4.35. Given a regular tetrahedron with edge length 1. A sphere is tangent to three of its edges, which emanate from one vertex, at their ends. Find the area of the part of the sphere's surface that is located inside the tetrahedron.
4.35. Consider a regular tetrahedron with edge length 2. The surface of the sphere that touches all its edges is divided by its surface into 4 equal curvilinear triangles, the area of each of which is the desired quantity, and 4 equal segments. Let \( x \) be the distance from the center of a face to a vertex, \( y \) be the distance from the center of the tetrahedron to a face, and \( z \) be the distance from the center of a face to an edge of that face. It is easy to verify that \( x = 2 / \sqrt{3} \) and \( z = 1 / \sqrt{3} \). Further, \( y = h / 4 \), where \( h = \sqrt{4 - x^2} = \sqrt{8 / 3} \) is the height of the tetrahedron, i.e., \( y = 1 / \sqrt{6} \). The radius \( r \) of the sphere is \( \sqrt{y^2 + z^2} = \sqrt{1 / 6 + 1 / 3} = 1 / \sqrt{2} \). The height of each of the four segments is \( r - y = 1 / \sqrt{2} - 1 / \sqrt{6} \). Therefore, the desired area is

\[
\frac{1}{4}\left(4 \pi \left(\frac{1}{\sqrt{2}}\right)^2 - 4 \cdot 2 \pi \frac{1}{\sqrt{2}} \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}\right)\right) = \pi \left(\frac{1}{\sqrt{3}} - \frac{1}{2}\right),
\]