Originally Posted by

**Malaclypse** Been beating my head against the wall on an integral for awhile now.... (Headbang)

The question is this:

Compute the surface area generated by revolving the curve $\displaystyle \[f(x) = 2\sqrt x ,\,\,{\text{for 1}} \leqslant x \leqslant 4\]$ about the x-axis.

I took the 1st derivative of the equation:

$\displaystyle \[f'(x) = \frac{1}{x}\]$

Easy enough. Then I went into my integral:

$\displaystyle \[2\pi \int_{\,1}^{\,4} {2\sqrt x \sqrt {1 + \frac{1}{{{x^2}}}} } dx\]$

That didn't look very pretty, and I didn't seem to be getting anywhere trying to solve it (no u-sub, and integration by parts didn't seem to be working), so I tried to manipulate the equation several times to see if a different variation would work...

$\displaystyle \[4\pi \int_{\,1}^{\,4} {\sqrt {x + \frac{1}{x}} } dx\]$

and

$\displaystyle \[4\pi \int_{\,1}^{\,4} {\left( {\sqrt {\frac{1}{x}} } \right)} \sqrt {\left( {{x^2} + 1} \right)} dx\]$

and

$\displaystyle \[4\pi \int_{\,1}^{\,4} {\left( {\sqrt {\frac{1}{x}} } \right)} \sqrt {{x^2}\left( {1 + \frac{1}{{{x^2}}}} \right)} dx\]$

and

$\displaystyle \[4\pi \int_{\,1}^{\,4} {x\sqrt {\left( {\frac{{{x^2} + 1}}{{{x^3}}}} \right)} } dx\]$

But I can't find one to work out. My professor stated on the assignment that the integral can be done exactly, otherwise I would have approximated it a long time ago.

Would one of you be able to possibly hint where I'm going wrong, or which form you'd take?