hi have 2 perpendicular cylinders that intersect (I read the resulting curve is called the Steinmetz curve).
$x^2+y^2=R_1^2$ and $y^2+z^2=R_2^2$, with $R_2\lt R_1$ and want to parametrize the length of their intersecting line as a function of $\theta$, where $\theta=arctan(z/y)$.
When my segment starts at $0$, I find I get the following integral to solve in order to get the right relationship.
$l= R_2\int_0^\theta{\sqrt{\dfrac{R_1^2-R_2^2\cos^4\theta}{R_1^2-R_2^2\cos^2\theta}}}$. I thought it might involve Incomplete elliptical integrals but can't seem to find a way to transform it into one.
Any ideas? I thought it must be a textbook problem but apparently not.
Thanks in advance.
The arclength of these Steinmetz curves can indeed be reduced to elliptic integrals (complete elliptic integrals, in fact), but this is a highly non-trivial fact to prove for cylinders of unequal radii.
We find ultimately,
Let $a>b>0$ and define the ratio $c:=\frac{b}{a}$. Note $0<c<1$.
$$\begin{align} \ell{\left(a,b\right)} &=b\int_{0}^{2\pi}\sqrt{\frac{a^2-b^2\cos^4{\left(\varphi\right)}}{a^2-b^2\cos^2{\left(\varphi\right)}}}\,\mathrm{d}\varphi\\ &=b\int_{0}^{2\pi}\sqrt{\frac{a^2-b^2\sin^4{\left(\varphi\right)}}{a^2-b^2\sin^2{\left(\varphi\right)}}}\,\mathrm{d}\varphi\\ &=b\int_{0}^{2\pi}\sqrt{\frac{1-c^2\sin^4{\left(\varphi\right)}}{1-c^2\sin^2{\left(\varphi\right)}}}\,\mathrm{d}\varphi\\ &=4b\int_{0}^{\frac{\pi}{2}}\sqrt{\frac{1-c^2\sin^4{\left(\varphi\right)}}{1-c^2\sin^2{\left(\varphi\right)}}}\,\mathrm{d}\varphi\\ &=4b\int_{0}^{1}\sqrt{\frac{1-c^2x^4}{\left(1-x^2\right)\left(1-c^2x^2\right)}}\,\mathrm{d}x;~~~\small{\left[\sin{\varphi}=x\right]}\\ &=\frac{4b}{\sqrt{c}}\int_{0}^{\sqrt{c}}\sqrt{\frac{1-y^4}{\left(1-\frac{1}{c}y^2\right)\left(1-cy^2\right)}}\,\mathrm{d}y;~~~\small{\left[\sqrt{c}\,x=y\right]}\\ &=4\sqrt{ab}\int_{0}^{\sqrt{c}}\sqrt{\frac{1-y^4}{1+y^4-2\mu\,y^2}}\,\mathrm{d}y;~~~\small{\left[2\mu:=c+\frac{1}{c}\right]}\\ &=\small{4\sqrt{ab}\int_{1}^{\sqrt{\frac{1-c}{1+c}}}\sqrt{\frac{2x^2}{\left(\mu+1\right)x^4-\mu+1}}\cdot\frac{(-2x)\,\mathrm{d}x}{\left(1+x^2\right)\sqrt{1-x^4}};~~~\small{\left[\sqrt{\frac{1-y^2}{1+y^2}}=x\right]}}\\ &=8\sqrt{2ab}\int_{\sqrt{\frac{1-c}{1+c}}}^{1}\frac{x^2\,\mathrm{d}x}{\left(1+x^2\right)\sqrt{1-x^4}\sqrt{\left(\mu+1\right)x^4-\left(\mu-1\right)}}\\ &=8\sqrt{2ab}\int_{\sqrt{\frac{1-c}{1+c}}}^{1}\frac{x^2\,\mathrm{d}x}{\left(1+x^2\right)\sqrt{1-x^4}\sqrt{\frac{(1+c)^2}{2c}x^4-\frac{(1-c)^2}{2c}}}\\ &=\frac{16ab}{a+b}\int_{\sqrt{\frac{a-b}{a+b}}}^{1}\frac{x^2\,\mathrm{d}x}{\left(1+x^2\right)\sqrt{1-x^4}\sqrt{x^4-\left(\frac{a-b}{a+b}\right)^2}}\\ &=\frac{16ab}{a+b}\int_{\kappa^2}^{1}\frac{x^2\,\mathrm{d}x}{\left(1+x^2\right)\sqrt{\left(1-x^4\right)\left(x^4-\kappa^8\right)}};~~~\small{\left[\kappa:=\sqrt[4]{\frac{a-b}{a+b}}\right]}\\ &=8\,H{\left(a,b\right)}\int_{\kappa^2}^{1}\frac{x^2\,\mathrm{d}x}{\left(1+x^2\right)\sqrt{\left(1-x^4\right)\left(x^4-\kappa^8\right)}}\\ &=:8\,H{\left(a,b\right)}\,I{\left(\kappa\right)},\\ \end{align}$$
where in the last line above we've introduced the auxiliary function $I{\left(\kappa\right)}$ defined for $0<\kappa<1$ by the hyper-elliptic integral
$$I{\left(\kappa\right)}:=\int_{\kappa^2}^{1}\frac{x^2\,\mathrm{d}x}{\left(1+x^2\right)\sqrt{\left(1-x^4\right)\left(x^4-\kappa^8\right)}}.$$
Also note that $H{\left(a,b\right)}$ stands here for the harmonic mean of $a$ and $b$.
