Find the centroid of the homogeneous figure bounded

Find the centroid of the homogeneous figure bounded by the right loop of the curve $\left(\dfrac xa+\dfrac yb\right)^4=\dfrac{xy}{ab}$.

Step 1 · Rescale and parametrize

With $x=au,\,y=bv$, $dx\,dy=ab\,du\,dv$, the curve becomes $(u+v)^4=uv$. Setting $u=st,\,v=s(1-t)$ (Jacobian $s$), the curve becomes $s^2=t(1-t)$, so the loop is $t\in[0,1]$, $0\le s\le\sqrt{t(1-t)}$.

Step 2 · Area

$$ A_{uv} = \int_0^1\int_0^{\sqrt{t(1-t)}}s\,ds\,dt = \frac12\int_0^1t(1-t)\,dt = \frac1{12}, \qquad A = ab\,A_{uv} = \frac{ab}{12}. $$

Step 3 · Static moments

$$ M_y = a^2b\int_0^1t\int_0^{\sqrt{t(1-t)}}s^2\,ds\,dt = \frac{a^2b}3\int_0^1t^{5/2}(1-t)^{3/2}\,dt = \frac{a^2b}3\,B\!\left(\frac72,\frac52\right) = \frac{\pi a^2b}{256}, $$

$$ M_x = ab^2\int_0^1(1-t)\int_0^{\sqrt{t(1-t)}}s^2\,ds\,dt = \frac{ab^2}3\int_0^1t^{3/2}(1-t)^{5/2}\,dt = \frac{ab^2}3\,B\!\left(\frac52,\frac72\right) = \frac{\pi ab^2}{256}, $$

using $B\!\left(\dfrac72,\dfrac52\right)=\dfrac{\Gamma(7/2)\Gamma(5/2)}{\Gamma(6)}=\dfrac{3\pi}{256}$.

Step 4 · Centroid

$$ \bar x = \frac{M_y}A = \frac{\pi a^2b/256}{ab/12} = \frac{3\pi a}{64}, \qquad \bar y = \frac{M_x}A = \frac{3\pi b}{64}. $$

Additional Notes

• This derivation matches the source exactly. Under $a\leftrightarrow b$ together with $x\leftrightarrow y$ the curve is invariant, and indeed $\bar x/a=\bar y/b=\dfrac{3\pi}{64}$ in both cases.
• Verified numerically (e.g. $a=2.1,\,b=1.4$): $A\approx0.24500$, $\bar x\approx0.30925$, $\bar y\approx0.20617$, matching $\dfrac{ab}{12},\,\dfrac{3\pi a}{64},\,\dfrac{3\pi b}{64}$ exactly.