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Find the length of the given curve $\displaystyle x=3y^{\frac{4}{3}}-\frac{3y^{\frac{2}{3}}}{32} $

where the bounds are given by $-8 \leq y \leq 343$.

I can solve for the positive part (0 to 343), but I am unsure of how I would go about integrating $\int_{-8}^{0} \sqrt{1+\left(\frac{dx}{dy}\right)^2}dy$

Thanks again for any help!

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    @Arturo, thanks it was my fault and I typed it wrong.Jonas was able to help me understand how to solve it.2011-03-10

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$f(y)=3y^{4/3}-\frac{3}{32}y^{2/3}$ is an even function of $y$, i.e., $f(-y)=f(y)$, and therefore f'(-y)=-f'(y), meaning that f'(y) is an odd function. This implies that f'(-y)^2=f'(y)^2. So if you know how to deal with positive $y$, then you also can deal with negative $y$, e.g., $\int_{-8}^0\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy=\int_0^8\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy.$

Even without noticing this, I don't see how you would be able to do the positive part if the negative part causes trouble.

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    @Finzz: There really is no "trick"; you can simply do the integral with limits from $-8$ to $343$ with absolutely no problem. I'm not sure why you think negative numbers are an issue: the function and the derivative require you to compute **cubic** roots, and there is no issue of positive or negative with cubic roots.2011-03-10
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the length of a curve with $r(t)=(x(t),y(t))$ can be found by \int^{t_{1}}_{t_{0}}|r'(t)|dt. In your problem you confused $\frac{dy}{dx}$ with $\frac{dx}{dy}$. That's why you cannot integrate.

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    The result is 232191/32.2011-03-10