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**icelated** Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x axis.

$\displaystyle y = \frac {x^3}{6} + \frac {1}{2x} $

Now, to find the derivative is easy.

$\displaystyle y \prime = \frac {x^2}{2} - \frac {1}{2x^2} $

Not sure why the solution manual shows a - instead of a plus!

Anyways, at this point we have the form

$\displaystyle 1 + (y \prime)^2 = ( \frac {x^2}{2} - \frac {1}{2x^2})^2 $

which i can follow along with..

What is going on here? because if i squared everything i would get

$\displaystyle \frac {x^4}{4} + \frac {1}{4x^2}$

no ... you're forgetting the middle term

However, the next step i dont see how the solutions manual got it..?

$\displaystyle S = 2 \pi \int (\frac {x^3}{6} + \frac {1}{2x}) ( \frac {x^2}{2} + \frac {1}{2x^2}) dx $

and why are they attaching y prime on the end? you dont need to explain the 2 pi