Calculate arc length of $\sqrt[3]{x^2} + \sqrt[3]{y^2} = \sqrt[3]{9}$ It is necessary to calculate the length of the arc of the  curve below. There  have been attempts to raise to the 3rd power, a complex derivative and integral are obtained
$$\sqrt[3]{x^2} + \sqrt[3]{y^2} = \sqrt[3]{9}$$
 A: Parametrize the curve as $x=3\cos^3t$, $y=3\sin^3t$. Then, the arc-length is
$$L=\int_0^{\pi/2} \sqrt{ (x’(t))^2 + (y’(t))^2}\>dt
= 9\int_0^{\pi/2}\sin t\cos t\>dt =\frac92
$$
A: Hint: Use
$$\int_0^3\sqrt{1+y'^2}dx$$
for $y=(9^{1/3} - x^{2/3})^{3/2}$.
A: The arc length integral can be found by doing $$L=2\int_{-3}^3 \sqrt {1+(\frac{d}{dx}((x^{2/3}-\sqrt[3]{9})^{3/2}))^2} dx$$ which can be thought of as a generalized pythagorean theorem. Also, the two haves of the shape have to be added hence the coefficient of 2.
Next, the simplification of the integral after deriving and taking the square root of the sum gives:
$$L=2\int_{-3}^3\sqrt{(\sqrt [3]3x^{-1/3})^2}dx=2\sqrt[3]3 \int_{-3}^3 x^{-1/3}dx=3^{4/3}x^{2/3}|_{-3}^3=9-\frac{9}{2}=\frac{9}{2}$$
Please give me feed back and notice that some simplification was made in each step. However, there seems to be a tiny mistake here, but the correct answer is still present! It is also interesting that the integral here could happen as usually only these types of functions of degree 2 or 1 can be evaluated and not so much of degree $\frac 23$.
A: The curve is an astroid:  finding its arc-length is a common textbook problem. (The calculation is also widely available.  Among numerous places, I recently read a discussion of it in Maor's The Pythagorean Theorem, Chapter 7.)
We can also get the necessary derivative by implicit differentiation and complete the integration that way:
$$ \frac{d}{dx} \ [x^{2/3} \ + \ y^{2/3}] \ \ = \ \ \frac{d}{dx} \ [a^{2/3}] \ \ \Rightarrow \ \ \frac23 · x^{-1/3} \ + \ \frac23 · y^{-1/3} · y'  \ \ = \ \ 0 \ \ \Rightarrow \ \ y' \ = \ -\frac{y^{1/3}}{x^{1/3}} \ \ ;  $$
the arc-length in the first quadrant (the curve is four-fold symmetric) is found from
$$ s \ \ = \ \ \int_0^a  \sqrt{1 \ + \ (y')^2} \ \ dx \ \ = \ \ \int_0^a  \sqrt{1 \ + \ \left(-\frac{y^{1/3}}{x^{1/3}} \right)^2} \ \ dx \ \ = \ \ \int_0^a  \sqrt{ \frac{x^{2/3} \ + \ y^{2/3}}{x^{2/3}}  } \ \ dx $$
$$ = \ \ \int_0^a  \sqrt{ \frac{a^{2/3} }{x^{2/3}}  } \ \ dx \ \ = \ \ a^{1/3}   \int_0^a  x^{-1/3}  \ \ dx \ \ = \ \ a^{1/3} \ · \frac32 \ a^{2/3} \ \ = \ \ \frac32 \ a \ \ . $$
For the specific astroid in question, $ \ a  =  3 \ \ , $ so the first-quadrant arc-length is $ \ s \ = \ \frac32 · 3 \ = \frac92 \ \ $ and the arc-length of the full curve is $ \ 4s \ = \ 18 \ \ . $
[The area enclosed by the curve and the surface area and volume of the solid of revolution are likewise reasonably simple to integrate.]
