MATH 104, Sections 6.1 and 6.2 | Source: Rimmer lecture notes

Difficulty: Intermediate | Prerequisites: Basic integration (antiderivatives, definite integrals, the Fundamental Theorem of Calculus)

Big Picture

This material sits at the start of second-semester calculus (Calculus II). You already know how to find the area under a curve using definite integrals. Now you are doing the same thing in three dimensions: instead of summing up thin rectangular strips to get area, you sum up thin slices (disks, washers, or cylindrical shells) to get volume. This is one of the most heavily tested application topics in any Calculus II course, so the formulas and the decision of which method to use for a given problem are both worth drilling.

TL;DR

To find the volume of a solid, slice it into thin pieces whose cross-sectional area you can write as a function, then integrate that area over the relevant interval. When the solid is formed by revolving a region around an axis, three standard methods apply: the disk method (no gap between the region and the axis), the washer method (there is a gap), and the shell method (you integrate parallel to the axis of rotation instead of perpendicular to it).

Key Terms

Cross-section

The flat face you see when you slice a solid with a plane perpendicular to a given axis. In simple terms, imagine cutting a loaf of bread: each cut surface is a cross-section.

Solid of revolution

A three-dimensional solid formed by rotating a two-dimensional region around an axis. Think of it as spinning a flat shape on a potter's wheel to create a symmetric 3D object.

Disk method

A volume technique used when the region being revolved touches the axis of rotation (no gap). Each cross-section is a full circular disk. In simple terms, you are stacking coins of varying radius along the axis.

Washer method

A volume technique used when there is a gap between the region and the axis of rotation. Each cross-section is a ring (a disk with a hole). Think of it as stacking flat washers, like the metal rings you put on a bolt.

Cylindrical shell method

A volume technique that wraps the region into nested thin cylinders (like Russian nesting dolls) instead of slicing it into disks. You integrate parallel to the axis of rotation rather than perpendicular to it.

Axis of rotation

The line around which a region is revolved to create a solid of revolution. It may be the x-axis, y-axis, or any other horizontal or vertical line.

Outer radius / Inner radius

In the washer method, the outer radius is the distance from the axis of rotation to the farther curve, and the inner radius is the distance to the nearer curve. The difference between their squared values gives the washer's area.

Core Content: General Volume by Cross-Sections

The central idea is to slice the solid into thin pieces perpendicular to an axis, treat each piece as a cylinder whose base is the cross-section and whose height is a small thickness dx (or dy), then add up all the volumes.

• Each slice has volume approximately equal to A(x*) multiplied by dx, where A(x*) is the area of the cross-section at that point.

• Summing all slices gives an approximation: V is approximately the sum from i = 1 to n of A(x_i*) dx.

• Taking the limit as the number of slices approaches infinity converts the sum into the integral:

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

This is the master formula. Every volume-by-slicing problem reduces to: (1) identify the shape of each cross-section, (2) write its area as a function of x (or y), and (3) integrate over the interval.

Core Content: Disk Method

Use the disk method when a region is revolved around an axis and the region touches that axis (no gap).

Every cross-section perpendicular to the axis of rotation is a full circle. The area of that circle is pi times the radius squared, where the radius is determined by the curve being revolved.

Horizontal axis of rotation (integrating with respect to x)

• Cross-sectional area: A(x) = pi [r(x)]^2, where r(x) is the distance from the axis to the curve.

• Volume formula:

V = \pi \int_a^b \bigl[r(x)\bigr]^2\, dx

Vertical axis of rotation (integrating with respect to y)

• Cross-sectional area: A(y) = pi [r(y)]^2, where r(y) is the distance from the axis to the curve.

• Volume formula:

V = \pi \int_a^b \bigl[r(y)\bigr]^2\, dy

The key step is always identifying the radius as a function of the integration variable. When the axis of rotation is not the x-axis or y-axis itself (for example, y = 3), the radius is the distance from the curve to that line, not just the function value.

Core Content: Washer Method

Use the washer method when the region being revolved does not touch the axis of rotation, so each cross-section has a hole in the middle.

Each cross-section is a ring (washer). Its area equals the area of the outer circle minus the area of the inner circle.

Horizontal axis of rotation (integrating with respect to x)

• Draw a radius from the axis to the outer curve: this is r_out(x).

• Draw a radius from the axis to the inner curve: this is r_in(x).

• Volume formula:

V = \pi \int_a^b \Bigl(\bigl[r_{\text{out}}(x)\bigr]^2 - \bigl[r_{\text{in}}(x)\bigr]^2\Bigr)\, dx

Vertical axis of rotation (integrating with respect to y)

• Same logic, but the radii are functions of y.

• Volume formula:

V = \pi \int_a^b \Bigl(\bigl[r_{\text{out}}(y)\bigr]^2 - \bigl[r_{\text{in}}(y)\bigr]^2\Bigr)\, dy

A common error is to subtract the radii first and then square. That is wrong. You must square each radius separately and then subtract.

Core Content: Cylindrical Shells Method

Sometimes the disk or washer method leads to integrals that are impossible or impractical to set up. The shell method offers an alternative by slicing the solid into thin cylindrical shells rather than flat disks.

A cylindrical shell is the region between two concentric cylinders. If the average radius is r, the height is h, and the thickness is dr, the volume of one shell is:

V_{\text{shell}} = 2\pi r \cdot h \cdot \Delta r

This comes from the formula: circumference times height times thickness.

