Correct method for finding arc length

Question

I was doing an exercise on Khan Academy about arc length, and i came upon this question. When I attempted it, I applied the arc length formula using an integration from x=-1 to x=1. However, the site shows that I have to take an integral from x=0 to x=1 then multiply that arc length by two to get both the left and right sides of the y-axis. The answer I got was 0 (which of course doesn't make sense) and the correct answer was 3.

Is the site correct, and why?

I suspect my answer was due to some sort of canceling out of the left and right sides of the y-axis, but I'm not sure how that works.

Edit: steps I took, as requested:

First, I solved the original equation for y (principal roots only). $y=(1-x^{2/3})^{3/2}$. I then took the derivative of this to obtain dy/dx to plug into the formula: $\frac{\sqrt{1-x^{2/3}}}{x^{1/3}}$. I then squared that and added one which yielded $x^{-1/3}$, square rooted that and then took the integral from -1 to 1. Indefinite integral: $\frac{3x^{2/3}}{2}$, evaluated at 1 and -1: $\frac{3}{2}-\frac{3}{2}$, which equals 0.

Answer 1

Nobody seems to have addressed the real problem. And that is that the equation at the Khan Academy is not the equation for the curve that is shown, because for negative $x$ you get complex numbers. The equation for the curve in the upper half-plane is

$$y=(1-|x|^{2/3})^{3/2}$$

and that is why recourse is made to the symmetry and integration in the first quadrant. And while it's true that $\sqrt{x^{−2/3}}=−x^{−1/3}$, it's still complex.

Answer 2

$$\begin{array}{rcl} x^{2/3} + y^{2/3} &=& 1 \\ \dfrac23 x^{-1/3} + \dfrac23 y^{-1/3} \dfrac{\mathrm dy}{\mathrm dx} &=& 0 \\ \dfrac{\mathrm dy}{\mathrm dx} &=& -\sqrt[3]{\dfrac yx} \\ &=& -\sqrt[3]{\dfrac{\left(1-x^{2/3}\right)^{3/2}}x} \\ &=& -\dfrac{\sqrt{1-x^{2/3}}}{\sqrt[3]{x}} \\ \text{Required arc length} &=& \displaystyle 2 \int_0^1 \sqrt{1+\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2} \ \mathrm dx \\ &=& \displaystyle 2 \int_0^1 \sqrt{1+\dfrac{1-x^{2/3}}{x^{2/3}}} \ \mathrm dx \\ &=& \displaystyle 2 \int_0^1 \sqrt{\dfrac1{x^{2/3}}} \ \mathrm dx \\ &=& \displaystyle 2 \int_0^1 x^{-1/3} \ \mathrm dx \\ &=& \displaystyle 2 \left[ \dfrac32 x^{2/3} \right]_0^1 \\ &=& \displaystyle 3 \\ \end{array}$$

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What is wrong with your approach is that when $x$ is negative, $\sqrt{x^{-2/3}}$ should be $-x^{-1/3}$ instead of $x^{-1/3}$.

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For your approach:

$$\begin{array}{rcl} \text{Required arc length} &=& \displaystyle \int_{-1}^1 \sqrt{1+\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2} \ \mathrm dx \\ &=& \displaystyle \int_{-1}^1 \sqrt{1+\dfrac{1-x^{2/3}}{x^{2/3}}} \ \mathrm dx \\ &=& \displaystyle \int_{-1}^1 \sqrt{\dfrac1{x^{2/3}}} \ \mathrm dx \\ &=& \displaystyle \int_{-1}^0 \sqrt{\dfrac1{x^{2/3}}} \ \mathrm dx + \int_0^1 \sqrt{\dfrac1{x^{2/3}}} \ \mathrm dx \\ &=& \displaystyle \int_{-1}^0 -x^{-1/3} \ \mathrm dx + \int_0^1 x^{-1/3} \ \mathrm dx \\ &=& \displaystyle \left[-\dfrac32 x^{2/3} \right]_{-1}^0 + \left[ \dfrac32 x^{2/3} \right]_0^1 \\ &=& \dfrac32 + \dfrac32 \\ &=& 3 \end{array}$$