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Use tensor-product quadrature with composite Simpson's rule in each coordinate. If there are $m$ subintervals in each direction, then $N=(m+1{)}^2$ and the quadrature rule error is $O(m^{-4})$: apply the one-dimensional error bound for composite Simpson's rule successively, using bounded fourth partial derivatives. Hence the error is $O(N^{-2})$ in big O notation. The rectangular domain makes the product grid convenient.

For a generic integrand in 20 dimensions, Monte Carlo integration gives root mean square error $O(N^{-1/2})$, whereas tensor-product quadrature with composite Simpson's rule gives only $O(N^{-4/20})=O(N^{-1/5})$. One can sample uniformly in the Euclidean unit ball, for example by an independent uniform direction and radius $U^{1/20}$ with $U$ uniform on $[0,1]$.

For this particular integrand, the most efficient elementary choice uses radial integration over a ball first. The area of the unit 19-sphere is $2{\pi }^{10}/9!$, as follows from the volume of an n-ball. HenceThe reduced integrand is smooth. Apply composite Simpson's rule to obtain $O(N^{-4})$ deterministic error in big O notation, with one original radial integrand evaluation per node. If the selection is restricted to direct multidimensional methods, Monte Carlo integration with $O(N^{-1/2})$ root mean square error is the appropriate choice.