The equation is: $x=\frac{1}{8}y^4 + \frac{1}{4}y^{-2},\qquad 1\leq y\leq 2.$
I have the formula. I'm not sure how to write it out but this is what it says:
Length is equal to the integral (with $b$ and $a$ for limits) of the square root of $1+(dy/dx)^2 dx$
So I first took the derivative of the equation and got $(1/2)y^3 - (1/2)y^{-3}$.
Now when I plug that back into the formula, I have to square it and I got $(1/4)y^6 - (1/4)y^{-6}$. I factored out a $1/4$ and turned the $y^{-6}$ into $1/y^6$. Just a mess at this point, that doesn't seem right. Can someone help me out with that? Haha I know it's ridiculous but my algebra skills are lacking.