How do I find the length of a curve using calculus?

[MozAngeles]

Homework Statement

Find the length of the curve
x=y^2/3, 1≤y≤8

Homework Equations

i want to see if i set this up right. and if so then how did i get started on this integral. seems really hard...

The Attempt at a Solution

L= ∫ (from 1to 8) √(1+2/3y^-1/3)dy

 

[hernando]

Homework Statement

Find the length of the curve
x=y^2/3, 1≤y≤8

Homework Equations

i want to see if i set this up right. and if so then how did i get started on this integral. seems really hard...

The Attempt at a Solution

L= ∫ (from 1to 8) √(1+2/3y^-1/3)dy

My boy, I've gone and went to wikipedia and found a certain equation for use

s = \sqrt{1+f'(x)^2}

So I went about applying that there formula

y = x^{\frac{3}{2}}
y' = \frac{3}{2} x^{\frac{1}{2}}
s = \sqrt{1+\frac{9}{4} x}

I figure you've best be able to know the next step from here. Remember, though, the integration goes from 1 to 4 for x instead of the 1 to 8 for y.

 

[hernando]

s = \int \sqrt{1+f'(x)^2}\, dx
s = \int \sqrt{1+\frac{9}{4} x} \, dx
I've gone and forgot my integrals, boy.

 

[MozAngeles]

well that was basically, i mean i already knew that formula, what i had only reversed for the x and y. i still can't figure the integral out :/

 

[hernando]

well that was basically, i mean i already knew that formula, what i had only reversed for the x and y. i still can't figure the integral out :/

Boy, I cannot understand your words too well, but I will tell you that I solved for y to make for an easier integral. Now, I'm sure your integral is possible, but it seems much harder than the one I've constructed, which can be solved by a single u substitution.

Though, if you truly want to suffer that there strife, I'll go on and give you a fixin'. See, you've not squared that there y. You're really lookin' for this answer

s = \int \limits_{1}^{8} \sqrt{1+\frac{4}{9}y^{-\frac{2}{3}}}

 

[MozAngeles]

sorry, its late. thanks for the help though. it just clicked.