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$y = \sqrt{4x+1}$ for $1 \leq x \leq 5$

I really have no idea what to do with this problem, I attempted something earlier which I will not type up because it took me two pages.

$y = \sqrt{4x+1}$

$\int 2 \pi \sqrt{4x+1} \sqrt{1 + \frac{4}{1+4x}}dx$

$2 \pi \int \sqrt{4x+1} \sqrt{1 + \frac{4}{1+4x}}dx$

Nothing really seems obvious at this point, I attempted a u substitution of $u = 1+4x$ but it does not help simplify this problem really.

$ \pi /2 \int \sqrt{u} \sqrt{1 + \frac{4}{u}}du$

I thought about making a wonky trig substitution but it didn't seem to help and was overly complicated.

  • 0
    Your subject says surface of revolution, but your question is for arc length. Please fix one or the other2012-06-11

1 Answers 1

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From the last step::

$ \frac{\pi}{2} \int \sqrt u \frac{\sqrt{4 +u}}{\sqrt u} du = \frac \pi 2 \int \sqrt{4 + u} du $

substituting $4 + u = p \implies du = dp \;\;$, we get

$ = \frac \pi 2 \int \sqrt p dp = \frac \pi 2 \frac{p^{3/2}}{3/2} = \frac \pi 3 (4+u)^{3/2} $

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    $ \frac 4 u + 1 = \frac{4 + u} u$2012-06-11