Disc Method Around the x-Axis (College Board AP® Calculus AB)

Study Guide

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Written by: Roger B

Reviewed by: Dan Finlay

Volume with disc method revolving around the x-axis

What is a volume of revolution around the x-axis?

A solid of revolution is formed when an area bounded by a function $y = f (x)$
(and other boundary equations) is rotated $2 π$ radians $(360 °)$ around the $x$ -axis
The volume of revolution is the volume of this solid

Example of a solid of revolution that is formed by rotating the area bounded by the function y=f(x), the lines x=a and x=b, and the x-axis about the x-axis

Be careful – the ’front’ and ‘back’ of this solid are flat
- they were created from straight (vertical) lines
- 3D sketches can be misleading

How can I use the disc method to calculate a volume of revolution around the x-axis?

For a continuous function $f$ , if the region bounded by
- the curve $y = f (x)$ and the $x$ -axis
- between $x = a$ and $x = b$
is rotated $2 π$ radians $(360 °)$ around the $x$ -axis, then the volume of revolution is
- $V = \int_{a}^{b} π y^{2} d x = π \int_{a}^{b} y^{2} d x$
  - Note that $y$ is a function of $x$
If $x = a$ and $x = b$ are not stated in a question, these boundaries could involve
- the $y$ -axis ( $x = 0$ )
- and/or an $x$ -intercept of $y = f (x)$
This method of finding volumes of revolution uses the idea of a definite integral as calculating an accumulation of change
- It is a special case of 'finding volumes from areas of known cross-sections'
- $π y^{2} \cdot ∆ x$ is the volume of a disc with
  - circular cross section of radius $|y|$
  - and length $∆ x$
- $π y^{2} d x$ is the limit of this volume element as $∆ x \to 0$
- The integral $\int_{a}^{b} π y^{2} d x$ sums up all these infinitesimal volume elements between $x = a$ and $x = b$

Examiner Tips and Tricks

If the given function involves a square root, the problem may seem daunting

But the square root will be 'squared away' when using the Volume of Revolution formula

If a diagram is not provided, sketching the curve, limits, etc. can really help

A graphing calculator can help with this

Worked Example

Let $R$ be the region enclosed by the graph of $f (x) = \sqrt{3 x^{2} + 2}$ , the $x$ - and $y$ -axes, and the vertical line $x = 3$ , as shown in the figure below.

Graph showing a shaded region R bounded by the curve y=sqrt(3x^2+2), the x- and y-axes, and the line x=3.

Find the volume of the solid generated when $R$ is rotated about the $x$ -axis. Give your answer as an exact value.

Answer:

Use $V = π \int_{a}^{b} y^{2} d x$

$\begin{array}{rcl} V & = & π \int_{0}^{3} {(\sqrt{3 x^{2} + 2})}^{2} d x \\ = & π \int_{0}^{3} (3 x^{2} + 2) d x \\ = & π {[x^{3} + 2 x]}_{0}^{3} \\ = & π [({(3)}^{3} + 2 (3)) - 0] \\ = & 33 π \end{array}$

The question doesn't specify units, so the units of volume will be ${units}^{3}$

33π units³

Last updated: 21 August 2024

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