Find the centroid of the homogeneous figure bounded

Find the centroid of the homogeneous figure bounded by the curves $x^2+y^2=13$ and $xy=6$ ($x>0$).

Step 1 · Intersection points

Substituting $y=6/x$ into $x^2+y^2=13$: $x^4-13x^2+36=0$, so $x^2\in\{4,9\}$. For $x,y>0$ this gives the points $(2,3)$ and $(3,2)$.

Step 2 · Symmetry

Both curves are invariant under $x\leftrightarrow y$, so the region between them (for $x>0$) is symmetric about $y=x$, giving $\bar x=\bar y$. The region splits into two mirror-image parts about $y=x$; it suffices to work with the part $2\le x\le3$, between $y=6/x$ (below) and $y=\sqrt{13-x^2}$ (above), and double.

Step 3 · Area

Using $\displaystyle\int\sqrt{13-x^2}\,dx=\frac{x\sqrt{13-x^2}}2+\frac{13}2\arcsin\frac x{\sqrt{13}}+C$ and $\arcsin\dfrac3{\sqrt{13}}+\arcsin\dfrac2{\sqrt{13}}=\dfrac\pi2$ (since $\left(\frac2{\sqrt{13}}\right)^2+\left(\frac3{\sqrt{13}}\right)^2=1$), together with $\arcsin\dfrac2{\sqrt{13}}=\arctan\dfrac23$:

$$ A = 2\int_2^3\left(\sqrt{13-x^2}-\frac6x\right)dx = 13\left(\frac\pi2-2\arctan\frac23\right) - 12\ln\frac32. $$

Step 4 · Static moment $M_y$ and centroid

With $\int_2^3x\sqrt{13-x^2}\,dx=\dfrac{19}3$ (substitution $u=13-x^2$) and $\int_2^36\,dx=6$:

$$ M_y = 2\int_2^3\left(x\sqrt{13-x^2}-6\right)dx = 2\left(\frac{19}3-6\right) = \frac23. $$

$$ \bar x = \bar y = \frac{M_y}A = \frac{2/3}{13\left(\dfrac\pi2-2\arctan\dfrac23\right)-12\ln\dfrac32}. $$

Additional Notes

• This derivation matches the source exactly. Numerically $A\approx0.266703$, so $\bar x=\bar y\approx2.4997$ (the source's own rough estimate of $\approx2.51$ is consistent, to its stated precision).
• Verified numerically by direct double integration: $A\approx0.266703$, $\bar x=\bar y\approx2.49966$, matching the boxed formula exactly.