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

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

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 y1=f(x) and y2=g(x)

    • between x=a and x=b

  • is rotated 2π radians (360°) around the line y=k, then the volume of revolution is

    •  V=πab((y2k)2(y1k)2) dx

      • Note that y1 and y2 are both functions of x

  • Make sure that y2 is the curve further away from the line y=k

    • and y1 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)=14x2 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=πab((y2k)2(y1k)2) dx

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

14x2=xx24x=0x(x4)=0

x=0  or  x=4

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 y1=x and y2=14x2

Set up and solve the integral

V=π04((14x25)2(x5)2) dx=π04(116x452x2+25(x210x+25)) dx=π04(116x472x2+10x) dx=π[180x576x3+5x2]04=π((180(4)576(4)3+5(4)2)0)=272π15=56.967546...

56.968 units cubed (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 x1=f(y) and x2=g(y)

    • between y=a and y=b

  • is rotated 2π radians (360°) around the line x=k, then the volume of revolution is

    •  V=πab((x2k)2(x1k)2) dy

      • Note that x1 and x2 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 x2 is the curve further away from the line x=k

    • and x1 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)=14x2 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 vertical line x=2.

Answer:

Use V=πab((x2k)2(x1k)2) dy

First rewrite the functions as functions of y

y=f(x)y=14x2x2=4yx=2y

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

y=g(x)y=xx=y

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 x1=y and x2=2y

To find a and b, solve x1(y)=x2(y) to find the y-coordinates of the points of intersection of the two curves

y=2yy2=4yy24y=0y(y4)=0

y=0  or  y=4

So a=0 and b=4

Set up and solve the integral

V=π04((2y(2))2(y(2))2) dy=π04((2y+2)2(y+2)2) dy=π04(4y+8y+4(y2+4y+4)) dy=π04(8yy2) dy=π04(8y12y2) dy=π[163y3213y3]04=π((163(4)3213(4)3)0)=64π3=67.020643...

67.021 units cubed (to 3 decimal places)

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Roger B

Author: Roger B

Expertise: Development Editor

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Portfolio Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.