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This questions is from Apostol's calculus 2:

Given the curve $x^{2/3} + y^{2/3} = 1, $ bounded by the lines $x=0$ and $y=0$, find its centroid in the first quadrant.

My attempt was first to find the function $y=f(x)$ and evaluate the double integral to find the area of the astroid curve. This is already a difficult part which i couldn't go any further, because evaluating the integral $\int (1-x^{2/3})^{3/2} dx$ isn't easy.

I may ask: is there any other way to make this process easier? Like, writing the curve into polar coordinates or something?

user2345678
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2 Answers2

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Hint. Let $x=\cos^3(t)$ for $t\in [0,\pi/2]$. Then the integrals you need to evaluate have the form $$\int_0^{\pi /2}\sin^m(t)\cos^n (t) d t=\begin{cases} \frac{(m-1)!!(n-1)!!}{(m+n)!!}\cdot\frac{\pi}{2} \quad\quad \text{if $m$ and $n$ are even},\\ \frac{(m-1)!!(n-1)!!}{(m+n)!!} \quad\qquad\text{ otherwise}.\\ \end{cases}$$ See Integral of $\int_0^{\pi/2} \ (\sin x)^7\ (\cos x)^5 \mathrm{d} x$ for a proof of the above formula.

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Robert Z
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Substitute $x=\cos^3t,y=\sin^3t$ so that

$$\int_0^1y\,dx=-3\int_{\pi/2}^0\sin^4t\cos^2t\,dt.$$

Then you can use a table https://en.wikipedia.org/wiki/List_of_definite_integrals#Definite_integrals_involving_trigonometric_functions or expand the complex forms $e^{i\theta}\pm e^{-i\theta}$.

The moments are obtained similarly

$$\int_0^1 xy\,dx=-3\int_{\pi/2}^0\sin^4t\cos^5t\,dt,$$ $$\int_0^1 yy\,dx=-3\int_{\pi/2}^0\sin^7t\cos^2t\,dt.$$ By symmetry, they will be equal.