The curve C has equation $$y=x^{1/2}-(1/3)x^{3/2}+λ$$ where $λ>0$ and $0\le x \le3$. The length of C is denoted by $s$ where $s=2√3$.

The area of the surface generated when C is rotated through one revolution about the x-axis is denoted by $S$. Find $S$ in terms of $λ$. (I've found that $S=3π+4π√3 λ$ and it's correct)

The y-coordinate of the centroid of the region bounded by $C$, the axes and the line $x=3$ is denoted by $h$. Given that the integration from 3 to 0 is

$$\int_0^3 y^2dx=3/4+8 \sqrt3 \cdot (λ/5) +3λ^2 $$

Show $$\lim_{\lambda \to \infty} \ \frac{S}{hs} = 4 \pi$$ This is a past exam question from Alevel Further math, I tried to find the $h$ by using the centroid formula but that makes $\frac{S}{hs}$ complicated..

Here's the suggestion given by the examiner report.

"Most of the survivors who got as far as this stage of the question evaluated $\int_0^3 ydx$ correctly in terms of λ and so, using the result given in the question, obtained a result for $h$. Almost without exception they went on to assemble an expression in terms of $λ$ for $S$ but then failed to establish the given limit. In fact all that was necessary was first to show in some intelligible notation that $h = λ/2 + O(1)$, $S/s = 2λπ + O(1)$ and hence that $\frac{S}{hs} = 4π + O(λ^{–1})$. The required result is then immediate."

Have no idea what's meant by $O(1)$..

  • $\begingroup$ I edited your question. Please check for mistakes, since it wasn't all as clear as I would like. $\endgroup$ – Bob Mar 22 '13 at 15:06
  • $\begingroup$ There are some problems with the display of x^(1/2) and x^(3/2). It means x to the power of (1/2) and (3/2). $\endgroup$ – Vic. Mar 22 '13 at 15:07
  • $\begingroup$ The powers are fine. $\endgroup$ – Bob Mar 22 '13 at 15:08
  • $\begingroup$ Thank you so much for helping me. but it should be (8√3)λ/5 :) other things are fine. $\endgroup$ – Vic. Mar 22 '13 at 15:10

The centroid here is given by

$$h = \frac{\frac{1}{2}\displaystyle\int_0^3 dx \: y^2}{\displaystyle\int_0^3 dx \: y}$$

Ultimately, you will end up with a ratio of two quadratics in $\lambda$. The ratio of the coefficients of $\lambda^2$ in numerator and denominator should be $4 \pi$. I get as that ratio:

$$\frac{S}{h s} = \frac{(3 \pi + 4 \pi \sqrt{3} \lambda) (\frac{4}{5} \sqrt{3} + 3 \lambda)}{2 \sqrt{3} \frac{1}{2} (\frac{3}{4} + \frac{8}{5} \sqrt{3} + 3 \lambda^2)}$$

The ratio of the quadratics is indeed $4 \pi$.

  • $\begingroup$ Are you sure? I think you missed a (1/2) in your formula for h.. $\endgroup$ – Vic. Mar 22 '13 at 15:45
  • $\begingroup$ @Vic. - i did, thanks. $\endgroup$ – Ron Gordon Mar 22 '13 at 15:52
  • $\begingroup$ Just about to check my answer for h on WolframAlpha.. Thank you so much! :) $\endgroup$ – Vic. Mar 22 '13 at 15:57

$\mathcal{O}(1)$ is called Landau notation. A function in terms of the Landau notation provides an upper bound for a function and makes an approximation, which usually is a slight difference (small number).

In Taylor's theorem you could see this approximation as well:

$$f(x)=f(a)+f'(a)(x-a)$$ $$=f(a)+\mathcal{O}(x-a)$$

So we see $\mathcal{O}(x-a)$ improves the approximation, but just slightly. Often negligible.

Still there can be slight different meanings for the notation, dependent on what's inside the brackets. $\mathcal{O}(1)$ is just referring to a small constant.


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