$$\begin{align} I{\left(\kappa\right)} &=\int_{\kappa^2}^{1}\frac{x^2\,\mathrm{d}x}{\left(1+x^2\right)\sqrt{\left(1-x^4\right)\left(x^4-\kappa^8\right)}}\\ &=\int_{\kappa^2}^{1}\frac{x^2\,\mathrm{d}x}{\left(1+x^2\right)\sqrt{-\kappa^8+\left(\kappa^8+1\right)x^4-x^8}}\\ &=\int_{\kappa^2}^{\frac{1}{\kappa^2}}\frac{\frac12\kappa\,\sqrt{w}\,\mathrm{d}w}{\left(1+\kappa^2w\right)\sqrt{-\kappa^4+\left(\kappa^8+1\right)w^2-\kappa^4w^4}};~~~\small{\left[x=\kappa\sqrt{w}\right]}\\ &=\frac{1}{4\kappa}\int_{\kappa^2}^{\frac{1}{\kappa^2}}\frac{\mathrm{d}w}{\left(1+\kappa^2w\right)\sqrt{w}\sqrt{\left(\frac{\kappa^8+1}{4\kappa^4}\right)-\left(\frac{w^4+1}{4w^2}\right)}}\\ &=\frac{1}{4\kappa}\int_{\kappa^2}^{\frac{1}{\kappa^2}}\frac{\mathrm{d}w}{\left(1+\kappa^2w\right)\sqrt{w}\sqrt{\left(\frac{\kappa^4+1}{2\kappa^2}\right)^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &=\frac{1}{4\kappa}\int_{\kappa^2}^{\frac{1}{\kappa^2}}\frac{\mathrm{d}w}{\left(1+\kappa^2w\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}};~~~\small{\left[\alpha:=\frac{\kappa^4+1}{2\kappa^2}\right]}\\ &=\frac{1}{4\kappa}\int_{\kappa^2}^{1}\frac{\mathrm{d}w}{\left(1+\kappa^2w\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &~~~~~+\frac{1}{4\kappa}\int_{1}^{\frac{1}{\kappa^2}}\frac{\mathrm{d}w}{\left(1+\kappa^2w\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &=\frac{1}{4\kappa}\int_{\kappa^2}^{1}\frac{\mathrm{d}w}{\left(1+\kappa^2w\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &~~~~~+\frac{1}{4\kappa}\int_{\kappa^2}^{1}\frac{\mathrm{d}v}{\left(v+\kappa^2\right)\sqrt{v}\sqrt{\alpha^2-\left(\frac{v^2+1}{2v}\right)^2}};~~~\small{\left[\frac{1}{w}=v\right]}\\ &=\frac{1}{4\kappa}\int_{\kappa^2}^{1}\frac{\mathrm{d}w}{\left(1+\kappa^2w\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &~~~~~+\frac{1}{4\kappa}\int_{\kappa^2}^{1}\frac{\mathrm{d}w}{\left(w+\kappa^2\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &=\frac{1}{4\kappa}\int_{\kappa^2}^{1}\frac{\left(1+\kappa^2\right)\left(1+w\right)\,\mathrm{d}w}{\left(1+\kappa^2w\right)\left(\kappa^2+w\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &=\frac{\kappa^2+1}{4\kappa}\int_{\kappa^2}^{1}\frac{\left(1+w\right)\,\mathrm{d}w}{\left(\kappa^2+w+\kappa^4w+\kappa^2w^2\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &=\frac{\kappa^2+1}{4\kappa}\int_{\kappa^2}^{1}\frac{\left(1+w\right)\,\mathrm{d}w}{\left(\left(\kappa^4+1\right)w+\kappa^2\left(w^2+1\right)\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &=\frac{\kappa^2+1}{4\kappa}\int_{\kappa^2}^{1}\frac{\left(1+w\right)\,\mathrm{d}w}{2\kappa^2w\left(\left(\frac{\kappa^4+1}{2\kappa^2}\right)+\left(\frac{w^2+1}{2w}\right)\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &=\frac{\kappa^2+1}{8\kappa^3}\int_{\kappa^2}^{1}\frac{\left(1+\frac{1}{w}\right)\,\mathrm{d}w}{\left(\alpha+\left(\frac{w^2+1}{2w}\right)\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &=\frac{\kappa^2+1}{8\kappa^3}\int_{\alpha-\sqrt{\alpha^2-1}}^{1}\frac{\left(1+\frac{1}{w}\right)\,\mathrm{d}w}{\left(\alpha+\left(\frac{w^2+1}{2w}\right)\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &=:\frac{\kappa^2+1}{8\kappa^3}\,J{\left(\alpha\right)}.\\ \end{align}$$
For $\alpha>1$,
$$\begin{align} J{\left(\alpha\right)} &=\int_{\alpha-\sqrt{\alpha^2-1}}^{1}\frac{\left(1+\frac{1}{w}\right)\,\mathrm{d}w}{\left(\alpha+\left(\frac{w^2+1}{2w}\right)\right)\sqrt{w}\sqrt{\alpha^2-\left(\frac{w^2+1}{2w}\right)^2}}\\ &=\frac{1}{\sqrt{2}}\int_{1}^{\alpha}\frac{\left(1+u+\sqrt{u^2-1}\right)\left(\frac{1}{\sqrt{u-1}}-\frac{1}{\sqrt{u+1}}\right)\,\mathrm{d}u}{\left(\alpha+u\right)\sqrt{\alpha^2-u^2}};~~~\small{\left[w=u-\sqrt{u^2-1}\right]}\\ &=\sqrt{2}\int_{1}^{\alpha}\frac{\mathrm{d}u}{\left(\alpha+u\right)\sqrt{u-1}\sqrt{\alpha^2-u^2}}\\ &=\sqrt{2}\int_{1}^{\alpha}\frac{\mathrm{d}u}{\sqrt{\left(\alpha-u\right)\left(u-1\right)\left(u+\alpha\right)^3}}\\ &=\sqrt{2}\,\mathcal{E}{\left(\alpha,1,-\alpha;1\right)},\\ \end{align}$$
where $\mathcal{E}{\left(A,B,C;U\right)}$ is an elliptic integral whose general evaluation in terms of standard elliptic integrals is given in the appendix below. One finds
$$\mathcal{E}{\left(A,B,C;U\right)}=\frac{2\,E{\left(\arcsin{\sqrt{\frac{A-U}{A-B}}},\sqrt{\frac{A-B}{A-C}}\right)}}{\left(B-C\right)\sqrt{A-C}}-\frac{2\sqrt{\frac{\left(A-U\right)\left(U-B\right)}{U-C}}}{\left(A-C\right)\left(B-C\right)}.$$
Setting $U=B=1\land A=\alpha\land C=-\alpha$, we have
$$\mathcal{E}{\left(\alpha,1,-\alpha;1\right)}=\frac{\sqrt{2}}{\left(1+\alpha\right)\sqrt{\alpha}}\,E{\left(\sqrt{\frac{\alpha-1}{2\alpha}}\right)}.$$
Hence,
