Discussion
A cylindrical rod of length \(h\) and diameter \(d\) is placed inside a cubic enclosure of side length \(L\). \(S\) denotes the inner surface of the cube. The view-factor \(F_{S-S}\) is
Answer: D
The view factor is \(1 - \dfrac{(\pi dh + \pi d^{2}/2)}{6L^{2}}\).
Call the rod's whole surface r. Its area is the curved surface plus two end discs, and the cube's inner area is six faces:
\[ \begin{aligned} A_{r} &= \pi dh + 2 \times \frac{\pi d^{2}}{4} \\ &= \pi dh + \frac{\pi d^{2}}{2} \\ A_{S} &= 6L^{2} \end{aligned} \]
The rod is convex and fully enclosed, so everything leaving it reaches the cube: \(F_{r-S} = 1\). Reciprocity gives \(A_{S}F_{S-r} = A_{r}F_{r-S}\), so \(F_{S-r} = A_{r}/(6L^{2})\). By the summation rule, \(F_{S-S} + F_{S-r} = 1\), so
\[ F_{S-S} = 1 - \frac{\pi dh + \pi d^{2}/2}{6L^{2}} \]
0 would hold for a flat or convex surface, but the inside of a cube sees itself. 1 would be correct only with no rod inside. \(\dfrac{(\pi dh + \pi d^{2}/2)}{6L^{2}}\) is \(F_{S-r}\), the fraction that reaches the rod, not the fraction that returns to the cube.
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