Applications of Integration: Volumes of Solids of Revolution (Disk and Washer Methods)

Applications of Integration

Volumes of Solids of Revolution: Disk and Washer Methods

In calculus, the disk and washer methods are powerful techniques for finding the volume of a solid generated by revolving a region around a given axis. These methods use definite integrals to sum up the volumes of infinitesimally thin disks or washers.

Key Concepts

• Disk Method: Used when the solid has no hole (the region is flush against the axis of revolution). The cross-section perpendicular to the axis is a disk.

• Washer Method: Used when the solid has a hole (the region does not touch the axis of revolution). The cross-section is a washer (a disk with a hole).

• Outer radius (R): The distance from the axis of revolution to the outer edge of the region.

• Inner radius (r): The distance from the axis of revolution to the inner edge (if present; for the disk method, r = 0).

• Volume Formulas:

  • Disk: $V = \pi \int_a^b [R(x)]^2 dx$

  • Washer: $V = \pi \int_a^b \left([R(x)]^2 - [r(x)]^2\right) dx$

Example 1: Region Bounded by $y = (x-3)^2$, $x = 0$, and $y = 0$

Let R be the region bounded by $y = (x-3)^2$, $x = 0$, and $y = 0$. We compute the volume of the solid obtained by revolving R about various axes.

Revolution About the x-axis

• Outer radius: $(x-3)^2$

• Inner radius: $0$

• Volume integral:

$V = \int_0^3 \pi \left((x-3)^2\right)^2 dx$

Revolution About the Line $y = 9$

• Outer radius: $9 - 0 = 9$

• Inner radius: $9 - (x-3)^2$

• Volume integral:

$V = \int_0^3 \left[\pi (9)^2 - \pi \left(9 - (x-3)^2\right)^2\right] dx$

Revolution About the y-axis

• We need $x$ as a function of $y$: $y = (x-3)^2 \implies x = 3 \pm \sqrt{y}$

• Based on the region, use $x = 3 - \sqrt{y}$

• Outer radius: $3 - \sqrt{y}$

• Inner radius: $0$

• Volume integral:

$V = \int_0^9 \pi \left(3 - \sqrt{y}\right)^2 dy$

Revolution About the Line $x = 4$

• Outer radius: $4 - 0 = 4$

• Inner radius: $4 - (3 - \sqrt{y}) = 1 + \sqrt{y}$

• Volume integral:

$V = \int_0^9 \left[\pi (4)^2 - \pi (1 + \sqrt{y})^2\right] dy$

Example 2: Triangular Region Bounded by $y = \frac{x}{2}$, $y = 15 - x$, and $x = -2$

Let R be the triangular region bounded by $y = \frac{x}{2}$, $y = 15 - x$, and $x = -2$. Find the volume of the solid of revolution obtained by revolving R about various lines.

Intersection Points

• Find where $y = 15 - x$ and $y = \frac{x}{2}$ intersect:

$15 - x = \frac{x}{2} \implies 15 = \frac{3}{2}x \implies x = 10$

Revolution About the Line $y = -1$

• Outer radius: $15 - x - (-1) = 16 - x$

• Inner radius: $\frac{x}{2} - (-1) = \frac{x}{2} + 1$

• Volume integral:

$V = \int_{-2}^{10} \left[\pi (16 - x)^2 - \pi \left(\frac{x}{2} + 1\right)^2\right] dx$

Revolution About the Line $y = 18$

• Outer radius: $18 - \frac{x}{2}$

• Inner radius: $18 - (15 - x) = 3 + x$

• Volume integral:

$V = \int_{-2}^{10} \left[\pi \left(18 - \frac{x}{2}\right)^2 - \pi (3 + x)^2\right] dx$

Revolution About the Line $x = 12$

• Express $x$ in terms of $y$:

  • $y = 15 - x \implies x = 15 - y$

  • $y = \frac{x}{2} \implies x = 2y$

• Find intersection points for $y$:

  • $15 - y = -2 \implies y = 17$

  • $15 - y = 2y \implies 3y = 15 \implies y = 5$

  • $2y = -2 \implies y = -1$

• For $-1 < y < 5$:

  • Outer radius: $12 - (-2) = 14$

  • Inner radius: $12 - 2y$

• For $5 < y < 17$:

  • Outer radius: $14$

  • Inner radius: $12 - (15 - y) = y - 3$

• Volume integral:

$V = \int_{-1}^{5} \left[\pi (14)^2 - \pi (12 - 2y)^2\right] dy + \int_{5}^{17} \left[\pi (14)^2 - \pi (y - 3)^2\right] dy$

Summary Table: Disk and Washer Method Setup

Axis of Revolution	Outer Radius	Inner Radius	Limits of Integration	Variable
x-axis	$(x-3)^2$	$0$	$x: 0 \to 3$	$x$
y = 9	$9$	$9 - (x-3)^2$	$x: 0 \to 3$	$x$
y-axis	$3 - \sqrt{y}$	$0$	$y: 0 \to 9$	$y$
x = 4	$4$	$1 + \sqrt{y}$	$y: 0 \to 9$	$y$

Example Application

• Physical context: These methods are used in engineering and physics to compute the volume of tanks, pipes, and other objects with rotational symmetry.

• Key steps:

  1. Sketch the region and axis of revolution.

  2. Express the outer and inner radii as functions of the variable of integration.

  3. Set up the definite integral using the disk or washer formula.

  4. Evaluate the integral to find the volume.

Additional info: The notes above expand on the original drill problems by providing general formulas, step-by-step setup, and a summary table for clarity. The examples illustrate how to handle both horizontal and vertical axes of revolution, and how to switch between $x$ and $y$ as the variable of integration as needed.