General shell formula:

V = \int_a^b 2\pi\,(\text{radius})\,(\text{height})\, dx

When to use shells vs. disks/washers:

• With the disk or washer method, the typical rectangle in your region is perpendicular to the axis of rotation.

• With the shell method, the typical rectangle is parallel to the axis of rotation.

Setting up a shell integral:

1. Identify the axis of rotation and draw a typical rectangle parallel to it.

1. The radius of the shell is the distance from the rectangle to the axis of rotation.

1. The height of the shell is determined by the function(s) bounding the region.

1. The thickness is dx or dy, depending on which variable you integrate with respect to.

Formulas at a Glance

Common Cross-Section Area Formulas

When the cross-sections are not circular (i.e. not a solid of revolution), you need the area formula for the given shape. These come up frequently in exam problems.

In each case, the side length s is typically determined by the distance between two curves at a given x-value, so s is a function of x.

Real-World Applications

Engineers use these volume calculations when designing containers, tanks, and vessels of non-standard shapes, for instance calculating the capacity of a tapered fuel tank or a dome-shaped reservoir. Manufacturing relies on the same principles to determine the amount of material needed to produce parts on a lathe, which literally works by revolving material around an axis, making the disk and shell methods a direct model of the physical process.

Common Misconceptions

• Students often subtract the radii before squaring in the washer method, writing pi times (r_out minus r_in)^2. This is incorrect. The correct formula squares each radius separately: pi times ([r_out]^2 minus [r_in]^2).

• When the axis of rotation is not the x-axis or y-axis, students frequently use the raw function value as the radius. The radius is the distance from the curve to the axis, so if you are revolving around y = 3, the radius from a curve y = f(x) is |f(x) minus 3|, not just f(x).

• Students sometimes set up a shell integral when the problem calls for a disk integral, or vice versa. The deciding factor is whether your typical rectangle is parallel or perpendicular to the axis of rotation.

• Forgetting to include the 2 pi factor in the shell method. The volume of a shell comes from 2 pi times radius times height times thickness, and leaving out 2 pi gives an answer that is off by a factor of 2 pi.

Why It Matters / Exam Flags

• Expect at least one problem requiring you to choose between disk/washer and shells. The exam often gives a region and axis where one method is straightforward and the other is messy.

• Problems with non-standard axes of rotation (e.g. revolving around x = 5 or y = -2) are a favourite test question because they force you to think about what the radius really is.

• Cross-section problems (square cross-sections, semicircular cross-sections, triangular cross-sections on a known base) appear regularly and test whether you know the area formulas.

• Setting up the integral correctly is typically worth more marks than evaluating it.

Quick Self-Test

1. True or False: In the washer method, you subtract the radii first and then square the result.

   • False. You square each radius separately, then subtract.

1. Fill in the blank: The volume of a cylindrical shell is 2 pi times ______ times ______ times ______.

   • radius, height, thickness

1. True or False: When using the shell method with a vertical axis of rotation, you integrate with respect to x.

   • True.

1. True or False: The disk method can be used when there is a gap between the region and the axis of rotation.

   • False. Use the washer method when there is a gap.

1. Fill in the blank: The area of an equilateral triangle cross-section with side length s is ______.

   • (sqrt 3 / 4) times s^2

Practice Q&A

Q: A solid has a circular base of radius 4. Every plane cross-section perpendicular to the x-axis is a square. Set up the integral for the volume.

A: The base is x^2 + y^2 = 16, so y = sqrt(16 minus x^2). The side length of the square at position x is 2y = 2 sqrt(16 minus x^2). The area is [2 sqrt(16 minus x^2)]^2 = 4(16 minus x^2). Volume = integral from -4 to 4 of 4(16 minus x^2) dx.

Q: Find the volume of the solid obtained by revolving y = sqrt(x), from x = 0 to x = 1, about the x-axis.

A: Using the disk method, the radius at position x is r(x) = sqrt(x). Volume = pi times integral from 0 to 1 of [sqrt(x)]^2 dx = pi times integral from 0 to 1 of x dx = pi times [x^2 / 2] from 0 to 1 = pi / 2.

Q: The region bounded by y = x^2 and y = x is revolved about the y-axis. Set up the volume using the shell method.

A: The curves intersect at x = 0 and x = 1. A typical vertical shell has radius x and height (x minus x^2). Volume = integral from 0 to 1 of 2 pi x(x minus x^2) dx = 2 pi times integral from 0 to 1 of (x^2 minus x^3) dx.

Q: When would you choose the shell method over the washer method?

A: When the washer method would require you to solve for x as a function of y (or vice versa) in a way that is difficult or impossible, or when the outer and inner radii come from the same curve so washer setup breaks down. Classic example: revolving y = sin(x^2) about the y-axis.

Connections to Other Topics

This material connects directly to Section 6.3 (Arc Length) and Section 6.4 (Surface Area of Revolution), which use the same slicing logic but measure length and surface area instead of volume. The shell method in particular reappears when studying centres of mass (Section 6.6), where you need to integrate radius times density over a region. All of these applications depend on the definite integral skills from Calculus I, so weak fundamentals there will compound here.

Related Terms / Search Tags

Volumes by slicing, disk method, washer method, cylindrical shells, shell method, solid of revolution, cross-sectional area, volume integral, revolving a region, axis of rotation, inner radius, outer radius, volume by cross-sections, stacking disks, nested cylinders, MATH 104, Calculus II volumes, Rimmer, Penn State Math 104