$$\begin{align} J{\left(\alpha\right)} &=\sqrt{2}\,\mathcal{E}{\left(\alpha,1,-\alpha;1\right)}\\ &=\frac{2}{\left(1+\alpha\right)\sqrt{\alpha}}\,E{\left(\sqrt{\frac{\alpha-1}{2\alpha}}\right)}.\\ \end{align}$$
And hence,
$$\begin{align} I{\left(\kappa\right)} &=\frac{\kappa^2+1}{8\kappa^3}\,J{\left(\alpha\right)}\\ &=\frac{\kappa^2+1}{8\kappa^3}\cdot\frac{2}{\left(1+\alpha\right)\sqrt{\alpha}}\,E{\left(\sqrt{\frac{\alpha-1}{2\alpha}}\right)}\\ &=\frac{\kappa^2+1}{4\kappa^3}\cdot\frac{2\kappa^2}{\left(\kappa^2+1\right)^2\sqrt{\frac{1+\kappa^4}{2\kappa^2}}}\,E{\left(\frac{1-\kappa^2}{\sqrt{2}\sqrt{1+\kappa^4}}\right)}\\ &=\frac{1}{\sqrt{2}\left(1+\kappa^2\right)\sqrt{1+\kappa^4}}\,E{\left(\frac{1-\kappa^2}{\sqrt{2}\sqrt{1+\kappa^4}}\right)}.\\ \end{align}$$
And finally,
$$\begin{align} \ell{\left(a,b\right)} &=8\,H{\left(a,b\right)}\,I{\left(\kappa\right)}\\ &=\frac{8\,H{\left(a,b\right)}}{\sqrt{2}\left(1+\kappa^2\right)\sqrt{1+\kappa^4}}\,E{\left(\frac{1-\kappa^2}{\sqrt{2}\sqrt{1+\kappa^4}}\right)}\\ &=\frac{16\,\frac{ab}{a+b}}{\sqrt{2}\left(1+\sqrt{\frac{a-b}{a+b}}\right)\sqrt{\frac{2a}{a+b}}}\,E{\left(\frac{1-\sqrt{\frac{a-b}{a+b}}}{\sqrt{2}\sqrt{\frac{2a}{a+b}}}\right)}\\ &=\frac{8\sqrt{a}b}{\sqrt{a+b}+\sqrt{a-b}}\,E{\left(\frac{\sqrt{a+b}-\sqrt{a-b}}{2\sqrt{a}}\right)}\\ &=4\sqrt{a}\left(\sqrt{a+b}-\sqrt{a-b}\right)\,E{\left(\frac{\sqrt{a+b}-\sqrt{a-b}}{2\sqrt{a}}\right)}.\blacksquare\\ \end{align}$$
Appendix:
For $A>U\ge B>C$,
$$\begin{align} \mathcal{E}{\left(A,B,C;U\right)} &=\int_{U}^{A}\frac{\mathrm{d}x}{\sqrt{\left(A-x\right)\left(x-B\right)\left(x-C\right)^3}}\\ &=\int_{0}^{A-U}\frac{\mathrm{d}y}{\sqrt{y\left(A-B-y\right)\left(A-C-y\right)^3}};~~~\small{\left[A-x=y\right]}\\ &=\small{\int_{0}^{\frac{A-U}{A-B}}\frac{\left(A-B\right)\,\mathrm{d}z}{\sqrt{\left(A-B\right)^2z\left(1-z\right)\left(A-C-\left(A-B\right)z\right)^3}}};~~~\small{\left[\frac{y}{A-B}=z\right]}\\ &=\frac{1}{\sqrt{\left(A-C\right)^3}}\int_{0}^{\frac{A-U}{A-B}}\frac{\mathrm{d}z}{\sqrt{z\left(1-z\right)\left(1-\kappa^2z\right)^3}};~~~\small{\left[\kappa:=\sqrt{\frac{A-B}{A-C}}\right]}\\ &=\frac{2}{\sqrt{\left(A-C\right)^3}}\int_{0}^{\sqrt{\frac{A-U}{A-B}}}\frac{\mathrm{d}w}{\sqrt{\left(1-w^2\right)\left(1-\kappa^2w^2\right)^3}};~~~\small{\left[\sqrt{z}=w\right]}\\ &=\small{\frac{2}{\sqrt{\left(A-C\right)^3}}\int_{0}^{\sin{\psi}}\frac{\mathrm{d}w}{\sqrt{\left(1-w^2\right)\left(1-\kappa^2w^2\right)^3}}};~~~\small{\left[\psi:=\arcsin{\sqrt{\frac{A-U}{A-B}}}\right]}\\ &=\frac{2}{\sqrt{\left(A-C\right)^3}}\int_{0}^{\psi}\frac{\mathrm{d}\varphi}{\sqrt{\left(1-\kappa^2\sin^2{\varphi}\right)^3}};~~~\small{\left[w=\sin{\varphi}\right]}.\\ \end{align}$$
