Washer Method Around Other Axes (College Board AP® Calculus AB)

Revision Note

Author

Roger B

Expertise

Maths

Volume with washer method revolving around other axes

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

Let $f$ and $g$ be continuous functions such that $|f (x) - k| < |g (x) - k|$ on the interval $[a, b]$
- I.e. $f (x)$ is closer to the horizontal line $y = k$ 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 line $y = k$ , then the volume of revolution is
- $V = π \int_{a}^{b} ({(y_{2} - k)}^{2} - {(y_{1} - k)}^{2}) d x$
  - Note that $y_{1}$ and $y_{2}$ are both functions of $x$
Make sure that $y_{2}$ is the curve further away from the line $y = k$
- and $y_{1}$ is the curve closer to the line $y = k$
  - 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

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 horizontal line $y = 5$ .

Answer:

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

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 $b = 4$

At $x = 4$ , $f (4) = g (4) = 4$ , so $y = 5$ is above region $R$ in the diagram

Therefore $g (x) = x$ is the function closest to the line $y = 5$ , so use $y_{1} = x$ and $y_{2} = \frac{1}{4} x^{2}$

Set up and solve the integral

$\begin{array}{rcl} V & = & π \int_{0}^{4} ({(\frac{1}{4} x^{2} - 5)}^{2} - {(x - 5)}^{2}) d x \\ = & π \int_{0}^{4} (\frac{1}{16} x^{4} - \frac{5}{2} x^{2} + 25 - (x^{2} - 10 x + 25)) d x \\ = & π \int_{0}^{4} (\frac{1}{16} x^{4} - \frac{7}{2} x^{2} + 10 x) d x \\ = & π {[\frac{1}{80} x^{5} - \frac{7}{6} x^{3} + 5 x^{2}]}_{0}^{4} \\ = & π ((\frac{1}{80} {(4)}^{5} - \frac{7}{6} {(4)}^{3} + 5 {(4)}^{2}) - 0) \\ = & \frac{272 π}{15} \\ = & 56.967546 . . . \end{array}$

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

56.968 units³ (to 3 decimal places)

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

Let $f$ and $g$ be continuous functions of $y$ such that $|f (y) - k| < |g (y) - k|$ on the interval $[a, b]$
- I.e. $f (y)$ is closer to the vertical line $x = k$ 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 line $x = k$ , then the volume of revolution is
- $V = π \int_{a}^{b} ({(x_{2} - k)}^{2} - {(x_{1} - k)}^{2}) d y$
  - 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 line $x = k$
- and $x_{1}$ is the curve closer to the line $x = k$
  - 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

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.

Find the volume of the solid generated when $R$ is rotated about the vertical line $x = - 2$ .

Answer:

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

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

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

The line $x = - 2$ is to the left of region $R$ in the diagram

Therefore $g (x) = x$ is the function closest to the line $x = - 2$ , 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

$\begin{array}{rcl} V & = & π \int_{0}^{4} ({(2 \sqrt{y} - (- 2))}^{2} - {(y - (- 2))}^{2}) d y \\ = & π \int_{0}^{4} ({(2 \sqrt{y} + 2)}^{2} - {(y + 2)}^{2}) d y \\ = & π \int_{0}^{4} (4 y + 8 \sqrt{y} + 4 - (y^{2} + 4 y + 4)) d y \\ = & π \int_{0}^{4} (8 \sqrt{y} - y^{2}) d y \\ = & π \int_{0}^{4} (8 y^{\frac{1}{2}} - y^{2}) d y \\ = & π {[\frac{16}{3} y^{\frac{3}{2}} - \frac{1}{3} y^{3}]}_{0}^{4} \\ = & π ((\frac{16}{3} {(4)}^{\frac{3}{2}} - \frac{1}{3} {(4)}^{3}) - 0) \\ = & \frac{64 π}{3} \\ = & 67.020643 . . . \end{array}$

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

67.021 units³ (to 3 decimal places)

Washer Method Around Other Axes (College Board AP® Calculus AB)

Revision Note

Volume with washer method revolving around other axes

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

Worked Example

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

Worked Example

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Author: Roger B