Activity 13.13.5

In some sense, this activity considers the reverse problem of that considered in this activity. Here, each part of the activity gives you an oriented simple closed curve $C$ in $\R^3$, and your task is to find

• a surface $S$ so that $C$ is the boundary of $S$ and
• a normal vector for the $S$ so that a person walking along $C$ in the direction of the given orientation with head pointing in the direction of your chosen normal vector would have their left hand over $S$.

You are encouraged to think about multiple possible answers, since as we saw in this activity, there may be more than one reasonable choice of a surface with a particular boundary.

1. The curve $C$ is the triangle with vertices $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ with orientation corresponding to the order the points are listed here.

2. The curve $C$ is the circle parameterized as $\vr(t) =\langle \sqrt{2}\cos(t), \sqrt{2}\cos(t), 2\sin(t)\rangle$ for $0\leq t\leq 2\pi$.

3. The curve $C$ consists of (given in order of the orientation)

   • the quarter-circle $C_1$of radius $2$ centered at the origin in the $xy$-plane from $(2,0,0)$ to $(0,2,0)$,
   • the line segment $C_2$ from $(0,2,0)$ to $(0,2,2)$,
   • the quarter-circle $C_3$ of radius $2$ centered at $(0,0,2)$ in the plane $z=2$ from $(0,2,2)$ to $(2,0,2)$, and
   • the line segment $C_4$ from $(2,0,2)$ to $(2,0,0)$.