# Disk integration

In mathematics, in particular integral calculus, disk integration (the "disk method") is a means of calculating the volume of a solid of revolution. This makes use of the so-called "representative disk". The idea is that a "representative rectangle" (used in the most basic forms of integration -- such as ∫ x dx) can be rotated about the axis of revolution; thus generating such a disk.

As volume is the antiderivative of area, the integral can be used to find the volume, V, of an integrated "family" of disks. The necessary equation, for calculating such a volume, is slightly different depending on which axis is serving as the axis of revolution. These equations note that the area of a disk (one which has no height, and no volume) equals: pi (π) multiplied by the disk's squared radius (r2). The volume of a representative disk equals: πr2which is in turn multiplied by the disk's length (dx) or height (dy), that being some number approaching zero.

Horizontal axis of revolution
[itex]V = \pi \int [R(x)]^2 dx[itex]
Vertical axis of revolution
[itex]V = \pi \int [R(y)]^2 dy[itex]

For instance, consider the function f(x) = √(sin x), as it exists between x = 0 and x = π. If one imagines this function being rotated around the x-axis (so as to create a solid of revolution); then, the radius of that solid (for any value, x) is equal to √(sin x). Using the above formula, one can determine the solid to have a volume of: π ∫ [√(sin x)]2 dx -- when evaluated from 0 to π. The solid has a volume of 2π.

## The "washer" method

More than one representative disk, each with a different radius and both being concentric, can be used to find the volume of more complex solids of revolution. Some mathematicians have observed that, when one deletes the area which is shared by both disks, the resulting annulus shape looks like a washer.

Consider the region bounded by R(x) = √x and r(x) = x2. One can calculate the volume of a sphere of revolution, which has a radius of √x, as shown above -- π ∫ (√x)2 dx = π / 2. One should then calculate the volume of the "inner" sphere of revolution, that having a radius of x2 -- π ∫ (x2)2 dx = π / 5. By subtracting the inner area, π / 2 - π / 5, one obtains the volume of the bounded area: 3π / 10.

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