# Math Help - Arc length

1. ## Arc length

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

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

Now, to find the derivative is easy.
$y \prime = \frac {x^2}{2} - \frac {1}{2x^2}$
Not sure why the solution manual shows a - instead of a plus!

Anyways, at this point we have the form
$1 + (y \prime)^2 = ( \frac {x^2}{2} - \frac {1}{2x^2})^2$
which i can follow along with..

What is going on here? because if i squared everything i would get
$\frac {x^4}{4} + \frac {1}{4x^2}$

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

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

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

2. ## Re: Arc length

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

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

Now, to find the derivative is easy.
$y \prime = \frac {x^2}{2} - \frac {1}{2x^2}$
Not sure why the solution manual shows a - instead of a plus!

Anyways, at this point we have the form
$1 + (y \prime)^2 = ( \frac {x^2}{2} - \frac {1}{2x^2})^2$
which i can follow along with..

What is going on here? because if i squared everything i would get
$\frac {x^4}{4} + \frac {1}{4x^2}$

no ... you're forgetting the middle term

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

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

and why are they attaching y prime on the end? you dont need to explain the 2 pi
$y = \frac{1}{6}x^3 + \frac{1}{2} x^{-1}$

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

... see why the second term of the derivative is negative?

$(y')^2 = \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2 = \frac{x^4}{4} - 2\frac{x^2}{2} \cdot \frac{1}{2x^2} + \frac{1}{4x^4} = \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}$

$A = 2\pi \int_a^b y\sqrt{1 + (y')^2} \, dx$

note ...

$1+(y')^2 =$

$1 + \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4} =$

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

$\left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2$

$A = 2\pi \int_a^b \left(\frac {x^3}{6} + \frac {1}{2x}\right) \cdot \sqrt{\left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2} \, dx$

you never did mention the limits of integration ... finish it.

3. ## Re: Arc length

wow, perfect... Yeah, i limits of integration...
Thank you so much!!!!!!!
I guess i just need more practice..