the curve lies below the axis (f(x) < k), then R = k − f(x). Since the radius is squared in the formula, [f(x) − k]² automatically equals [k − f(x)]², so the sign issue resolves itself algebraically. However, when you set up the integral, it helps to think clearly about which quantity is larger to avoid confusion.

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Common Mistake Alert

Students often forget to adjust the radius and simply use f(x) as though the axis were y = 0. Always ask yourself: "What is the distance from the curve to the axis?" That distance is your radius.

SECTION 5

Setting Up the Radius for Different Axes

The most important step when revolving around a non-standard axis is correctly identifying the radius. Let's look at a specific example: the curve y = √x on the interval [0, 4], revolved around different axes. The diagram below shows how the radius changes in each scenario.

The green curve y = √x is shown with three possible axes of revolution. The dashed vertical lines at x ≈ 2.5 show the radius from the curve to each axis. For y = 0 the radius is √x. For y = 3 (above the curve) the radius is 3 − √x. For y = −1 (below the curve) the radius is √x + 1.

Axis of Revolution	Radius R	Volume Integral
y = 0 (x-axis)	R = √x	V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx
y = 3 (above curve)	R = 3 − √x	V = π ∫₀⁴ (3 − √x)² dx
y = −1 (below curve)	R = √x − (−1) = √x + 1	V = π ∫₀⁴ (√x + 1)² dx
x = 5 (vertical, right)	R = 5 − y² (since x = y²)	V = π ∫₀² (5 − y²)² dy

How the radius and integral change for the curve y = √x depending on the axis of revolution

SECTION 6

Worked Example

Let's work through a complete example. We'll find the volume of the solid formed by revolving the curve y = x² on the interval [0, 2] about the line y = −1 using the disc method. Here, the axis y = −1 itself forms the inner boundary of every disc, so the radius of each disc runs directly from the axis up to the curve — no hollow center, no gap between the axis and the region being swept.

Volume by Disc Method — Revolving y = x² Around y = −1

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Step 1 — Identify the Problem Setup

We are revolving the curve y = x² (for 0 ≤ x ≤ 2) around the horizontal line y = −1. The axis y = −1 lies below the entire curve (since x² ≥ 0 > −1 for all x). Because the axis is the lower boundary of every disc — with the curve y = x² as the only upper boundary and no separate lower curve between the axis and y = x² — each cross-section is a solid disc (not a washer). The radius of each disc runs from the axis y = −1 straight up to the curve y = x².
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Step 2 — Determine the Radius

For the disc method, each disc extends from the axis y = −1 up to the curve y = x². The radius of each disc at position x is the distance from the axis to the curve:

R(x) = x² − (−1) = x² + 1
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Step 3 — Write the Volume Integral

Using the disc method formula V = π ∫ₐᵇ R(x)² dx, we substitute our radius and limits of integration (x goes from 0 to 2):

V = π ∫₀² (x² + 1)² dx
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Step 4 — Expand the Integrand

We expand (x² + 1)² using the algebraic identity (a + b)² = a² + 2ab + b²:

(x² + 1)² = x⁴ + 2x² + 1
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Step 5 — Integrate Term by Term

Now we integrate each term from 0 to 2: ∫₀² x⁴ dx = [x⁵/5]₀² = 32/5 ∫₀² 2x² dx = [2x³/3]₀² = 16/3 ∫₀² 1 dx = [x]₀² = 2

Sum = 32/5 + 16/3 + 2 = 96/15 + 80/15 + 30/15 = 206/15
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Step 6 — Multiply by π for Final Answer

Multiply the result by π to obtain the volume:

V = 206π/15 ≈ 43.14 cubic units

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Check Your Intuition

Compare this to revolving the same curve around y = 0 (the x-axis). Around y = 0, the radius would be just x², giving V = π ∫₀² x⁴ dx = 32π/5 ≈ 20.11. Our answer of 206π/15 ≈ 43.14 is larger, which makes sense — the axis is farther away, so the discs have bigger radii and the solid is fatter.

