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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.

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you seem to have used the wrong formula. $S=2\pi\int_a^b y\sqrt{1+y'^2}dx$ gives the area of the surface of revolution. you need $L=\int_a^b \sqrt{1+y'^2}dx$ – Valentin Jun 11 '12 at 22:09
I meant to do area of a surface of revolution, is it too late to change the title? – user138246 Jun 11 '12 at 22:10
$1+{4\over1+4x}={4x+5\over4x+1}$. Make this simplification and cancel square roots ($\sqrt{4x+1}$) afterwards. – David Mitra Jun 11 '12 at 22:12
Your subject says surface of revolution, but your question is for arc length. Please fix one or the other – Thomas Andrews Jun 11 '12 at 22:46
up vote 2 down vote accepted

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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I don't see how you get a square root of u in the denominator. – user138246 Jun 11 '12 at 22:24
$ \sqrt{ \frac{4 + u}{u}} = \frac{\sqrt{4 + u}}{\sqrt{u}} $ – Santosh Linkha Jun 11 '12 at 22:28
I don't see how you get 4+u – user138246 Jun 11 '12 at 22:30
$ \frac 4 u + 1 = \frac{4 + u} u$ – Santosh Linkha Jun 11 '12 at 22:32

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