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

Revision Note

Author

Roger B

Expertise

Maths

Volume with washer method revolving around the x-axis

When should I use the washer method for a volume of revolution around the x-axis?

The washer method should be used when there is a gap between a region being rotated and the line it is being rotated around
For example, in the following diagram consider rotating the shaded area between the two curves around the $x$ -axis
- There would be a gap between the $x$ -axis and the curve $y_{1} = f (x)$

A graph showing a shaded region between the curves y_1=f(x) and y_2=g(x), and the lines x=a and x=b.

At a point $x$ between $a$ and $b$ , the cross section of the solid of revolution would look like this:

A cross section of the solid of rotation following from the previous diagram

I.e., it would have the shape of a washer, with
- $Area = π ({(g (x))}^{2} - {(f (x))}^{2})$

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

Let $f$ and $g$ be continuous functions such that $|f (x)| < |g (x)|$ on the interval $[a, b]$
- I.e. $f (x)$ is closer to the $x$ -axis than $g (x)$ is on that interval
If the region bounded by
- the curves $y_{1} = f (x)$ and $y_{2} = g (x)$
- between $x = a$ and $x = b$
is rotated $2 π$ radians $(360 °)$ around the $x$ -axis, then the volume of revolution is
- $V = \int_{a}^{b} π ({y_{2}}^{2} - {y_{1}}^{2}) d x = π \int_{a}^{b} ({y_{2}}^{2} - {y_{1}}^{2}) d x$
  - Note that $y_{1}$ and $y_{2}$ are both functions of $x$
Make sure that $y_{2}$ is the curve further away from the $x$ -axis
- and $y_{1}$ is the curve closer to the $x$ -axis
  - If the curves 'swap places' over the interval
    - then split the calculation into separate integrals
If $x = a$ and $x = b$ are not stated in a question, these boundaries could involve
- the $y$ -axis ( $x = 0$ )
- and/or point(s) of intersection of the two curves
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'
- $π ({y_{2}}^{2} - {y_{1}}^{2}) \cdot ∆ x$ is the volume of a washer with
  - inner radius $|y_{1}|$
  - outer radius $|y_{2}|$
  - and length $∆ x$
- $π ({y_{2}}^{2} - {y_{1}}^{2}) d x$ is the limit of this volume element as $∆ x \to 0$
- The integral $\int_{a}^{b} π ({y_{2}}^{2} - {y_{1}}^{2}) d x$ sums up all these infinitesimal volume elements between $x = a$ and $x = b$

Exam Tip

Be careful not to confuse $({y_{2}}^{2} - {y_{1}}^{2})$ with ${(y_{2} - y_{1})}^{2}$

These are not equal!
- ${(y_{2} - y_{1})}^{2} = {y_{2}}^{2} - 2 y_{1} y_{2} + {y_{1}}^{2}$

Worked Example

Let $R$ be the region enclosed by the graphs of $f (x) = \frac{1}{4} x^{2}$ and $g (x) = x$ , as shown in the figure below.

Graph showing a shaded region R enclosed by the curves y=x and y=x^2/4

Find the volume of the solid generated when $R$ is rotated about the $x$ -axis.

Answer:

Use $V = π \int_{a}^{b} ({y_{2}}^{2} - {y_{1}}^{2}) d x$

$f (x) = \frac{1}{4} x^{2}$ is the function closest to the $x$ -axis, so use $y_{1} = \frac{1}{4} x^{2}$ and $y_{2} = x$

To find $a$ and $b$ , solve $f (x) = g (x)$ to find the $x$ -coordinates of the points of intersection of the two curves

$\begin{array}{rcl} \frac{1}{4} x^{2} & = & x \\ x^{2} - 4 x & = & 0 \\ x (x - 4) & = & 0 \end{array}$

$\begin{array}{rcl} x & = & 0 or x = 4 \end{array}$

So $a = 0$ and $b = 4$

Set up and solve the integral

$\begin{array}{rcl} V & = & π \int_{0}^{4} ({(x)}^{2} - {(\frac{1}{4} x^{2})}^{2}) d x \\ = & π \int_{0}^{4} (x^{2} - \frac{1}{16} x^{4}) d x \\ = & π {[\frac{1}{3} x^{3} - \frac{1}{80} x^{5}]}_{0}^{4} \\ = & π ((\frac{1}{3} {(4)}^{3} - \frac{1}{80} {(4)}^{5}) - 0) \\ = & \frac{128 π}{15} \\ = & 26.808257 . . . \end{array}$

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

26.808 units³ (to 3 decimal places)

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