Forgetting to square y in a volume of revolution

The area under a curve uses $\int y\,dx$, and students meet it first. When the same region is rotated about the x-axis to make a solid, the formula changes to $V=\pi\int y^{2}\,dx$, but the habit from area work is strong, so many carry the plain $y$ across and integrate that instead.

The square is the whole point: each slice of the solid is a thin disc whose area is $\pi r^{2}$ with radius $r=y$, so the integrand must be $y^{2}$, not $y$. It is a tempting slip because the $\pi$ is usually remembered while the square, tucked quietly onto the $y$, is dropped.

Miss it and the answer is not merely a bit off; it is a different quantity entirely, you have found something with the units of area multiplied by $\pi$, not a volume. The clue that you have gone wrong is often the units: a genuine volume comes out in cubic units, so if your working never squared anything the dimensions will not add up.

Rotate the region under $y=2x$ between $x=0$ and $x=1$ about the x-axis. The tempting move copies the area integrand and keeps a bare $y$:

The disc radius is $y=2x$, so the integrand must be $y^{2}=(2x)^{2}=4x^{2}$:

1. 1

   Write the formula first

   Put down $V=\pi\int_a^b y^{2}\,dx$ before anything else, with the square already in place.

2. 2

   Express y in terms of x

   Rearrange the curve so $y$ is a formula in $x$.

3. 3

   Square the whole thing

   Compute $y^{2}$, square the entire expression, including any coefficient and sign, not just one term.

4. 4

   Integrate the squared expression

   Integrate $y^{2}$ term by term, then apply the limits $a$ and $b$.

5. 5

   Keep π at the front

   Multiply by $\pi$ and, unless told otherwise, leave the answer in terms of $\pi$.

Our teachers make the disc picture the first thing you draw: one thin slice, radius $y$, area $\pi y^{2}$. Once you see where the square comes from, it stops being a formula to memorise and becomes something you can reconstruct under pressure.

In a one-to-one lesson we insist you write $V=\pi\int y^{2}\,dx$ with the square before substituting, so the mistake has no room to appear. Because the paper awards method marks, a correctly set up volume integral protects your score even if the final arithmetic slips.

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