One picture replaces the derivation, and the expansion that quietly loses the question.
Spin it before you write the formula.
The solid builds as a stack of discs, with the leading disc drawn solid and its radius marked on the curve. Ask what the radius of each disc is: someone will say y, and the formula is then obvious rather than given.
Students who meet πy² as symbols never picture a disc, and then cannot work out what to do when the rotation is about the other axis.
Squaring y carelessly. If y = 2x + 1 then y² = 4x² + 4x + 1, and a student who writes 4x² + 1 has lost the question with everything else perfect.
Insist that y² is written as its own line before integrating. That one habit prevents it.
| y = √x about the x-axis, 0 to 4 | π∫x dx = π[x²/2] = 8π ≈ 25.13. |
| Area about the y-axis | ∫₀² y² dy = 8/3 ≈ 2.667. |
| Signed against unsigned | ∫−11x³dx = 0, but the area is ½. |
| 1. The volume | 25.13. |
| 2. About the y-axis | B, ∫πx² dy. |
| 3. The area | ½. |
1 markWriting V = π∫y²dx with correct limits, before substituting.
1 markSquaring and simplifying correctly.
1 markEvaluating, including the π, and with the right units if a context is given.
1 markFor an area question where the curve crosses the axis, splitting the integral.
| They give | What it means |
|---|---|
| 8 (Q1) | Lost the π. Very common, because the integral itself is the interesting part. |
| 16.76 (Q1) | Integrated root x rather than its square. They did not register that y² cancels the root. |
| ∫πy²dx for the y-axis (Q2) | Swapped one thing and not the other. The radius AND the thickness both change. |
| 0 (Q3) | Gave the integral, which the question had already given them. Area and integral are different instructions. |
| ¼ (Q3) | Only did one half. Two pieces, and areas add. |
"Why πy² and not 2πy?" 2πy is a circumference. A disc has area πr² and here r is y. Pointing at the solid disc in the figure settles it faster than saying it.
"What if it is rotated about y = 2?" Beyond what is set here, and worth saying so rather than improvising. The radius would be the distance to that line.
"Can the volume be negative?" No. y² is never negative, so the integral cannot be. If a volume comes out negative the limits are the wrong way round.
| What is happening | |
|---|---|
| 1 | Spin the solid. Establish that each slice is a disc of radius y. |
| 2 | Build the formula from the disc, then the worked example in full. |
| 3 | Rotation about the y-axis, with everything rewritten in terms of y first. |
| 4 | Signed integrals against areas, with a sketch and a split. |
| 5 | Questions 1 to 3. |
| 6 | A reminder that the setup line is where the marks are. |
Do not present the formula first and the picture afterwards. The derivation is one disc, it takes thirty seconds, and without it the y-axis case has to be memorised rather than reasoned.
Do not let a volume question be attempted without y² written as a separate line. That is where the losable expansion lives.
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.
The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.
The same skill inside a real situation, where the first job is working out what is being asked.
Reasoning, working backwards, or spotting an error. These are where the top grades are decided.
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.