Ask Me Help Desk - The arclength of the ellipse

Finding the arc length of ellipses can be difficult. They involve Elliptic Integrals, hence the name.

The ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ can be expressed parametrically:

The total arc length of the ellipse $x=a\cdot cos(t), \;\ y=bsin(t)$ is given by

$4a\int_{0}^{\frac{\pi}{2}}\sqrt{1-e^{2}cos^{2}(t)}dt$

Where $e=\frac{\sqrt{a^{2}-b^{2}}}{a}$ . This e means eccentricity, not to be confused with the exponential e.

But, with yours maybe we can solve for y and use the arc length formula.

Knowing that arc length is given by $\int_{a}^{b}\sqrt{1+[y']^{2}}dx$

$y=\pm\sqrt{\frac{1-x^{2}}{4}}$

$y'=\frac{-x}{2\sqrt{1-x^{2}}}$

We could also try the polar and parametric arc length formulas:

In parametric: $\int_{a}^{b}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{d t})^{2}}dt$

Polar: $\int_{\alpha}^{\beta}\sqrt{r^{2}+(\frac{dr}{d \theta})^{2}}d{\theta}$

There you go. There is enough info to solve it one way or another.