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

Revision Note

Author

Roger B

Expertise

Maths

Volume with washer method revolving around the y-axis

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

This is very similar to the washer method for volumes of revolution around the $x$ -axis
- Use this method when there is a gap between the region to be rotated and the $y$ -axis
Let $f$ and $g$ be continuous functions of $y$ such that $|f (y)| < |g (y)|$ on the interval $[a, b]$
- I.e. $f (y)$ is closer to the $y$ -axis than $g (y)$ is on that interval
If the region bounded by
- the curves $x_{1} = f (y)$ and $x_{2} = g (y)$
- between $y = a$ and $y = b$
is rotated $2 π$ radians $(360 °)$ around the $y$ -axis, then the volume of revolution is
- $V = \int_{a}^{b} π ({x_{2}}^{2} - {x_{1}}^{2}) d y = π \int_{a}^{b} ({x_{2}}^{2} - {x_{1}}^{2}) d y$
  - Note that $x_{1}$ and $x_{2}$ are both functions of $y$
  - If the functions are given as functions of $x$
    - e.g $y = p (x)$ and $y = q (x)$
  - then you will need to rewrite them as functions of $y$
    - See the Worked Example
  - Also note that the integration is done with respect to $y$
Make sure that $x_{2}$ is the curve further away from the $y$ -axis
- and $x_{1}$ is the curve closer to the $y$ -axis
  - If the curves 'swap places' over the interval
    - then split the calculation into separate integrals
If $y = a$ and $y = b$ are not stated in a question, these boundaries could involve
- the $x$ -axis ( $y = 0$ )
- and/or point(s) of intersection of the two curves

Exam Tip

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

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

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 $y$ -axis.

Answer:

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

First rewrite the functions as functions of $y$

$\begin{array}{rcl} y & = & f (x) \\ y & = & \frac{1}{4} x^{2} \\ x^{2} & = & 4 y \\ x & = & 2 \sqrt{y} \end{array}$

Note that $x = 2 \sqrt{y}$ is used instead of $x = - 2 \sqrt{y}$ because it can be seen from the graph that the $x$ values of the relevant part of the curve are positive

$\begin{array}{rcl} y & = & g (x) \\ y & = & x \\ x & = & y \end{array}$

$g (x) = x$ is the function closest to the $y$ -axis, so use $x_{1} = y$ and $x_{2} = 2 \sqrt{y}$

To find $a$ and $b$ , solve $x_{1} (y) = x_{2} (y)$ to find the $y$ -coordinates of the points of intersection of the two curves

$\begin{array}{rcl} y & = & 2 \sqrt{y} \\ y^{2} & = & 4 y \\ y^{2} - 4 y & = & 0 \\ y (y - 4) & = & 0 \end{array}$

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

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

Set up and solve the integral

$\begin{array}{rcl} V & = & π \int_{0}^{4} ({(2 \sqrt{y})}^{2} - {(y)}^{2}) d y \\ = & π \int_{0}^{4} (4 y - y^{2}) d y \\ = & π {[2 y^{2} - \frac{1}{3} y^{3}]}_{0}^{4} \\ = & π ((2 {(4)}^{2} - \frac{1}{3} {(4)}^{3}) - 0) \\ = & \frac{32 π}{3} \\ = & 33.510321 . . . \end{array}$

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

33.510 units³ (to 3 decimal places)

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