Find the centroid of the area of the ellipse (x²)/(a²)+

Find the centroid of the area of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ lying in the first quadrant.

Step 1: Area

\[A=\frac{\pi ab}4\]

Step 2: Moment about the $y$-axis

\[M_y=\int_0^ax\cdot b\sqrt{1-\frac{x^2}{a^2}}\,dx\]

With $u=1-\frac{x^2}{a^2}$: \[M_y=\frac{a^2b}2\int_0^1\sqrt u\,du=\frac{a^2b}2\cdot\frac23=\frac{a^2b}3\]

Step 3: Moment about the $x$-axis

\[M_x=\int_0^a\frac{b^2}2\left(1-\frac{x^2}{a^2}\right)dx=\frac{b^2}2\left[x-\frac{x^3}{3a^2}\right]_0^a=\frac{b^2}2\cdot\frac{2a}3=\frac{ab^2}3\]

Step 4: Centroid

\[\bar x=\frac{M_y}A=\frac{a^2b/3}{\pi ab/4}=\frac{4a}{3\pi},\qquad \bar y=\frac{M_x}A=\frac{ab^2/3}{\pi ab/4}=\frac{4b}{3\pi}\]

\[\boxed{\left(\frac{4a}{3\pi},\frac{4b}{3\pi}\right)\approx(0.4244a,\,0.4244b)}\]

Step 5: Special case

When $a=b=r$ (a circle), this reduces to the familiar quarter-circle centroid $\left(\frac{4r}{3\pi},\frac{4r}{3\pi}\right)$.