Let $M=\{ (x_1,x_2,x_3) \in \mathbb{S}^2 \, | \, x_3 \ge 0 \}$ be the closed upper-hemisphere in $\mathbb{R}^3$.
Is there a nice way to calculate $\int_M(1-x_3)^2d\sigma$, where $d\sigma$ is the standard spherical measure?
(without using spherical coordinates).
My idea was $$\int_M(1-x_3)^2d\sigma=\int_M 1 d\sigma+\int_M x_3^2 d\sigma-2\int_M x_3 d\sigma=2\pi+\frac{1}{2}\int_{\mathbb{S}^2} x_3^2 d\sigma-2\int_M x_3 d\sigma. \tag{1}$$
By symmetry, $$\int_{\mathbb{S}^2} x_3^2 d\sigma=\frac{1}{3} \int_{\mathbb{S}^2} x_1^2+x_2^2+x_3^2 d\sigma=\frac{1}{3} \int_{\mathbb{S}^2} 1 d\sigma=\frac{4\pi}{3}. \tag{2}$$
Combining equations $(1),(2)$ we get $$ \int_M(1-x_3)^2d\sigma=\frac{8\pi}{3}-2\int_M x_3 d\sigma.$$
So, I am asking essentially if there is an elegant way to compute $\int_M x_3 d\sigma$.
Consider that on $M$, $$\hat n = \frac{\hat{e_1}x_1+\hat{e_2}x_2+\hat{e_3}x_3}{\sqrt{x_1^2+x_2^2+x_3^2}}=\hat{e_1}x_1+\hat{e_2}x_2+\hat{e_3}x_3$$ and $\vec{\nabla}x_3=\hat{e_3}$ so $\vec{\nabla}x_3\cdot\hat n=x_3$. Then if $\Omega$ is the solid upper hemisphere $x_1^2+x_2^2+x_3^2\le1$, $x_3\ge0$, $$\begin{align}\int_{\Omega}\nabla^2x_3d\tau&=\int_{\Omega}0d\tau=0=\int_{\partial\Omega}\vec{\nabla}x_3\cdot\hat nd\sigma\\ &=\int_M\vec{\nabla}x_3\cdot\hat nd\sigma+\int_N\vec{\nabla}x_3\cdot\hat nd\sigma\\ &=\int_Mx_3d\sigma-\int_Nd\sigma=\int_Mx_3d\sigma-\pi\end{align}$$ Where $N$ is the bottom surface of the hemisphere, $x_1^2+x_2^2\le1$, $x_3=0$. So $$\int_Mx_3d\sigma=\pi$$ A simpler way to look at it is that $x_3d\sigma$ is the projection of the areal element of $M$ onto $N$, so the answer is just the area of $N$, which is $\pi$. $$$$