How to find the length of a wire making a spherical spiral?

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Suppose you have a Christmas ball, which has a decorative lining around it so that it forms a spherical spiral around it. I want to find how long that decorative lining can be given some parameters. This is more or less the shape I am looking at. This was taken from Wikipedia at this link.

The parametric equations of this were also provided:

$$x = r \cdot \sin(\theta) \cdot \cos(c\theta)$$ $$y = r \cdot \sin(\theta) \cdot \sin(c\theta)$$ $$z = r \cdot \cos(\theta)$$ Where $0 \le \theta \le \pi$.

$c$ is twice the number of turns, and in my specific case, I set $c=8$, just like in the picture, for simplicity sakes. $r$ is the radius, and in my specific case the radius is $r=4cm$. What I am looking for is the length of the red line. How do I find it?

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I'm going to use the parameter $t$ because I'm more comfortable with it. Our parametric equations are: $$\begin{bmatrix} x( t)\\ y( t)\\ z( t) \end{bmatrix} =r\begin{bmatrix} \sin( t)\cos( ct)\\ \sin( t)\sin( ct)\\ \cos( t) \end{bmatrix}$$ The formula for the arc length is $$s(t)=\int_0^t \sqrt{\dot{x}(t')^2+\dot{y}(t')^2+\dot{z}(t')^2}~\mathrm{d}t'$$ We can compute $$\begin{bmatrix} \dot{x}( t)\\ \dot{y}( t)\\ \dot{z}( t) \end{bmatrix} =r\begin{bmatrix} \cos( t)\cos( ct) -c\sin( ct)\sin( t)\\ \cos( t)\sin( ct) +c\cos( ct)\sin( t)\\ -\sin( t) \end{bmatrix}$$ Now I hope you can trust me when I say that $\dot{x}(t)^2+\dot{y}(t)^2+\dot{z}(t)^2=r^2(c^2\sin^2(t)+1).$ (You can verify the algebra yourself if you want.) Therefore, taking the endpoint of integration to be $\pi$, our integral is $$s(\pi)=\int_0^\pi r\sqrt{c^2\sin^2(t)+1}~\mathrm{d}t$$ Unfortunately this integral doesn't have any nice expressions in terms of elementary functions, but we can write it as $$s(\pi)=r\left(E(-c^2)+\sqrt{c^2+1}E\left(\frac{c^2}{c^2+1}\right)\right)$$ Where $E$ denotes the complete elliptic integral of the second kind. You can evaluate this numerically for different values of $c$ and $r$ if you like.

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Starting from @K.defaoite's answer $$s=r\Big[E(-c^2)+\sqrt{c^2+1}E\left(\frac{c^2}{c^2+1}\right)\Big]$$ you could get a quite accurate approximation of the quantity in brackets.

It would write $$2 c+\frac{2\log (4c)+1}{2c}-\frac{4 \log (4c)-3}{32 c^3}+\frac{3\log (4c)-3}{64 c^5}+O\left(\frac{1}{c^ 7}\right)$$ and it is quite good as soon as $c>1$.

Just a few numbers for checking $$\left( \begin{array}{ccc} c & \text{approximation} & \text{exact} \\ 1 & 3.82487 & 3.82020 \\ 2 & 5.27053 & 5.27037 \\ 3 & 6.98722 & 6.98721 \\ 4 & 8.81428 & 8.81428 \\ 5 & 10.6969 & 10.6969 \\ 6 & 12.6116 & 12.6116 \\ 7 & 14.5465 & 14.5465 \\ 8 & 16.4951 & 16.4951 \\ 9 & 18.4532 & 18.4532 \\ 10 & 20.4185 & 20.4185 \end{array} \right)$$

Applied to your case $(r=4,c=8)$ the exact solution is $$L=8 E(-64)\approx 65.98022986\text{ cm}$$ while the given formula leads to $$L \sim \frac{5 (6737177+261635 \log (2))}{524288}\approx 65.98022997\text{ cm}$$