Disk Method
In short: Disk method measures the volume of a solid made by spinning a flat region about a line, treating the solid as a stack of ultra-thin round slices whose radius is the curve’s height.
AP Calculus ABCalculusAP Calculus AB Unit 8
What is Disk Method?
The disk method turns a spinning region into a stack of coins. When the area under \(y = f(x)\) from \(x = a\) to \(x = b\) is rotated about the x-axis, a slice at position \(x\) becomes a thin disk of radius \(f(x)\) and thickness \(dx\).
A disk has volume \(\pi r^2 \cdot \text{thickness}\), so adding all of them gives
\[V = \pi\int_a^b \left[f(x)\right]^2 dx\]
The radius is squared before integrating, and the factor \(\pi\) comes out front.
How to spot it
- A region beneath a graph is rotated about the x-axis, and the capacity of the resulting shape is wanted.
- The solid looks like a vase, a bowl, a horn or a cone and has no hole in the middle, so the disk method fits.
- A formula involves pi times an integral of a squared function.
- The curve touches the axis of rotation, so every cross-section is a full disk.
The method
- Sketch the region and the axis of rotation; the radius is the gap between the axis and the curve.
- Write the radius as a function, such as \(r = f(x)\), and square it.
- Set up the disk method integral \(V = \pi\int_a^b r^2\,dx\) using the interval of the region.
- Integrate, evaluate, multiply by \(\pi\), and state the cubic units.
Disk Method: solved example
Problem
A glassblower’s vase profile is the curve \(y = x^2\) for \(0 \le x \le 2\) (in inches), turned about the x-axis. How much liquid fits inside?
Solution
The radius is \(x^2\), so \(r^2 = x^4\).
\(V = \pi\int_0^2 x^4\,dx = \pi\left[\frac{x^5}{5}\right]_0^2 = \pi\cdot\frac{32}{5}\).
The volume is \(\frac{32\pi}{5}\) cubic units, about 20.11.
Try it
Question 1
A rectangle with corners at \((0, 0)\), \((5, 0)\), \((5, 3)\) and \((0, 3)\) is spun around the x-axis. Name the solid and compute its volume with an integral.
Show the answer
It makes a can of radius 3 and length 5. \(V = \pi\int_0^5 9\,dx = 45\pi\), the same as \(\pi r^2 h\).
Question 2
Spin the area beneath \(y = \sqrt{x}\), for \(x\) from 0 to 9, around the x-axis. What volume results?
Show the answer
\(V = \pi\int_0^9 x\,dx = \pi\cdot\frac{81}{2} = 40.5\pi\) cubic units.
Disk Method vs. Area Between Curves
The area formula subtracts curves and has no \(\pi\); its result is flat, measured in square units. The disk method squares one radius, multiplies by \(\pi\), and yields a three-dimensional volume. A hollow solid would need an outer disk minus an inner one.
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