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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 $\int_M x_3 d\sigma$.

Asaf Shachar
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2 Answers2

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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 $$\int_M x_3\>d\sigma=2\pi\int_0^1 x_3\>dx_3=\pi\ .$$

Christian Blatter
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  • Thanks. Can you please elaborate on how exactly do you use Fubini's theorem here? I am afraid I don't see how exactly did you get the first equality. – Asaf Shachar Feb 25 '18 at 13:12
  • I still don't get it. Eventually you are using some "integration-by-substitution" scheme, or "pullback-invariance" or some other measure-theoretic machinery. (Or indeed some version of Fubini on product spaces). Intuitively, your argument seems fine, but I don't see it rigorously yet. – Asaf Shachar Feb 25 '18 at 13:57
  • (+1) When I saw this question just now, this is how I thought to do the integral. – robjohn Feb 25 '18 at 15:00
  • @AsafShachar: what Christian Blatter is using is the fact that the surface area of the unit sphere between the planes $z=a$ and $z=b$ is $2\pi,| ,a-b,|$. See this answer. – robjohn Feb 25 '18 at 15:03
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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 $\pi$. $$$$

user5713492
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