# Intersection and Curvature of Surfaces

a) Describe the intersection $(C)$ of sphere $x^2 + y^2 + z^2 = 1$ and the elliptic cylinder $x^2 + 2z^2 = 1$, and find out the total arc-length of this intersection.

b) Determine the points on the curve $(C)$, where the curvature have the maximum value.

I've been tackling this problem for quite awhile now, going through my notes, researching, reading my textbook but I have not gotten anything productive. Any form of help or solution would be greatly appreciated.

There are two curves. They can be parametrized as follows.

$\gamma_1 :[0,2\pi]\to\mathbb{R}^3$ given by $\gamma_1(t)=(\cos t,\frac{1}{\sqrt{2}}\sin t,\frac{1}{\sqrt{2}}\sin t)$, and

$\gamma_2 :[0,2\pi]\to\mathbb{R}^3$ given by $\gamma_2(t)=(\cos t,\frac{1}{\sqrt{2}}\sin t,-\frac{1}{\sqrt{2}}\sin t)$.

So $C=\gamma_1([0,2\pi])\cup \gamma_2([0,2\pi])$. Length of $\gamma_1$ is the same of $\gamma_2$, then the length of $C$ is \begin{align*} s&=2\int_0^{2\pi}\sqrt{(-\sin t)^2+2\left(\frac{1}{\sqrt{2}}\cos t\right)^2}\mathrm dt\\ &=2\int_0^{2\pi}\mathrm dt\\ &=4\pi \end{align*}

Also, it can be proved that $\gamma_1([0,2\pi])$ and $\gamma_2([0,2\pi])$ are circles of radius $r=1$ in the planes $z=y$ and $z=-y$, respectively. And is a circle in the plane. Therefore, the curvature is constant, $\kappa=1$, for every point of $C$.

• Thank you! This helps a lot. I was wondering though if you'd be able to explain how you got your parametrized equations for the two curves? I understand why you do it, just not entirely how. And one last thing, before you found the length of C, you took the derivative, why? By definition are we meant to? Commented Nov 7, 2014 at 15:25
• Sorry for the constant questions, but for b) to find the points on C with the maximum curvature, how would I go about it? Would I have to individually make $\gamma_1 = 0$ and $\gamma_2 = 0$, isolate $t$ and find that coordinate? Commented Nov 7, 2014 at 15:42
• Hello, there are many parametrizations, I prefer these ones that make easier to obtain the arc length, using trigonometric functions. Also, since $C\subset \mathbb{R}^3$ is the set of points that satisfy both equations $x^2+y^2+z^2=1$ and $x^2+2z^2=1$ it follows $y^2=z^2$ or $y=\pm z$. Then, $C$ is the set of points $(x,y,y)$ and $(x,y,-y)$ such that $x^2+2y^2=1$, also I knew this equation describes an ellipse in the plane, then I took the parametrization $x=\cos t, y=\frac{1}{\sqrt{2}}\sin t$ $t\in[0,2\pi]$, I knew this parametrization works! Finally, I gave $\gamma_1$ and $\gamma_2$ as above Commented Nov 7, 2014 at 19:09

From $x^2+y^2+z^2=1$ and $x^2+2z^2=1$ it immediately follows that $z^2-y^2=0$, or $z=\pm y$. Conversely, $x^2+y^2+z^2=1$ and $z=\pm y$ imply $x^2+2z^2=1$. Therefore the intersection of $S^2$ and the elliptical cylinder coincides with the intersection of $S^2$ with the two planes $z=\pm y$, i.e. with the union of two unit circles intersecting at $(1,0,0)$. The total arc length is $4\pi$, and the curvature along these circles is of course constant.