It is known that a regular tetrahedron can be inscribed in a cube such that two vertices of the tetrahedron are on opposite faces of the cube, and the edges of the tetrahedron form the diagonals of the cube's faces (Figure 1). From this, it is evident that the opposite edges of a regular tetrahedron are perpendicular to each other, and if the edge length of the tetrahedron is \(a\), then the length of the segment connecting the midpoints of two opposite edges is \(\frac{a \sqrt{2}}{2}\), and this segment is perpendicular to the planes that are parallel to both edges.

Using these properties, we can easily determine the dimensions of the cylinder. Let the vertices \(A\) and \(B\) of the tetrahedron be on the base of the cylinder, and the vertices \(C\) and \(D\) be on the top face. The perpendicular projection of the edge \(CD\) onto the base is \(C' D'\) (Figure 2). Since the cylinder is right, \(C'\) and \(D'\) are on the circumference of the base, and \(C' D' = CD = AB\). In the tetrahedron, \(AB\) and \(CD\) are opposite edges, so the segment connecting the midpoints of \(AB\) and \(CD\), \(E\) and \(F\), is perpendicular to the planes that are parallel to both \(AB\) and \(CD\); hence, it is perpendicular to the base and the top face of the cylinder. This means that \(EF = \frac{2 \sqrt{2}}{2} = \sqrt{2}\) is the height of the cylinder, and the perpendicular projection of \(F\) onto the base, \(F'\), coincides with \(E\). Since the perpendicular projection of the midpoint of a segment is the midpoint of the perpendicular projection of the segment, \(E\) also bisects \(C' D'\). Therefore, the points \(A, B, C'\), and \(D'\) on the circumference of the base circle \(k\) are such that \(AB = C' D'\), \(AB \perp C' D'\), and the midpoints of \(AB\) and \(C' D'\) (i.e., \(E\) and \(F'\)) coincide (not as shown in Figure 2). Since the midpoint of a chord of a circle lies on the perpendicular from the center of the circle to the chord, \(AB\) and \(CD\) must be diameters of the base and top circles, respectively.

Thus, the radius of the base circle of the cylinder is \(r = \frac{AB}{2} = 1\) unit, and the height of the cylinder is \(m = \sqrt{2}\) units, so the volume of the cylinder is \(V = r^2 \pi m = \sqrt{2} \pi \approx 4.44\) cubic units.
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A regular tetrahedron with 2 unit edges has two of its vertices on the circumference of the base and the top of a right circular cylinder. What is the volume of the cylinder?