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Lesson 2: Volumes: Disk and Washer Methods

Estimated time: 45-55 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Solids of Revolution

A solid of revolution is formed when a plane region is revolved around a line (the axis of rotation). Think of a potter's wheel: a 2D profile spins to create a 3D shape.

To find the volume of such a solid, we slice it into thin cross-sections, compute the area of each slice, and integrate.

The Disk Method

When a region between y = f(x) and the x-axis is revolved around the x-axis, each cross-section is a circular disk.

Disk Method (rotation about the x-axis)

If the region under y = f(x) on [a, b] is revolved about the x-axis:

V = π ∫ab [f(x)]² dx

Each disk has radius r = f(x) and thickness dx. Area of disk = πr².

Example 1: Cone via Disk Method

Find the volume of the solid obtained by revolving y = x on [0, 3] about the x-axis.

Solution: V = π ∫03 x² dx = π [x³/3]03 = π(9) = .

This is indeed the volume of a cone with radius 3 and height 3: (1/3)πr²h = (1/3)π(9)(3) = 9π.

Example 2: Paraboloid

Revolve y = √x on [0, 4] about the x-axis.

Solution: V = π ∫04 (√x)² dx = π ∫04 x dx = π [x²/2]04 = .

The Washer Method

When there is a gap between the region and the axis of rotation (or the region is between two curves), the cross-sections are washers (disks with holes).

Washer Method (rotation about the x-axis)

If the region between y = f(x) (outer curve) and y = g(x) (inner curve), with f(x) ≥ g(x) ≥ 0, is revolved about the x-axis:

V = π ∫ab {[f(x)]² − [g(x)]²} dx

Outer radius R = f(x), inner radius r = g(x). Washer area = π(R² − r²).

Example 3: Washer Method

Find the volume when the region between y = x² and y = x on [0, 1] is revolved about the x-axis.

Step 1: On [0, 1], x ≥ x², so outer radius R = x, inner radius r = x².

Step 2: V = π ∫01 [x² − x4] dx = π [x³/3 − x5/5]01 = π(1/3 − 1/5) = 2π/15.

Rotation About the y-Axis

For rotation about the y-axis, express x as a function of y and integrate with respect to y.

Disk/Washer Method (rotation about the y-axis)

V = π ∫cd [f(y)]² dy  (disk)

V = π ∫cd {[f(y)]² − [g(y)]²} dy  (washer)

Example 4: Rotation About the y-Axis

Find the volume when y = x² (i.e., x = √y) on [0, 4] is revolved about the y-axis.

Solution: The region from y = 0 to y = 4. Radius = x = √y.

V = π ∫04 (√y)² dy = π ∫04 y dy = π [y²/2]04 = .

Rotation About Other Lines

When the axis of rotation is not the x-axis or y-axis, you must adjust the radii by measuring the distance from the curve to the axis of rotation.

Example 5: Rotation About y = −1

Find the volume when y = √x on [0, 4] is revolved about y = −1.

Solution: The outer radius from y = −1 to y = √x is R = √x + 1. The inner radius from y = −1 to y = 0 is r = 1.

V = π ∫04 [(√x + 1)² − 1²] dx = π ∫04 [x + 2√x + 1 − 1] dx

= π ∫04 (x + 2√x) dx = π [x²/2 + (4/3)x3/2]04 = π(8 + 32/3) = 56π/3.

Setting Up Volume Integrals: A Checklist

  1. Sketch the region and identify the axis of rotation.
  2. Draw a representative slice perpendicular to the axis of rotation.
  3. Identify R and r (outer and inner radius) as distances from the axis to the curves.
  4. Set up the integral: π ∫ (R² − r²) with appropriate variable and limits.
  5. Evaluate the integral.

Key Takeaways

Check Your Understanding

1. Find the volume when y = x² on [0, 2] is revolved about the x-axis.

Answer: V = π ∫02 x4 dx = π [x5/5]02 = 32π/5.

2. Find the volume when the region between y = x and y = x² on [0, 1] is revolved about the x-axis.

Answer: Washer: V = π ∫01 (x² − x4) dx = π(1/3 − 1/5) = 2π/15.

3. Set up the integral for the volume when y = 4 − x² (for y ≥ 0) is revolved about the y-axis.

Answer: x = √(4 − y), y ranges from 0 to 4. V = π ∫04 (4 − y) dy = π[4y − y²/2]04 = .

Ready for More?

Next Lesson

Lesson 3 introduces the shell method, an alternative approach to volumes of revolution.

Start Lesson 3