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