2
$\begingroup$

$y = \sin(\pi x), 0 \le x \le 1$

y' = \pi \cos(\pi x)

(y')^2 = \pi^2\cos^2(\pi x)

$ds = \sqrt{1 + \pi^2 \cos^2(\pi x)}$

$r = y = \sin(\pi x)$

$S = \displaystyle\int_0^1 2 \pi\sin(\pi x)\sqrt{1 + \pi^2 \cos^2(\pi x)} dx$

I'm a little lost as to what to do next. Should I have simplified ds more or do I need to do a substitution with $u = \pi^2\cos^2(\pi x)$?

  • 2
    Your questions can be much easier to read if you $\LaTeX$ them. If you right click and Show Source you can see how, then enclose it in dollar signs.2011-03-07

1 Answers 1

1

Hint: I think a substitution of $u=\pi \cos(\pi x)$, will help as you have $du$ available.

  • 0
    ahh now i see! thanks2011-03-08