Find the centroid of the homogeneous upper half
Find the centroid of the homogeneous upper half of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\le1,\ y\ge0$.
Explanation of Method
Symmetry about the $y$-axis gives $\bar x=0$ immediately; the static moment $M_x$ is then computed both directly and via the elliptic substitution $x=ar\cos\theta,\,y=br\sin\theta$, as a cross-check.
- Use symmetry to get $\bar x=0$.
- Compute the area $A$.
- Compute $M_x$ directly.
- Verify $M_x$ via the elliptic substitution.
- Divide to get $\bar y$.
Step 1 · Symmetry and area
The region is symmetric about the $y$-axis, so $\bar x=0$. The area is $A=\dfrac12\pi ab$.
Step 2 · Static moment $M_x$ directly
$$ M_x = \iint_Dy\,dA = \int_{-a}^{a}\int_0^{b\sqrt{1-x^2/a^2}}y\,dy\,dx = \frac{b^2}2\int_{-a}^{a}\left(1-\frac{x^2}{a^2}\right)dx = \frac{b^2}2\cdot2\cdot\frac{2a}3 = \frac{2ab^2}3. $$
Step 3 · Verification via elliptic substitution
With $x=ar\cos\theta,\,y=br\sin\theta$, Jacobian $abr$, $0\le r\le1$, $0\le\theta\le\pi$:
$$ M_x = \int_0^\pi\int_0^1br\sin\theta\cdot abr\,dr\,d\theta = ab^2\int_0^\pi\sin\theta\,d\theta\cdot\int_0^1r^2\,dr = ab^2\cdot2\cdot\frac13 = \frac{2ab^2}3.\ \checkmark $$
Step 4 · Centroid coordinate $\bar y$
$$ \bar y = \frac{M_x}A = \frac{2ab^2/3}{\pi ab/2} = \frac{4b}{3\pi}. $$
Additional Notes
- This derivation matches the source exactly. For $a=b$ (a semicircle of radius $a$), this correctly reduces to $\bar y=\dfrac{4a}{3\pi}$, matching doubleint-255.
- Verified numerically (e.g. $a=1.7,\,b=2.9$): $\bar y\approx1.23080$, matching $\dfrac{4b}{3\pi}$ exactly.
Result
The centroid of the upper half-ellipse lies on the axis of symmetry at: