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Disk Method

MATH162
Reference: Stewart 5.2  •  Chapter: 5  •  Section: 2

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Disk Method

Spinning Regions into Solids

Take a flat region and spin it around an axis. What do you get? A solid of revolution—and remarkably, we can compute its volume using a simple formula.

When you rotate a region around an axis, every cross-section perpendicular to that axis is a circle (a disk). Since we know the area of a circle is $\pi r^2$, and we know how to integrate cross-sectional areas, we get an elegant volume formula.

The key insight: The radius of each circular cross-section comes from the function you're rotating. The function value becomes the radius.

Prerequisite Map

Prerequisites
Volume by SlicingArea of Circle
→
This skill
Disk Method
→
Unlocks
Washer MethodShell Method

Before You Start

Self-check: Can you answer these questions? If not, review the linked prerequisites first.

Question If you struggle...
Explain why $V = \int_a^b A(x)\,dx$ gives the volume of a solid Review Volume by Slicing
What is the area of a circle with radius $r$? This is essential—the answer is $\pi r^2$
If $f(x) = \sqrt{x}$, what is $[f(x)]^2$? This simplification is used constantly in disk problems
Refresh: The General Slicing Formula

From Volume by Slicing:

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

where $A(x)$ is the cross-sectional area at position $x$. The disk method is the special case where every cross-section is a circle.

Refresh: Why Squaring Matters

If the radius of a circular cross-section is $r = f(x)$, then the area is:

$$A(x) = \pi r^2 = \pi [f(x)]^2$$

Common simplifications:

Quick Reference

Property Value
Concept Applications of Integration
Course MATH162
Section Stewart 5.2
Difficulty Intermediate
Time ~20 minutes

Key Concepts

The Disk Method Formula

When a region is rotated about the $x$-axis, each cross-section is a disk with radius equal to the function value $f(x)$:

$$\boxed{V = \int_a^b \pi [f(x)]^2\,dx = \pi\int_a^b [f(x)]^2\,dx}$$

    y
    |
f(x)+---.      Rotate about x-axis
    |   |      ─────────────────→
    |   |
    +---|---→ x               ┌──────────┐
        x                     │  ○───○   │  Solid of revolution
                              │ / disk \ │
                              │○   ●   ○│  ← disk has radius f(x)
                              │ \     / │
                              │  ○───○   │
                              └──────────┘

Why the Formula Works

Why does this simple formula capture the volume of such complex shapes? From the general slicing principle $V = \int_a^b A(x)\,dx$:

  1. Each cross-section perpendicular to the $x$-axis is a circle
  2. The radius of the circle at position $x$ is $r = f(x)$
  3. The area of the circle is $A(x) = \pi r^2 = \pi[f(x)]^2$
  4. Substituting into the slicing formula gives the disk formula

Rotation About the $y$-Axis

If rotating about the $y$-axis, express everything in terms of $y$:

$$V = \pi\int_c^d [g(y)]^2\,dy$$

where $g(y)$ gives the horizontal distance from the $y$-axis to the curve.

The Algorithm

  1. Sketch the region and identify the axis of rotation
  2. Identify the radius function (distance from axis to curve)
  3. Find the limits (where the solid begins and ends along the axis)
  4. Set up the integral $V = \pi\int (\text{radius})^2\,d(\text{axis variable})$
  5. Evaluate the integral

Common Setups

Axis of Rotation Radius Volume Formula
$x$-axis $r = f(x)$ $V = \pi\int_a^b [f(x)]^2\,dx$
$y$-axis $r = g(y)$ $V = \pi\int_c^d [g(y)]^2\,dy$
Line $y = k$ $r = f(x) - k$ or $k - f(x)$ $V = \pi\int_a^b [f(x) - k]^2\,dx$
Line $x = h$ $r = g(y) - h$ or $h - g(y)$ $V = \pi\int_c^d [g(y) - h]^2\,dy$

Note: The radius is always the distance from the axis to the curve. Use absolute value thinking: if the curve is above the axis, radius = curve − axis; if below, radius = axis − curve.

Common Pitfalls

Mistake Why It's Wrong Correction
Forgetting to square Area is $\pi r^2$, not $\pi r$ Always write $[f(x)]^2$ explicitly
Using diameter instead of radius The formula uses radius, not diameter If given diameter $d$, use $r = d/2$
Wrong limits when rotating about $y$-axis Limits should be $y$-values, not $x$-values Convert the problem to functions of $y$
Using disks when there's a hole If region doesn't touch axis, solid has a hollow core Use washers instead (see Washer Method)

Practice Problems

Level 1 Basic Disk Setup

Find the volume of the solid obtained by rotating the region under $y = x$ from $x = 0$ to $x = 2$ about the $x$-axis.

Thought Process

The region is a triangle under the line $y = x$. When rotated about the $x$-axis, each vertical slice becomes a disk with radius $r = x$. Apply the disk formula directly.

Show Answer

The radius at position $x$ is $r = x$.

$$V = \pi\int_0^2 x^2\,dx = \pi\left[\frac{x^3}{3}\right]_0^2 = \pi \cdot \frac{8}{3} = \boxed{\frac{8\pi}{3}}$$

Level 2 Region Under a Root Function

Find the volume of the solid obtained by rotating the region under $y = \sqrt{2x}$ from $x = 0$ to $x = 2$ about the $x$-axis.

Thought Process

The radius at position $x$ is $r = \sqrt{2x}$. When we square this for the area formula, we get $r^2 = 2x$, which is easy to integrate. The factor of 2 inside the root stays as a coefficient.

Show Answer

The radius is $r = \sqrt{2x}$, so $r^2 = 2x$.

$$V = \pi\int_0^2 (\sqrt{2x})^2\,dx = \pi\int_0^2 2x\,dx = 2\pi\left[\frac{x^2}{2}\right]_0^2 = 2\pi \cdot 2 = \boxed{4\pi}$$

Sanity check: The region has width 2 and maximum height $\sqrt{4} = 2$. A cylinder with radius 2 and height 2 has volume $\pi(4)(2) = 8\pi$. Our parabolic shape fills about half that space, so $4\pi$ is reasonable.

Level 3 Rotation About the $y$-Axis

Find the volume of the solid obtained by rotating the region bounded by $y = x^2$, $y = 9$, and $x = 0$ (first quadrant only) about the $y$-axis.

Thought Process

Since we're rotating about the $y$-axis, we should integrate with respect to $y$. Solve for $x$ in terms of $y$: from $y = x^2$, we get $x = \sqrt{y}$ (taking positive root for first quadrant). The radius of each horizontal disk is $r = x = \sqrt{y}$.

Show Answer

Step 1: Express $x$ in terms of $y$: $x = \sqrt{y} = y^{1/2}$.

Step 2: The region extends from $y = 0$ to $y = 9$.

Step 3: The radius at height $y$ is $r = y^{1/2}$.

Step 4: Apply the disk formula: $$V = \pi\int_0^9 (y^{1/2})^2\,dy = \pi\int_0^9 y\,dy$$

$$= \pi\left[\frac{y^2}{2}\right]_0^9 = \pi \cdot \frac{81}{2} = \boxed{\frac{81\pi}{2}}$$

Sanity check: This is a paraboloid. At height 9, the radius is 3, so the "bounding cylinder" has volume $\pi(9)(9) = 81\pi$. Our paraboloid fills exactly half of it—which confirms the factor of $\frac{1}{2}$ in any paraboloid rotated about its axis.

Level 4 Rotation About a Horizontal Line

Find the volume of the solid obtained by rotating the region bounded by $y = x^2$ and $y = 4$ about the line $y = 5$.

Thought Process

Key observation: The axis of rotation ($y = 5$) is above the region. Since the region doesn't touch the axis, the solid will have a hollow core—this is actually a washer problem, not a pure disk problem.

Why this matters: The region is bounded by $y = x^2$ (bottom) and $y = 4$ (top). When rotated about $y = 5$:

  • The outer radius is the distance from $y = 5$ to the lower boundary ($y = x^2$)
  • The inner radius is the distance from $y = 5$ to the upper boundary ($y = 4$)

Decision: This problem intentionally transitions from the disk method to recognize when washers are needed. The key insight is: if the region doesn't touch the axis, there's a hole in the solid.

Show Answer

Important observation: This problem actually requires the washer method because the region doesn't touch the axis of rotation. Here's why—and how to solve it correctly.

The region is bounded by $y = x^2$ (parabola, opens up) and $y = 4$ (horizontal line). They intersect where $x^2 = 4$, so $x = \pm 2$.

