Using Simpson's formula, determine the arc length

Using Simpson's formula, determine the arc length of the ellipse $\dfrac{x^2}{100}+\dfrac{y^2}{36}=1$ above the $x$-axis, from $x=0$ to $x=8$.

Step 1: Solve for $y(x)$ and its derivative

With $a=10,\ b=6$: $y=\dfrac35\sqrt{100-x^2}$, so $\dfrac{dy}{dx}=-\dfrac{3x}{5\sqrt{100-x^2}}$.

Step 2: Arc-length integrand

\[1+\left(\frac{dy}{dx}\right)^2=1+\frac{9x^2}{25(100-x^2)}=\frac{2500-16x^2}{25(100-x^2)},\]

so with $f(x)=\sqrt{\dfrac{2500-16x^2}{100-x^2}}$, the arc length is $L=\dfrac15\displaystyle\int_0^8f(x)\,dx$.

Step 3: Simpson's formula, $n=8$, $h=1$

Weighted sum $S\approx127.538579$, so $I=\dfrac{1}{3}\times127.538579\approx42.512860$.

Step 4: Arc length

\[L=\frac{1}{5}I\approx\frac{42.512860}{5}\approx8.502572.\]

\[\boxed{L\approx8.503}\]

Step 5: Sanity check

At $x=0$, $y=6$ (the top of the ellipse); at $x=8$, $y=\tfrac35\sqrt{100-64}=3.6$. Direct high-precision numerical integration of the same arc-length integral confirms $L\approx8.502$, matching Simpson's estimate to within the expected discretization error of a step size $h=1$ — a reminder that the ellipse's arc length has no elementary closed form (it is an elliptic integral) and must be evaluated numerically.