Find Length of the Curve

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What is the increment of length, in terms of dx and dy?

I can somewhat solve it, I just do not know how to go about starting the problem when x and y are switched.
X = y^4/8 + 1/(4y^2)
dx/dy = y^3/2-1/(2y^3)

Length = integral(sqrt(1+(dx/dy)^2))

would the integral be from 1 to 2?
i do not remember if you need to solve for x, then do the integral. or just use 1 and 2?
 
foiling dx/dy)^2 for the length formula. i get y^6/4 + 1/4y^6 - 1/2

now add 1 and quantity squared would give the length formula.
when i add one to that trinomial, the -1/2 goes to 1/2.
then I can (un)foil that trinomial to (y^3/2 + 1/2y^3)^2
basically just switching the original from subtraction to addition
the sqrt and ^2 cancel, leaving me with integral of y^3/2 + 1/2y^3
 
finishing this length integral - the integral(y^3/2 + 1/2y^3) gives the answer y^4/8 - 1/4y^2, almost like the original equation ( - not + )

but still unsure of what to plug into the solved integral. 1,2 or find x from these y values and plug those in.
 
I can somewhat solve it, I just do not know how to go about starting the problem when x and y are switched.
X = (y^4)/8 + 1/(4y^2)
dx/dy = (y^3)/2 - 1/(2y^3)

Length = integral[1,2](sqrt(1+(dx/dy)^2)) dy

would the integral be from 1 to 2?
i do not remember if you need to solve for x, then do the integral. or just use 1 and 2?
The way the problem was stated, it is MUCH easier to integrate in the y direction. "x" and "y" are just dummy names anyway.
foiling dx/dy)^2 for the length formula. i get (y^6)/4 + 1/(4y^6) - 1/2

now add 1 and quantity squared would give the length formula.
when i add one to that trinomial, the -1/2 goes to 1/2.
then I can (un)foil that trinomial to [(y^3)/2 + 1/(2y^3)]^2
basically just switching the original from subtraction to addition
the sqrt and ^2 cancel, leaving me with integral of (y^3)/2 + 1/(2y^3)
When typing inline, you must use enough parentheses to make the order of operations correct.
finishing this length integral - the integral(y^3/2 + 1/2y^3) gives the answer y^4/8 - 1/4y^2, almost like the original equation ( - not + )

but still unsure of what to plug into the solved integral. 1,2 or find x from these y values and plug those in.
The limits of the integration are values of y, from 1 to 2. Since the integration is in terms of y, you are ready to evaluate and take the difference.
 
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