TITLE: Finding the curve length
QUESTION [0 upvotes]: I am solving some curve length questions and came across this question:
$$ x = \frac{y^3}{6} + \frac{1}{2y} ;  2<= y <=3 $$
I know how to solve for the curve length when they provide y equals to something, however, i am not sure on how to solve x equal to something. I have tried moving around the values and made y equal to x. However, then i am not sure on how to change the boundary value 2<= y <=3. 
How do we go about solving for the curve length in such scenarios? 
Hope I am clear with my question, if not, please drop me a comment and i will try to improve on it. Any and all help is greatly appreciated!!!
Thank you very much!

REPLY [1 votes]: Well, you should probably know the formula for arc length. There are multiple versions, of course, most commonly:
$$s=\int \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx$$
$$=\int \sqrt{1+\left(\frac{dx}{dy}\right)^2} dy$$
$$=\int \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt$$
There are a lot of other ones... We can go a little crazier, of course....
$$s=\int \sqrt{\left(\frac{d\rho}{dt}\right)^2+\rho^2\sin^2\varphi\left(\frac{d\theta}{dt}\right)^2+\rho^2\left(\frac{d\varphi}{dt}\right)^2}$$
But yeah, don't worry about that. As for your problem, we'll just use the second equation (you can use the first, but $\frac{dx}{dy}$ is easier to obtain and you have an interval of $y$ so you clearly want the second one).
$$x = \frac{y^3}{6} + \frac{1}{2y}$$
Step one, find $\frac{dx}{dy}$. Simple enough:
$$\frac{dx}{dy}=\frac{y^2}{2}-\frac{1}{2y^2}$$
Now use the equation... Our bounds are $2 \le y \le 3$... Hence:
$$s=\int_2^3 \sqrt{1+\left(\frac{y^2}{2}-\frac{1}{2y^2}\right)^2} dy$$
Integrating that is a big mess... Don't think any teacher would make you do that. Use a calculator to evaluate this and you'll get:
$$\therefore s=3.25$$
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