SECTION 7

Disc Method vs. Other Approaches

When revolving around non-standard axes, the disc method is just one option. It's important to understand when it works well and when another technique might be easier. The two main alternatives are the washer method (an extension of the disc method for hollow solids) and the shell method (which uses cylindrical shells instead of discs).

Feature	Disc / Washer Method	Shell Method
Slice orientation	Perpendicular to axis of revolution	Parallel to axis of revolution
Best when...	Curve is easily expressed as a function of the variable along the axis	Curve is hard to solve for the "other" variable, or two functions enclose a region
Radius for other axes	R = \|f(x) − k\| or \|g(y) − k\|	r = \|x − k\| or \|y − k\|
Revolving around y = k (horizontal)	Integrate with respect to x	Integrate with respect to y
Hollow centers?	Yes — use washer (subtract inner radius²)	Handled naturally without subtraction

Comparison of disc/washer and shell methods for volumes of revolution

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WHEN TO CHOOSE DISC

Use the disc method when the region being revolved has the axis of revolution as one of its boundaries — meaning the radius runs from the axis directly to a single bounding curve, with no separate inner boundary creating a hollow center. If the region has a gap between itself and the axis (i.e., there is a second boundary curve between the axis and the outer curve), you'll need the washer method instead. Think of it like this: discs are solid circles (like hockey pucks), while washers are circles with holes (like actual metal washers). Same setup idea — just subtract the inner radius squared from the outer radius squared.

SECTION 8

Connection to Advanced Techniques

The disc method with non-standard axes is a stepping stone toward more advanced integration techniques you'll encounter in Calculus 2 and beyond. Understanding how to adjust the radius for different axes builds the geometric intuition needed for multivariable calculus, where volumes are computed using double and triple integrals over complex regions.

This Lesson	What Comes Next
Disc method: revolving around y = k or x = k	Washer method: revolving regions with gaps around any axis
Single-function boundaries	Regions bounded by two or more curves
Horizontal and vertical axes	Pappus' theorem: revolving around any line using centroids
Integration in one variable	Volumes via double/triple integrals in Calc 3

How the disc method connects to future topics

The key takeaway for your future studies is that the disc method teaches you a powerful pattern: break a complicated 3D shape into simple pieces, compute each piece's contribution, and add them up with an integral. This same philosophy applies to computing surface areas, arc lengths, moments of inertia, and much more. Mastering the radius adjustment for non-standard axes is excellent practice for the kind of geometric reasoning that makes all of these applications possible.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

When revolving the curve y = x³ around the line y = 5 on the interval [0, 1], what is the radius of each disc? Explain why the radius is not simply x³.

Show answer

The radius is R = 5 − x³. Since the axis y = 5 is above the curve y = x³ on [0, 1] (where x³ ranges from 0 to 1), the distance from the curve to the axis is 5 − x³, not x³ itself. The value x³ would only be the radius if the axis were y = 0. The radius must always measure the distance from the curve to the axis of revolution, which is |f(x) − k| where k is the y-value of the axis.

Question 2: BASIC CALCULATION

Find the volume of the solid formed by revolving y = 2x on the interval [0, 3] around the line y = −2.

Show answer

The radius at each x is R(x) = 2x − (−2) = 2x + 2. Using the disc formula: V = π ∫₀³ (2x + 2)² dx = π ∫₀³ (4x² + 8x + 4) dx. Integrating term by term: π [4x³/3 + 4x² + 4x]₀³ = π [36 + 36 + 12] = 84π. The volume is 84π cubic units, approximately 263.9 cubic units.

Question 3: INTERMEDIATE

Find the volume of the solid generated by revolving the region bounded by y = √x, y = 0, and x = 4 about the line y = 3.

