Expected value of first coordinate of a vector uniformly sampled from a hemisphere I'm looking for a way to compute the following: Defining $H$ to be the  hemisphere $H=\left\{  x\in S^{d-1} \mid x_1>0 \right\}$, what is $\mathbb{E}[x_1]$ when $x$ is sampled uniformly from $H$?
Another way to look at this that may be easier on the eyes, is to ask what is $\mathbb{E}\left[ \mathbb{1}_{[x_1 > 0]} \cdot x_1\right]$ when $x$ is sampled uniformly from $S^{d-1}$?
Thanks in advance!
 A: Let's first consider when $d=3$, and then we'll generalize. Define the function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ by
$$f(x,y,z) = \left\{ \begin{array}{ll} \frac{1}{2\pi}: \quad (x,y,z)\in H \\ 0: \quad \text{ else} \end{array} \right. $$
Then $f$ corresponds to the pdf for the random vector $(X,Y,Z)\sim \mathcal{U}(H)$. The expected value of $X$ is the surface integral $$\mathbb{E}(X)=\int_{H}xf(x,y,z)\mathrm{d}S$$ We can parameterize $H$ by $$\vec{r}(u,v)=\Big<\sqrt{1-u^2-v^2},u,v\Big>:u^2+v^2<1$$ Notice how $$\mathrm{d}S=\|\vec{r}_u \times  \vec{r}_v\|\mathrm{d}u\mathrm{d}v=\frac{1}{\sqrt{1-u^2-v^2}}$$ So we get $$\mathbb{E}(X)=\int_{\{u^2+v^2<1\}}\frac{\mathrm{d}u\mathrm{d}v}{2\pi}=\frac{\pi}{2\pi}=\frac{1}{2}$$ This generalizes as follows. Re$-$define $f:\mathbb{R}^d\rightarrow \mathbb{R}$ by $$f(\vec{x}) = \left\{ \begin{array}{ll} \frac{\Gamma(d/2)}{\pi^{d/2}}: \quad \vec{x}\in H \\ 0: \quad \text{ else} \end{array} \right. $$ Just as before, $f$ corresponds to a pdf associated with $\vec{X}\sim \mathcal{U}(H)$, and we're tasked to evaluate the surface integral $$\mathbb{E}(X_1)=\int_{H}x_1f(\vec{x})\mathrm{d}S_{d-1}$$ By invoking a similar parameterizaion $$\vec{r}(x_1,\ldots,x_{d-1})=\Big<\sqrt{1-x_1^2-\dots -x_{d-1}^2},x_1,\ldots,x_{d-1}\Big>$$ defined on the domain $$x_1^2+\dots + x_{d-1}^2<1$$ you'll get that \begin{eqnarray*}
\mathbb{E}(X_1) & = & \frac{\Gamma(d/2)}{\pi^{d/2}}\int_{H}x_1\mathrm{d}S_{d-1}
\\ & = & \frac{\Gamma(d/2)}{\pi^{d/2}}\int_{\{x_1^2+\dots +x_{d-1}^2<1\}}\mathrm{d}V_{d-1} \\
 & = & \frac{\Gamma(d/2)}{\pi^{d/2}} \cdot \frac{\pi^{\frac{d-1}{2}}}{\Gamma\big(\frac{d+1}{2}\big)} \\ & = & \frac{\Gamma(d/2)}{\sqrt{\pi}\Gamma\big(\frac{d+1}{2}\big)}
\end{eqnarray*}
