# Thread: Arc Length and Surfaces of Revolution

1. ## Arc Length and Surfaces of Revolution

Fing the arc length of the graph of the function over the indicated interval.

y = (x^4)/(8) - (1/(4x^2))

on interval [1,2]

2. I think ya know what the arc-lenght formula is, so post what you've done to verify where you're getting stuck.

3. $\displaystyle y'= \frac {1}{2}x^3 - \frac {1}{2x^3}$ on int [1,2]

$\displaystyle = \int \sqrt{1+ (\frac {1}{2}x^3 - \frac {1}{2x^3})^2}$

I don't know how to square this and combine it with the 1. I understand how to integrate it from there but, combining this is really stumping me.

4. Originally Posted by saiyanmx89
$\displaystyle y'= \frac {1}{2}x^3 - \frac {1}{2x^3}$ on int [1,2]
Well that's actually a $\displaystyle +\frac1{2x^3}.$ Now, do the algebra in the radicand, what do you get?

5. $\displaystyle \int \sqrt{1 + \frac{(x^6+1)^2}{4x^6}}$

is this right?