Disc Method Around the y-Axis (College Board AP® Calculus AB): Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Volume with disc method revolving around the y-axis

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

  • This is very similar to 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 2π radians (360°) around the y-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 y=a and y=b, and the y-axis about the y-axis

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

  • For a continuous function f, if the region bounded by

    • the curve y=f(x) and the y-axis

    • between y=a and y=b

  • is rotated 2π radians (360°) around the y-axis, then the volume of revolution is

    •  V=abπx2 dy=πabx2 dy

      • Note that x here is a function of y

        • This will mean rewriting y=f(x) in the form x=g(y)

      • Also note that the integration is done with respect to y

  • If y=a and y=b are not stated in a question, these boundaries could involve

    • the x-axis (y=0)

    • and/or a y-intercept of y=f(x)

Examiner Tips and Tricks

Do not get confused about the orientation of the revolution.

If it is a revolution around the x-axis, then use |y| as the radius of the discs and dx as the thickness.

If it is a revolution around the y-axis, then use |x| as the radius of the discs and dy as the thickness.

Worked Example

Let R be the region enclosed by the graph of f(x)=arcsin(2x+1), the negative x-axis, and the positive y-axis, as shown in the figure below.

Graph showing the function y = arcsin(2x+1), shading the area (R) enclosed by the curve, the negative x-axis, and the positive y-axis

Find the volume of the solid generated when R is rotated about the y-axis.  Give your answer correct to 3 decimal places.

Answer:

Use V=πabx2 dy

First rewrite the function as a function of y

y=arcsin(2x+1)siny=2x+1x=siny12

Now that can be put into the integral

Note that the integration will be along the y-axis, from y=0 to y=π2

The integral can be evaluated using your calculator

V=π0π2(siny12)2 dy=0.279754...

0.280 units cubed (to 3 decimal places)

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Build on this topic

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.