How do you find the lengths of the curve #(3y-1)^2=x^3# for #0<=x<=2#?

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Answer 1 · 2018-06-28T05:49:23

Isolate #y# then apply the arclength formula.

#(3y-1)^2=x^3#, #x in [0,2]#

Isolate #y#:

#y=1/3(x^(3/2)+1)#

Take the derivative with respect to #x#:

#y'=1/2sqrtx#

Arc length is given by:

#L=int_0^2sqrt(1+x/4)dx#

Integrate directly:

#L=8/3[(1+x/4)^(3/2)]_0^2#

Insert the limits of integration:

#L=8/3((3/2)^(3/2)-1)#