When rotated about $y = 5$, at each $x$-value:

  • The outer radius is from $y = 5$ to the parabola: $R = 5 - x^2$
  • The inner radius is from $y = 5$ to the line $y = 4$: $r = 5 - 4 = 1$

The cross-sectional area (a washer) is: $$A(x) = \pi R^2 - \pi r^2 = \pi(5 - x^2)^2 - \pi(1)^2$$

$$V = \pi\int_{-2}^{2}\left[(5 - x^2)^2 - 1\right]\,dx$$

Expanding $(5 - x^2)^2 = 25 - 10x^2 + x^4$:

$$V = \pi\int_{-2}^{2}(25 - 10x^2 + x^4 - 1)\,dx = \pi\int_{-2}^{2}(24 - 10x^2 + x^4)\,dx$$

Since the integrand is even: $$V = 2\pi\int_0^2(24 - 10x^2 + x^4)\,dx = 2\pi\left[24x - \frac{10x^3}{3} + \frac{x^5}{5}\right]_0^2$$

$$= 2\pi\left(48 - \frac{80}{3} + \frac{32}{5}\right) = 2\pi\left(\frac{720 - 400 + 96}{15}\right) = 2\pi \cdot \frac{416}{15} = \boxed{\frac{832\pi}{15}}$$

Takeaway: When the region doesn't touch the axis, you need washers, not disks.

Level 5 Deriving the Cone Volume Formula

A right circular cone has base radius $R$ and height $h$.

  1. Set up the cone as a solid of revolution by identifying an appropriate linear function and axis of rotation.
  2. Use the disk method to derive the formula $V = \frac{1}{3}\pi R^2 h$.
  3. Explain geometrically why the factor of $\frac{1}{3}$ appears (hint: compare to a cylinder).
Thought Process

Position the cone with vertex at the origin and axis along the $x$-axis. The radius increases linearly from 0 (at vertex) to $R$ (at height $h$). Find the linear function $r(x)$ that describes this, then apply the disk formula.

Show Answer

(a) Setting up the solid:

Place the vertex at the origin with the axis along the positive $x$-axis. The cone extends from $x = 0$ to $x = h$.

At $x = 0$: radius = 0 At $x = h$: radius = $R$

The radius increases linearly, so: $$r(x) = \frac{R}{h}x$$

This is a line through the origin with slope $\frac{R}{h}$.

The cone is obtained by rotating the line $y = \frac{R}{h}x$ (for $0 \leq x \leq h$) about the $x$-axis.

(b) Applying the disk method:

$$V = \pi\int_0^h [r(x)]^2\,dx = \pi\int_0^h \left(\frac{R}{h}x\right)^2\,dx$$

$$= \pi \cdot \frac{R^2}{h^2}\int_0^h x^2\,dx = \frac{\pi R^2}{h^2}\left[\frac{x^3}{3}\right]_0^h$$

$$= \frac{\pi R^2}{h^2} \cdot \frac{h^3}{3} = \boxed{\frac{1}{3}\pi R^2 h}$$

(c) Why the factor of $\frac{1}{3}$:

A cylinder with the same base and height has volume $V_{\text{cyl}} = \pi R^2 h$.

The cone has volume $V_{\text{cone}} = \frac{1}{3}\pi R^2 h = \frac{1}{3}V_{\text{cyl}}$.

Geometric explanation: The radius of the cone grows linearly from 0 to $R$, while the cylinder's radius is constant at $R$. When we square the radius for the disk areas, we get:

  • Cylinder: constant $R^2$
  • Cone: grows as $x^2$ (from squaring $\frac{R}{h}x$)

Integrating $x^2$ gives $\frac{x^3}{3}$—the cubic growth with the $\frac{1}{3}$ factor is exactly what accounts for the cone having one-third the volume.

Alternatively: the average value of $x^2$ over $[0, h]$ is $\frac{h^2}{3}$, compared to $h^2$ for a constant. This ratio of $\frac{1}{3}$ appears in all "pyramid-like" solids that taper to a point.

Mastery Checklist

Mental Model

The Stack of Coins: Imagine stacking infinitely many coins (circular disks) of varying radii. The radius of each coin is determined by the function value at that position. The volume is the "sum" of the volumes of all these infinitesimally thin coins: $\pi r^2 \cdot dx$ for each coin, integrated over the entire stack.

Connections

Looking back:

Looking ahead:

When to use disks:

Common error: Forgetting to square the radius. If $r = \sqrt{x}$, then $r^2 = x$. Don't write the integrand as $\pi\sqrt{x}$ when it should be $\pi x$.

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Last updated: 2026-01-23