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Disk Method

Disk Method – definition and example, AP Calculus AB math glossary – Mathovore

Key takeaway: The disk method finds the volume of a solid formed by rotating a region about an axis by adding up thin circular disks, each with area pi times radius squared.

Grade 12 – AP Calculus ABAP Calculus AB Unit 8Calculus

Definition of Disk Method

Definition

The disk method computes the volume of a solid of revolution. Rotate a region around a line that forms one of its edges, and slice the resulting solid perpendicular to the axis. Every slice is a thin circular disk, a cylinder with tiny thickness.

A disk’s volume is the area of its circular face times its thickness. Adding all disks along the axis is a definite integral. The radius of each disk is the distance from the axis to the curve, so it changes with position.

Formula and notation

Rotating \(y=R(x)\) about the \(x\)-axis between \(x=m\) and \(x=n\):

\[ V=\pi\int_{m}^{n}\big[R(x)\big]^2\,dx. \]

The radius \(R(x)\) is squared inside the integral, and \(\pi\) comes outside. When the region does not touch the axis, the washer method subtracts a hole.

r = sqrt(3x) = 3y = sqrt(3x)r = 4.24disk at x = 3
Rotating the curve about the x-axis builds a solid made of thin disks whose radius is the curve height.

Key facts

  • The radius is the distance from the axis of rotation to the curve, not the curve’s height in general.
  • In the disk method, square the radius before integrating, never after.
  • Use dy and functions of y when rotating around a vertical line.
  • Volume units are cubic units; the integral of an area gives a volume.

Disk Method example

Problem

The region under \(y=\sqrt{3x}\) from \(x=0\) to \(x=6\) is rotated about the \(x\)-axis. Find the volume of the solid.

  1. The radius of each disk is \(R(x)=\sqrt{3x}\), so \([R(x)]^2=3x\).

  2. Set up: \(V=\pi\int_{0}^{6}3x\,dx\).

  3. Integrate: \(\pi[\frac{3x^2}{2}]_{0}^{6}=\pi\cdot\frac{3\cdot 36}{2}=54\pi\).

The volume is \(54\pi\approx 169.6\) cubic units.

Common mistake

Watch out

A student forgets to square the radius and evaluates \(\pi\int_{0}^{6}\sqrt{3x}\,dx=12\sqrt{2}\,\pi\approx 53.3\).

Why it is wrong

Each disk has an area \(\pi r^2\), so the radius must be squared. Without the square, the integrand has the wrong units for an area and the volume is far too small. With \(r^2=3x\), the disk method gives \(54\pi\approx 169.6\).

Related terms and lessons

Browse the glossary: A–Z list · AP Calculus AB vocabulary · Calculus

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