Volume of rotation about y-axis

Is this a standard formula that can be used as an alternative to the common A-level method?

Yes, it's a standard formula, though I've rarely seen it used.

The usual formula arises from considering elementary discs perpendicular to the axis about which you're rotating, hence the $\pi y^2\; dx$, etc.

The formula used in this question uses concentric elementary annuli of thickness $\delta x$, with the volume of the annulus being given by, circumference x height x thickness, i.e. $2\pi x\times y\times \delta x$, hence the integral of $2\pi xy\;dx$

I'm having difficulty visualising the annuli. Is the word elementary in reference to the way that the volume consists of a row of infinitely thin "segments" side by side of different heights, as found generally in integration from first principles?

Would it be possible to use the usual method for discs to solve the problem?

Thanks.

Would it be possible to use the usual method for discs to solve the problem?

Thanks.

Yes.

I think Ghostwalker has overlooked a small point:

the volume of revolution is a disc of which the radius is 1, height is 1/2, plus a dome shape.
volume of dome shape above the disc is given by 'usual' formula, which will result in

Volume of dome = pi x [ F(1) - F(1/2) ]
where F(y) is the integrated x^2 expression

ie F(y) = lny - y

Good spot. I was looking at the curve over the whole of the reals.

PRSOM.