Volumes of Revolution (Disk/Washer)

Overview

When a region is rotated around an axis, it creates a solid of revolution. The disk and washer methods calculate volume by slicing perpendicular to the axis of rotation.

Disk Method

For rotation around the x-axis (no hole in the middle):

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

where $R(x)$ is the distance from the curve to the axis.

For rotation around the y-axis:

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

Washer Method

When there's a hole (region between two curves):

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

where:

• $R(x)$ = outer radius (distance to outer curve)
• $r(x)$ = inner radius (distance to inner curve)

Key Formulas

Rotation around x-axis

$V = \pi \int_a^b (\text{outer}^2 - \text{inner}^2) \, dx$

Rotation around y-axis

$V = \pi \int_c^d (\text{outer}^2 - \text{inner}^2) \, dy$

Rotation around $y = k$ (horizontal line)

$V = \pi \int_a^b \left[(f(x) - k)^2 - (g(x) - k)^2\right] dx$

Rotation around $x = k$ (vertical line)

$V = \pi \int_c^d \left[(f(y) - k)^2 - (g(y) - k)^2\right] dy$

Examples

Example 1: Disk method around x-axis

Find the volume when $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated around the x-axis.

$V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx$

$= \pi \left[\frac{x^2}{2}\right]_0^4 = \pi (8 - 0) = 8\pi$

Example 2: Disk method around y-axis

Find the volume when $x = y^2$ from $y = 0$ to $y = 2$ is rotated around the y-axis.

$V = \pi \int_0^2 (y^2)^2 \, dy = \pi \int_0^2 y^4 \, dy$

$= \pi \left[\frac{y^5}{5}\right]_0^2 = \pi \left(\frac{32}{5}\right) = \frac{32\pi}{5}$

Example 3: Washer method

Find the volume when the region between $y = x^2$ and $y = x$ is rotated around the x-axis.

Intersection: $x^2 = x \Rightarrow x = 0, 1$

Outer radius: $R = x$ (top curve) Inner radius: $r = x^2$ (bottom curve)

$V = \pi \int_0^1 (x^2 - x^4) \, dx = \pi \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1$

$= \pi \left(\frac{1}{3} - \frac{1}{5}\right) = \pi \left(\frac{2}{15}\right) = \frac{2\pi}{15}$

Example 4: Rotation around $y = -1$

Rotate the region bounded by $y = x^2$, $y = 0$, $x = 1$ around $y = -1$.

Outer radius: $R = x^2 - (-1) = x^2 + 1$ Inner radius: $r = 0 - (-1) = 1$

$V = \pi \int_0^1 \left[(x^2 + 1)^2 - 1^2\right] dx$

$= \pi \int_0^1 [x^4 + 2x^2 + 1 - 1] \, dx = \pi \int_0^1 (x^4 + 2x^2) \, dx$

$= \pi \left[\frac{x^5}{5} + \frac{2x^3}{3}\right]_0^1 = \pi \left(\frac{1}{5} + \frac{2}{3}\right) = \pi \left(\frac{13}{15}\right) = \frac{13\pi}{15}$

Example 5: Rotation around $x = 2$

Rotate the region bounded by $y = x$, $y = 0$, $x = 1$ around $x = 2$.

Convert to functions of $y$: $x = y$, from $y = 0$ to $y = 1$

Outer radius: $R = 2 - 0 = 2$ (distance to $x = 0$) Inner radius: $r = 2 - y$ (distance to $x = y$)

$V = \pi \int_0^1 [4 - (2-y)^2] \, dy = \pi \int_0^1 [4 - 4 + 4y - y^2] \, dy$

$= \pi \int_0^1 (4y - y^2) \, dy = \pi \left[2y^2 - \frac{y^3}{3}\right]_0^1$

$= \pi \left(2 - \frac{1}{3}\right) = \frac{5\pi}{3}$

Example 6: Sphere

Rotate $y = \sqrt{r^2 - x^2}$ (semicircle) around x-axis.

$V = \pi \int_{-r}^r \left(\sqrt{r^2 - x^2}\right)^2 dx = \pi \int_{-r}^r (r^2 - x^2) \, dx$

$= \pi \left[r^2 x - \frac{x^3}{3}\right]_{-r}^r = \pi \left[\left(r^3 - \frac{r^3}{3}\right) - \left(-r^3 + \frac{r^3}{3}\right)\right]$

$= \pi \left[2r^3 - \frac{2r^3}{3}\right] = \pi \left(\frac{4r^3}{3}\right) = \frac{4}{3}\pi r^3$