Defining $\kappa^{\prime}:=\sqrt{1-\kappa^2}$,
$$\begin{align} \mathcal{E}{\left(A,B,C;U\right)} &=\frac{2}{\sqrt{\left(A-C\right)^3}}\int_{0}^{\psi}\frac{\mathrm{d}\varphi}{\sqrt{\left(1-\kappa^2\sin^2{\varphi}\right)^3}};~~~\small{\left[w=\sin{\varphi}\right]}\\ &=\small{\frac{2}{\sqrt{\left(A-C\right)^3}}\int_{0}^{\psi}\left[\frac{\sqrt{1-\kappa^2\sin^2{\varphi}}}{\kappa^{\prime\,2}}-\frac{\kappa^2}{\kappa^{\prime\,2}}\frac{1-2\sin^2{\varphi}+\kappa^2\sin^4{\varphi}}{\sqrt{\left(1-\kappa^2\sin^2{\varphi}\right)^3}}\right]\,\mathrm{d}\varphi}\\ &=\frac{2}{\sqrt{\left(A-C\right)^3}}\frac{1}{\kappa^{\prime\,2}}\int_{0}^{\psi}\sqrt{1-\kappa^2\sin^2{\varphi}}\,\mathrm{d}\varphi\\ &~~~~~-\frac{2}{\sqrt{\left(A-C\right)^3}}\frac{\kappa^2}{\kappa^{\prime\,2}}\int_{0}^{\psi}\frac{1-2\sin^2{\varphi}+\kappa^2\sin^4{\varphi}}{\sqrt{\left(1-\kappa^2\sin^2{\varphi}\right)^3}}\,\mathrm{d}\varphi\\ &=\frac{2}{\sqrt{\left(A-C\right)^3}}\frac{1}{\kappa^{\prime\,2}}\,E{\left(\psi,\kappa\right)}\\ &~~~~~-\frac{2}{\sqrt{\left(A-C\right)^3}}\frac{\kappa^2}{\kappa^{\prime\,2}}\int_{0}^{\psi}\frac{d}{d\varphi}\left[\frac{\sin{\varphi}\cos{\varphi}}{\sqrt{1-\kappa^2\sin^2{\varphi}}}\right]\,\mathrm{d}\varphi\\ &=\frac{2}{\sqrt{A-C}}\frac{1}{\kappa^{\prime\,2}}\left[E{\left(\psi,\kappa\right)}-\frac{\kappa^2\sin{\psi}\cos{\psi}}{\sqrt{1-\kappa^2\sin^2{\psi}}}\right]\\ &=\frac{2}{\sqrt{\left(A-C\right)^3}}\cdot\frac{A-C}{B-C}\left[E{\left(\psi,\kappa\right)}-\frac{\left(\frac{A-B}{A-C}\right)\sqrt{\frac{A-U}{A-B}}\sqrt{\frac{U-B}{A-B}}}{\sqrt{1-\left(\frac{A-B}{A-C}\right)\left(\frac{A-U}{A-B}\right)}}\right]\\ &=\small{\frac{2}{\left(B-C\right)\sqrt{A-C}}\left[E{\left(\arcsin{\sqrt{\frac{A-U}{A-B}}},\sqrt{\frac{A-B}{A-C}}\right)}-\frac{\sqrt{A-U}\sqrt{U-B}}{\sqrt{A-C}\sqrt{U-C}}\right]}\\ &=\frac{2\,E{\left(\arcsin{\sqrt{\frac{A-U}{A-B}}},\sqrt{\frac{A-B}{A-C}}\right)}}{\left(B-C\right)\sqrt{A-C}}-\frac{2\sqrt{\frac{\left(A-U\right)\left(U-B\right)}{U-C}}}{\left(A-C\right)\left(B-C\right)}.\\ \end{align}$$