AA Topic 5.17 · teacher page · HL

Running volumes of revolution

The same curve about two axes, and why the formula has to change twice.

The one thing to do with the figure

Switch the axis mid-sentence.

Spin about the x-axis, then switch. The curve has not moved and the solid is completely different: a shallow bowl becomes a deep one, and the volume goes from 32π/5 to 8π.

Students who have memorised π∫y²dx cannot see why anything should change. Watching it change is faster than explaining.

Both parts swap, not one

About the y-axis the radius becomes x AND the thickness becomes dy. Swapping one and not the other is the standard error.

Say it as a pair every time: radius and thickness, both perpendicular to the axis you are spinning about.

The answers

y = x² about x, 0 to 2π∫x⁴dx = 32π/5 ≈ 20.11.
Same curve about y, 0 to 4π∫y dy = 8π ≈ 25.13.
Area about the y-axisx = y², y from 0 to 2, gives 8/3 ≈ 2.667.
1, 2, 320.11; B, πx² dy; 25.13.

Where the marks go

1 markThe correct formula with the correct variable of integration.

1 markLimits in the right variable.

1 markRearranging the equation before integrating.

1 markEvaluating, keeping the π.

What each wrong answer tells you

They giveWhat it means
π∫y²dySwapped the thickness but not the radius. The commonest single error here.
x limits in a dy integralA structural error worth catching early; the numbers will be wrong by a lot.
Lost πRoutine, because the integral is the interesting part and the π feels like decoration.
Forgot to rearrangeTrying to integrate x² with respect to y without writing x² = y.

Other things they will say

"Why πy² and not 2πy?" 2πy is a circumference. A disc has area πr², and here r is the height. Point at the drawn disc.

"What if it is rotated about another line?" Beyond what is set here. Say so rather than improvising; the radius would be the distance to that line.

"Can the volume be negative?" No. If it comes out negative the limits are reversed.

A possible order

 What is happening
1Spin about the x-axis. Establish the disc and its radius.
2Build the formula from one disc.
3Switch the axis. Both parts swap, slowly.
4Rearranging into the right variable, with the worked example.
5Areas against the y-axis.
6Questions 1 to 3.

Two things not to say

Do not give the two formulas as a pair to memorise. Derive the second from the first by asking what a horizontal slice looks like.

Do not let a y-axis question be attempted before the equation is rearranged. That one line prevents most of the errors.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

  1. About the x axisRotate y = x² from 0 to 2 about the x axis and find the volume.
    π times the integral of x⁴ = 32π/5 = 20.1.
  2. A trigonometric oneRotate y = sin x from 0 to π about the x axis.
    π times the integral of sin²x = π²/2 = 4.93.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

  1. About the y axisRotate the same region under y = x², 0 to 4 in y, about the y axis.
    The radius is x, so π times the integral of y dy from 0 to 4 = 8π = 25.1.
  2. Why sin squaredExplain why rotating y = sin x needs the integral of sin²x and how to integrate it.
    Each disc has area πy², so the square is forced by the geometry. Use the identity sin²x = (1 − cos 2x)/2, since sin²x has no elementary antiderivative in that form.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

  1. Which axis changes whatExplain what must change when the axis of rotation changes from x to y.
    The radius of a disc, the variable of integration and the limits. The curve must be rearranged so the radius is expressed in the integration variable, and the limits converted. It is a different solid with a different volume, not a relabelling.
  2. Check against a coneRotate y = x from 0 to 3 about the x axis and verify against the cone formula.
    π times the integral of x² = 9π, and (1/3)πr²h with r = h = 3 gives 9π. A known solid is the cheapest check that the method is being applied correctly.

Practicalities

Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.