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I need a fast and accurate method to calculate 3d spherical volume integrals.

I have pre-calculated data of high precision that just needs a few trivial manipulations before each integration step - in other words function calls are mostly just array look ups and multiplication and so are relatively cheap.

The data is evenly spaced in spherical polar coordinates: 4 samples per degree for theta and phi, 500 radial samples.

I need good precision, but I also need to perform several million such integrals, so speed is an important concern.

What algorithm is most suitable for this problem?

Bonus points for a link to an implementation in C++!

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    $\begingroup$ Do you need your data evenly spaced? Probably that's not the best way to perform the integrals. Have you checked on Lebedev quadrature? There is a C implementation in Github. $\endgroup$ – nicoguaro Jun 16 '17 at 14:40
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    $\begingroup$ ... or, a bit simpler than the Lebedev stuff, use Legendre nodes for the polar angle. Equidistant in $\phi$ plus Legendre in $\theta$ are optimal by themselves, Lebedev adds the two-dimensional view. When I used it, I concluded for myself that the reduction in gridpoints is not worth the complication of using the Lebedev points). $\endgroup$ – davidhigh Jun 16 '17 at 15:01
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So you want to evaluate

$$ F=\int_0^r \int_0^{2\pi} \int_{0}^\pi f(r,\phi,\theta) \, r^2 \, \sin (\theta) \ dr \, d\phi \, d\theta $$ and you have your function $f$ available on equidistant gridpoints $r_i$, $\phi_j$ and $\theta_k$. Call

$$ f_{ijk} = f(r_i, \phi_j, \theta_k) $$

The plain simple approach is then just to use the trapezoidal rule (I'll omit the one-halves at the endpoints):

$$ F \approx \sum_{ijk}f_{ijk} \, r_i^2 \, \sin(\theta_i) $$

... or, if you're feeling for higher order, use another Newton-Cotes formula.

Just to note that once. For other approaches, see the comment section.

EDIT: I've just noticed you were asking for the "best" method, so this will probably be quite boring for you.

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If the subvolumes are evenly distributed and you can assume the function is constant inside them you probably only need a lookup table for the volume size in the radius and do a weighted sum. If the assumptions are not true it will probably not be very accurate though :)

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