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Suppose I have a function $f(\mathbf{x})$ across the surface of the unit hypersphere, where $\mathbf{x}=(x_1,...,x_d)'$ are the hyperspherical coordinates of a point on the surface of the unit hypersphere. I want to integrate this function over the positive orthant $\mathcal{S}^{p-1}_+$: $I=\int_{\mathcal{S}^{p-1}_+}f(\mathbf{x})d\mathbf{x}$. Let's assume this integral is analytically intractable, so I need to use numerical techniques, and $d$ is large, so Monte Carlo integration is more attractive that numerical quadrature approaches. Thus, I want to approximate the integral $I$ using Monte Carlo integration.

From Wikipedia, I will first sample $N$ points uniformly on the surface of the positive orthant. I am doing this by normalizing a vector of independent standard normal samples, then taking the absolute value of the normalized $d$-dimensional vector; this produces a uniform sample in $\mathcal{S}^{p-1}_+$. Then, I evaluate $f(\cdot)$ at each of these random samples and sum all the output together. Finally, I would need to multiple the sum by $V$ and divide by $N$, where $V$ is the volume of $\mathcal{S}^{p-1}_+$.

If everything looks right so far, then I just have one question: what is $V$? Is it the volume of the positive orthant, or the volume of the surface of the positive orthant? Technically, the point $\mathbf{x}$ can only lay on the surface of the sphere, so I'm tempted to thing $V$ is the "volume" (area) of the positive orthant, but I'm not too sure.

Ron Snow
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    Just follow Wikipedia: V is the volume of whatever you integrate over. You integrate over $\mathcal{S}^{p-1}_+$ so this is the volume you need to compute. Don't get hung up with words ("volume" vs "area") just follow the definitions. tl;dr: Yes it is the "volume of the surface". There is also no shame in checking this kind of thing experimentally: Just take a function with known integral and compare results. – g g May 10 '24 at 08:05

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