The Disc Method The Disc Method

Concept

We will now calculate the volumes of solids formed by rotating a function around the x or y-axis.

Rotating around an axis creates disc cross-sections, each with a radius of

f(x)
.
\begin{align*} A &= \pi r^2\\ &= \pi f(x)^2 \\ \end{align*}
For example, here is a volume formed by rotating
y=x^2
around the x-axis:

Examples

Example 1: Revolving Around the x-Axis
Find the volume of the solid formed by rotating

y = 2x
around the x-axis on
x=[0,1]
.
\begin{align*} \text{Volume} &= \int_{0}^{1} A \cdot d \\ &= \int_{0}^{1} \pi(2x)^2 \, dx \\ &= 4\pi \int_{0}^{1} x^2 \, dx \\ &= 4\pi \left[ \frac{x^3}{3} \right]_0^1 \\ &= 4\pi \left( \frac{1}{3} - 0 \right) = \frac{4\pi}{3} \approx 4.19 \, \text{units}^3 \end{align*}

Example 2: Revolving Around the y-Axis
Find the volume of the solid formed by rotating
x = \sqrt{y}
around the y-axis on
y=[0,4]
.
\begin{align*} \text{Volume} &= \int_{0}^{1} A \cdot d \\ &= \int_{0}^{4} \pi(\sqrt{y})^2 \, dy \\ &= \pi \int_{0}^{4} y \, dy \\ &= \pi \left[ \frac{y^2}{2} \right]_0^4 \\ &= \pi \left( \frac{16}{2} - 0 \right) = 8\pi \approx 25.13 \, \text{units}^3 \end{align*}


Example 3: Cylinder Verification
Rotate
y = 5
around the x-axis from
x = 0
to
x = 4
. Verify the volume matches the cylinder formula.
\begin{align*} \text{Volume} &= \int_{0}^{1} A \cdot d \\ &= \int_{0}^{4} \pi \cdot 5^2 \, dx \\ &= 25\pi \int_{0}^{4} dx \\ &= 25\pi (4 - 0) = 100\pi \approx 314.16 \, \text{units}^3 \end{align*}
This matches the formula for the volume of a cylinder:
\begin{align*} V &= \pi r^2 h \\ &= \pi (5)^2 (4) \\ &= 100\pi \end{align*}

Revolving Around Other Axes

So far, we have revolved our discs around the x and y-axes. However, we can also calculate the volumes of discs around any line parallel to the axes.

The simplest way to do so is to shift the function up or down so that the axis of rotation is the x or y-axis. Then, we can use the same formulas as before.

Horizontal Line: If we are revolving around

y = k
:
V = \pi \int_{a}^{b} (f(x) - k)^2 \, dx
Vertical Line: If we are revolving around
x = h
:
V = \pi \int_{c}^{d} (g(y) - h)^2 \, dy

Examples

Example 4: Revolving Around y = -1
Rotate

y = x^2
around
y = -1
on
[0,1]
:
\begin{align*} V &= \pi \int_{0}^{1} (x^2 - (-1))^2 \, dx \\ &= \pi \int_{0}^{1} (x^4 + 2x^2 + 1) \, dx \\ &= \pi \left[ \frac{x^5}{5} + \frac{2x^3}{3} + x \right]_0^1 \\ &= \frac{28\pi}{15} \approx 5.89 \, \text{units}^3 \end{align*}
Example 5: Revolving Around x = 2
Rotate
x = \sqrt{y}
around
x = 2
on
[0,4]
:
\begin{align*} V &= \pi \int_{0}^{4} (2 - \sqrt{y})^2 \, dy \\ &= \pi \int_{0}^{4} (4 - 4\sqrt{y} + y) \, dy \\ &= \pi \left[ 4y - \frac{8}{3}y^{3/2} + \frac{y^2}{2} \right]_0^4 \\ &= \frac{8\pi}{3} \approx 8.38 \, \text{units}^3 \end{align*}
Example 6: Crossing the Axis
Rotate
y = 2x
around
y = 1
from
x = 0
to
x = 1
:
\begin{align*} V &= \pi \int_{0}^{1} (2x - 1)^2 \, dx \\ &= \pi \int_{0}^{1} (4x^2 - 4x + 1) \, dx \\ &= \pi \left[ \frac{4x^3}{3} - 2x^2 + x \right]_0^1 \\ &= \pi \left( \frac{4}{3} - 2 + 1 \right) \\ &= \frac{\pi}{3} \approx 1.05 \, \text{units}^3 \end{align*}