Find the centroid of the homogeneous figure bounded

Find the centroid of the homogeneous figure bounded by the right loop of the curve $x^4+y^4=x^2y$.

Step 1 · Polar form

With $x=r\cos\theta,\,y=r\sin\theta$, for $r\ne0$: $r=\dfrac{\cos^2\theta\sin\theta}{\cos^4\theta+\sin^4\theta}$. The right loop ($x\ge0$) corresponds to $\theta\in\left[0,\dfrac\pi2\right]$, closing at the origin at both endpoints.

Step 2 · Area

With $t=\tan\theta$, then $u=t^2$, then $u=\tan\psi$, the area reduces to a Beta-function integral:

$$ A = \frac12\int_0^{\pi/2}r^2\,d\theta = \frac14\int_0^{\pi/2}\sin^{1/2}\psi\cos^{3/2}\psi\,d\psi = \frac18\,B\!\left(\frac34,\frac54\right) = \frac{\Gamma(3/4)\Gamma(1/4)}{32} = \frac{\pi\sqrt2}{32}. $$

Step 3 · Static moments

The same substitution chain gives, after reducing to elementary and Beta-function integrals:

$$ M_y = \frac13\int_0^{\pi/2}r^3\cos\theta\,d\theta = \frac13\int_0^\infty\frac{t^3}{(1+t^4)^3}\,dt = \frac1{24}, $$

$$ M_x = \frac13\int_0^{\pi/2}r^3\sin\theta\,d\theta = \frac1{12}\int_0^\infty\frac{s^{1/4}}{(1+s)^3}\,ds = \frac{B(5/4,7/4)}{12} = \frac{\pi\sqrt2}{128}. $$

Step 4 · Centroid

$$ \bar x = \frac{M_y}A = \frac{1/24}{\pi\sqrt2/32} = \frac{4}{3\pi\sqrt2} = \frac{2\sqrt2}{3\pi}, \qquad \bar y = \frac{M_x}A = \frac{\pi\sqrt2/128}{\pi\sqrt2/32} = \frac14. $$

Additional Notes

• This derivation matches the source exactly.
• Verified numerically by direct integration in polar coordinates: $A\approx0.138840$, $\bar x\approx0.300105$, $\bar y=0.25$, all matching the closed forms $\dfrac{\pi\sqrt2}{32},\,\dfrac{2\sqrt2}{3\pi},\,\dfrac14$ exactly.