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Active Calculus - Multivariable

Activity 13.12.5.
In some sense, this activity considers the reverse problem of that considered in Activity 13.12.4. Here, each part of the activity gives you an oriented simple closed curve \(C\) in \(\R^3\text{,}\) 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 \(S\text{.}\)
You are encouraged to think about multiple possible answers, since as we saw in Preview Activity 13.12.1, there may be more than one reasonable choice of a surface with a particular boundary.
(a)
The curve \(C\) is the triangle with vertices \((1,0,0)\text{,}\) \((0,1,0)\text{,}\) and \((0,0,1)\) with orientation corresponding to the order the points are listed here.
(b)
The curve \(C\) is the circle parameterized as \(\vr(t) =\langle \sqrt{2}\cos(t), \sqrt{2}\cos(t), 2\sin(t)\rangle\) for \(0\leq t\leq 2\pi\text{.}\)
(c)
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 \((0,2,0)\text{,}\)
  • the line segment \(C_2\) from \((0,2,0)\) to \((0,2,2)\text{,}\)
  • the quarter-circle \(C_3\) of radius \(2\) centered at \((0,0,2)\) in the plane \(z=2\) from \((0,2,2)\) to \((2,0,2)\text{,}\) and
  • the line segment \(C_4\) from \((2,0,2)\) to \((2,0,0)\text{.}\)