Find the centroid of the homogeneous figure bounded

Find the centroid of the homogeneous figure bounded by the curve $\sqrt x+\sqrt y=\sqrt a$ and the coordinate axes $x=0,\,y=0$.

Step 1 · Symmetry and setup

The curve is symmetric under $x\leftrightarrow y$, so $\bar x=\bar y$. From $\sqrt y=\sqrt a-\sqrt x$: $y=\left(\sqrt a-\sqrt x\right)^2=a-2\sqrt{ax}+x$, for $0\le x\le a$.

Step 2 · Area

$$ A = \int_0^a\left(a-2\sqrt a\sqrt x+x\right)dx = a^2\left(1-\frac43+\frac12\right) = \frac{a^2}6. $$

Step 3 · Static moment $M_y$

$$ M_y = \int_0^ax\left(a-2\sqrt a\sqrt x+x\right)dx = a^3\left(\frac12-\frac45+\frac13\right) = \frac{a^3}{30}. $$

Step 4 · Centroid

$$ \bar x = \frac{M_y}A = \frac{a^3/30}{a^2/6} = \frac a5 = \bar y. $$

(Verification: the substitution $x=a\cos^4t,\,y=a\sin^4t$ satisfies $\sqrt x+\sqrt y=a\cos^2t+a\sin^2t=a$.)

Additional Notes

• This derivation matches the source exactly.
• Verified numerically (e.g. $a=3.7$): $A\approx2.28167$, $\bar x\approx0.74$, matching $\dfrac{a^2}6$ and $\dfrac a5$ exactly.