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

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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That's not the right formula if you are integrating with respect to $y$; you would need to integrate $$\sqrt{1 + \left(\frac{dx}{dy}\right)^2},$$because here $x$ is a function of $y$, not the other way around. So perhaps you did not get the right answer when $0\leq y\leq 343$ either... –  Arturo Magidin Mar 10 '11 at 4:16
@Arturo, thanks it was my fault and I typed it wrong.Jonas was able to help me understand how to solve it. –  Finzz Mar 10 '11 at 4:21

$f(y)=3y^{4/3}-\frac{3}{32}y^{2/3}$ 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., $$\int_{-8}^0\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy=\int_0^8\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy.$$
@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. –  Arturo Magidin Mar 10 '11 at 4:26
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.