Is there a nice way to calculate $\int (1-x_3)^2$ over the hemisphere? 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 $\int_M x_3 d\sigma$.
 A: As is well known the area of an "infinitesimal lampshade" on $S^2$ between two horizontal planes at distance $dx_3>0$ is $2\pi\, dx_3$.  It  follows that your original integral is given by
$$\int_M(1-x_3)^2\>d\sigma=2\pi\int_0^1(1-x_3)^2\>dx_3={2\pi\over3}\ ,$$
and the integral in your last displayed formula is
$$\int_M x_3\>d\sigma=2\pi\int_0^1 x_3\>dx_3=\pi\ .$$
A: 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 $\pi$.
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