How to find the arc length of an ellipse

How to find the arc length of an ellipse

This is difficult. This is why Elliptic Integrals were 'invented'.

Do a search on this site. It seems to me I answered something like this in the past.

See here for a nice paper on the topic:

http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2012%20arc%20length%20ellip se.pdf

I will add a few more thoughts to the topic.

For an ellipse c=\sqrt{a^{2}-b^{2}} and the eccentricity, e, is e=\frac{\sqrt{a^{2}-b^{2}}}{a}

The equation for the ellipse can be solved for y.
y = \frac{b}{a}sqrt{a^{2}-x^{2}}, then we can differentiate and get y'=\frac{-bx}{a}\sqrt{a^{2}-x^{2}}.

Square this and it in the arc length formula.

\int\sqrt{1+(\frac{-bx}{a}\sqrt{a^{2}-x^{2}})^{2}}dx

The substitution u=\frac{x}{a} gives
s = a\int_{0}^{\frac{x}{a}}\frac{\sqrt{1 - e^{2}u^{2}}}{\sqrt{1 - u^{2}}} du,

This is the result for the arc length from x = 0 to x/a in the first

quadrant, beginning at the point (0,b) on the y-axis. We have the whole

arc for x/a = 1, and four times this is the circumference of the ellipse.

This does not appear to be a particularly difficult integral, but all

attempts to express it in terms of elementary functions fail. It must be

considered as defining a new function, the elliptic integral of the second

kind. Often we express it in a different form using the substitution

u = sin(t), where t is called the amplitude, and k = e the modulus

of the function E(t,k). In this case, s = a\int_{0}^{sin^{-1}(\frac
{x}{a})}\sqrt{1 - e^{2}sin^{2}(t)}dt.

The whole arc corresponds to the upper limit \frac{\pi}{2}, and E(\frac{\pi}{2},e) = E(e), the complete

elliptic integral of the second kind. We found an integral for the arc

length, defined a new function, and expressed the arc length in terms of

this new function. The reason for the name "elliptic".