Show answer

Here the axis y = 3 is above the curve y = √x (which reaches a maximum of 2 at x = 4). The outer radius from the axis to y = 0 is R_outer = 3, and the inner radius from the axis to y = √x is R_inner = 3 − √x. Since there's a gap between the curve and the axis, this is technically a washer problem: V = π ∫₀⁴ [3² − (3 − √x)²] dx = π ∫₀⁴ [9 − (9 − 6√x + x)] dx = π ∫₀⁴ (6√x − x) dx = π [6 · (2/3)

x^{3/2}

− x²/2]₀⁴ = π [4(8) − 8] = π [32 − 8] = 24π ≈ 75.4 cubic units.

Question 4: APPLIED

A decorative bowl is modeled by revolving the curve y = x² (for 0 ≤ x ≤ 3) around the vertical line x = −1. Set up (but do not evaluate) the integral that gives the volume of this solid using the disc method with dy.

Show answer

Since the axis of revolution is vertical (x = −1), we integrate with respect to y. We rewrite the curve as x = √y (taking the positive root since 0 ≤ x ≤ 3). When x ranges from 0 to 3, y ranges from 0 to 9. The radius of each disc at height y is R(y) = √y − (−1) = √y + 1 (the distance from the curve to the axis x = −1). The volume integral is V = π ∫₀⁹ (√y + 1)² dy. This approach correctly uses the disc method because the curve is the only boundary on the right side, and the axis x = −1 is to the left.

Question 5: CRITICAL THINKING

Consider revolving y = 1/x on [1, 4] about y = −1 versus about y = 2. Without computing both volumes exactly, explain which revolution produces the larger solid. Then set up both integrals and verify your reasoning by comparing the integrands at x = 2.

Show answer

At x = 2, f(2) = 1/2. For the axis y = −1: R = 1/x − (−1) = 1/x + 1, so R(2) = 3/2, and R² = 9/4. For the axis y = 2: R = 2 − 1/x, so R(2) = 3/2, and R² = 9/4. Interestingly, at x = 2 the integrands are equal! However, at x = 1, for y = −1: R = 2, R² = 4; for y = 2: R = 1, R² = 1. At x = 4, for y = −1: R = 5/4, R² = 25/16; for y = 2: R = 7/4, R² = 49/16. The integrals are V₁ = π ∫₁⁴ (1/x + 1)² dx and V₂ = π ∫₁⁴ (2 − 1/x)² dx. Near x = 1, the first integral dominates; near x = 4, the second dominates. By expanding and integrating, V₁ = π ∫₁⁴ (1/x² + 2/x + 1) dx and V₂ = π ∫₁⁴ (4 − 4/x + 1/x²) dx. The first gives π[−1/x + 2 ln x + x]₁⁴ and the second gives π[4x − 4 ln x − 1/x]₁⁴. Computing: V₁ = π(3/4 + 2 ln 4 + 3) = π(15/4 + 4 ln 2) and V₂ = π(12 − 4 ln 4 + 3/4) = π(51/4 − 8 ln 2). Since 51/4 − 8 ln 2 ≈ 7.20 and 15/4 + 4 ln 2 ≈ 6.52, the revolution around y = 2 produces the larger solid because the curve is closer to y = −1 on average over [1,4], making discs around y = 2 generally wider.

SUMMARY

Lesson Summary

Summary

The disc method computes volumes of revolution by stacking thin circular discs perpendicular to the axis of revolution. When revolving around a horizontal line y = k, the radius becomes R(x) = |f(x) − k|, and the volume formula is V = π ∫ₐᵇ [f(x) − k]² dx. For a vertical line x = k, rewrite the curve as x = g(y) and use R(y) = |g(y) − k| with integration in dy.

The critical step is always correctly identifying the radius as the distance from the curve to the axis — not just the function value itself. When the axis is below the region, the radius is f(x) − k (which will be positive). When the axis is above the region, the radius is k − f(x). Since the radius is squared in the integral, both expressions give the same result algebraically, but careful setup prevents errors in more complex problems involving washers or multiple curves.