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nicnicman
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Homework Statement
Find the length of the curve
[itex]x = 3 y^{4/3}-\frac{3}{32}y^{2/3}, \quad -64\le y\le 64[/itex]
Homework Equations
Integral for arc length (L):
[itex]L = \int_a^b \sqrt{1 + (\frac{dy}{dx})^{2}} dx[/itex]
The Attempt at a Solution
Using symmetry of the interval and the above integral for arc length I got
[itex]L = 2\int_0^{64} \sqrt{1 + (4y^{\frac{1}{3}-}\frac{1}{16y^{\frac{1}{3}}})^{2}} dy[/itex]
Unfortunately, I'm having trouble integrating this beast. Using Wolfram Alpha as a last resort, I found the integrand can be simplified to this:
[itex]4y^{\frac{1}{3}}+\frac{1}{16y^{\frac{1}{3}}}[/itex]
However, I can't see how Worlfram made this simplification. (A step-by-step solution was not available.)
Could someone show me how the the original integrand could be simplified to this form? After that I should be fine.
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