After years of doing things the wrong way, the wireless industry has recently realized that the traditional summation formulation for a spherical surface integral gives the wrong answer for the low resolution spherical measurements typically used for evaluation of total radiated power and receiver sensitivity. That's because the sin(theta) weighting from the dOmega term of the integral makes it appear that the points measured at the pole have zero weight. Certainly in the infinitesimal limit of a spherical integral summation, that may be true, but for relatively large step sizes (e.g. 15 degrees or more as typically used) the points at top and bottom of the sphere cannot be assumed to have zero weight.

In bringing this problem to light, I demonstrated a solution using surface segment weighting that resolves the problem nicely. However, subsequently it was also demonstrated that the Clenshaw-Curtis integral does an admirable job of integrating the surface in theta and produces weighting coefficients that, while generally only differing from the spherical surface weighting by tenths to hundredths of a percent, invariably tend to produce a slightly better fit than the spherical surface weighting. Thus, the current consensus is to adopt Clenshaw-Curtis for spherical integration.

Unfortunately, we've hit a snag in that in addition to full spherical integrals, there are various metrics based on partial surface integrals such as upper hemisphere or +/- 45 degrees from the equatorial (theta = 90 degree) cut. While the upper hemisphere is a trivial problem, determining the proper weighting coefficients for other partial surface endpoints such as theta = 90 +/- 30 or 45 degrees has proven to be more of a challenge. The solution in the spherical segment weighting approach is quite straightforward, but I must admit that I'm not enough of a pure mathematician to recreate the summation formulas for the Clenshaw-Curtis weights when endpoints other than theta = 0 and 180 (+/- 1 in the cos(theta) substituted transformation) are used. I've seen papers that indicate that "other intervals are trivial" but so far I have not found that to be the case, an no one has bothered to illustrate this triviality beyond only showing the "simple" [-1,1] case!

Thus, I've decided to post this here in the hopes that some mathematics expert who does find this trivial can help guide us to the correct answer. Please note that the intervals of the sampled data will remain the same (e.g. 15 or 30 degree steps from 0-180 degrees in theta) but only the portion of the surface integrated over will change. Thus, the weighting of all points fully represented within the bounds can be expected to be the same as for the full integral, those fully outside the bounds are zero, and the point representing the portion of the surface intersected by the boundary will be reduced by the appropriate portion used. And note that this is not as trivial as using 1/2 of the weight for the point that happens to lie on the boundary. That's because the partial surface area inside the boundary (for a partial surface including the equator) is larger than the partial surface outside the boundary.

At any rate, thanks for any help or guidance!

Sincerely,

Dr. Michael D. Foegelle