Find the centroid of the area bounded by the parabola

Find the centroid of the area bounded by the parabola $y=-ax^2+b$ and the $x$-axis (where $a>0,\ b>0$).

Step 1: Intersections and area

$y=0$ at $x=\pm\sqrt{b/a}$. By symmetry:

\[A=2\int_0^{\sqrt{b/a}}(-ax^2+b)\,dx=2\left[-\frac{ax^3}3+bx\right]_0^{\sqrt{b/a}}=\frac{4b}3\sqrt{\frac ba}=\frac{4}{3}\sqrt{\frac{b^3}a}\]

Step 2: Moment about the $y$-axis

By symmetry, $M_y=0$.

Step 3: Moment about the $x$-axis

\[M_x=2\int_0^{\sqrt{b/a}}\frac{(-ax^2+b)^2}2\,dx = \int_0^{\sqrt{b/a}}(a^2x^4-2abx^2+b^2)\,dx\]

\[=\left[\frac{a^2x^5}5-\frac{2abx^3}3+b^2x\right]_0^{\sqrt{b/a}}=\frac{b^{5/2}}{\sqrt a}\left(\frac15-\frac23+1\right)=\frac{8b^{5/2}}{15\sqrt a}\]

Step 4: Centroid

\[\bar y=\frac{M_x}A=\frac{8b^{5/2}/(15\sqrt a)}{4b^{3/2}/(3\sqrt a)}=\frac{2b}5\]

\[\boxed{\left(0,\frac{2b}5\right)}\]

Step 5: Note

This $\frac25$-of-height result for a parabolic segment is a classical fact traceable to Archimedes' method of exhaustion, and holds regardless of the specific values of $a$ and $b$.