Math Help - Curve Lengths

1. Curve Lengths

Don't have any real idea how to go about this one, even thought its probably really easy... thnanks for the help

Find the length of the curve:
x=[3y^(4/3)]-(3/32)y^(2/3)

-27<y<27 (both those are less than or equal to)

thanks

2. Hello, N736RA!

Find the length of the curve: . $
x\:=\:3y^{\frac{4}{3}} - \frac{3}{32}y^{\frac{2}{3}} \qquad -27 \:\leq\: y\:\leq \:27$
We will use this formula: . $L \;=\;\int^b_a\sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy$

We have: . . . . . $x \;=\;3y^{\frac{4}{3}} - \frac{3}{32}y^{\frac{2}{3}}$

Differentiate: . $\frac{dx}{dy} \;=\;4y^{\frac{1}{3}} - \frac{1}{16}y^{-\frac{1}{3}}$

. . . . . . . . $\left(\frac{dx}{dy}\right)^2 \;=\;16y^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{256}y^{-\frac{2}{3}}$

. . . . . . $1 + \left(\frac{dx}{dy}\right)^2 \;=\;1 + \left(16y^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{256}y^{-\frac{2}{3}}\right) \;=\;16y^{\frac{2}{3}} + \frac{1}{2} + \frac{1}{256}y^{-\frac{2}{3}} \;=\;\left(4y^{\frac{1}{3}} + \frac{1}{16}y^{-\frac{1}{3}}\right)^2
$

. . . . $\sqrt{1 + \left(\frac{dx}{dy}\right)^2} \;=\;4y^{\frac{1}{3}} + \frac{1}{16}y^{-\frac{1}{3}}$

You must integrate and evaluate: . $L \;=\;\int^{27}_{\text{-}27}\left(4y^{\frac{1}{3}} + \frac{1}{16}y^{-\frac{1}{3}}\right)\,dy$

3. thanks for the reply! Although, when I do that integral I get 0 (which makes sense), yet the program I'm using tells me that 0 isnt the right answer, any ideas? Edit: nevermind, I figured it out, thanks for your help!