Integration of a coordinate function over a hemisphere

Question

Let $\mathbb{S}^n = \{ (x_0, x_1, \dots, x_n) \in \mathbb{R}^{n+1} : x_0^2 + \cdots + x_n^2 = 1 \}$ be the unit sphere in $\mathbb{R}^{n+1}$ and let $\mathbb{S}^{n}_+ = \mathbb{S}^n \cap \{x_0 \geq 0 \}$ be a closed hemisphere. Could you help me to compute the integrals

$$I = \int_{\mathbb{S}^{n}_+} x_0 \, \mathrm{d}A$$

and

$$J = \int_{\mathbb{S}^{n}_+} (x_0)^3 \, \mathrm{d}A$$ please? I thought that polar coordinates might be useful...

Here is my try for $I$. Consider the map $f : A:= [0, \pi/2] \times \mathbb{S}^{n-1} \to \mathbb{S}^n_+$ given by

$$f(\theta,x) = (\sin \theta, \cos \theta x).$$

The Jacobian determinant of $f$ is easily seen to be $\cos^{n+1} \theta$. So,

$$I = \int_{f(A)} x_0 \, \mathrm{d}A = \int_0^{\pi/2} \int_{\mathbb{S}^{n-1}} \sin \theta \cos^{n+1} \theta \, \mathrm{d} A_{n-1} \mathrm{d} \theta = \operatorname{vol}(\mathbb{S}^{n-1}) \int_0^{\pi/2} \cos^{n+1} \theta \sin \theta \, \mathrm{d} \theta \\ = - \frac{\operatorname{vol}(\mathbb{S}^{n-1})}{n+2} \cos^{n+2} \theta \big\vert_0^{\pi/2} = \frac{\operatorname{vol}(\mathbb{S}^{n-1})}{n+2}.$$

Is it correct? What about $J$?

Answer 1

The surface area form for the part of the sphere $\mathbb{S}^n$ in $\mathbb{R}^{n+1}$ with $x_0>0$ is $$ \frac{1}{x_0}dx_1\cdots dx_n. $$

As a sanity check, we can use this form to compute the length/surface area of half a sphere. In the case where $n=1$, then $x_1$ varies between $-1$ and $1$, and $$ \int_{x_0\geq 0}dA=\int_{-1}^1\frac{1}{x_0}dx_1=\int_{-1}^1\frac{1}{\sqrt{1-x_1^2}}dx_1=\left.\arcsin(x_1)\right|_{-1}^1=\frac{\pi}{2}-\frac{-\pi}{2}=\pi, $$ which is half the circumference of the unit circle.

In the case where $n=2$, then $x_1$ and $x_2$ vary over the unit disk in $\mathbb{R}^2$, so $$ \int_{x_0\geq 0}dA=\int_{x_1^2+x_2^2\leq 1}\frac{1}{x_0}dx_1dx_2=\int_{x_1^2+x_2^2\leq 1}\frac{1}{\sqrt{1-x_1^2-x_2^2}}dx_1dx_2. $$ Switching to polar, this becomes $$ \int_0^{2\pi}\int_{r\leq 1}\frac{rdrd\theta}{\sqrt{1-r^2}}=\left.-2\pi\sqrt{1-r^2}\right|_{r=0}^{r=1}=2\pi, $$ which is half the surface area of a sphere.

Now that we've done the sanity check, let's try your cases. Let $D^n$ be the $n$-dimensional unit disk in $\mathbb{R}^n$. Then $$ \int_{\mathbb{S}^n_+}x_0dA=\int_{D^n}dx_1\cdots dx_n, $$ which is the area of a disk in $\mathbb{R}^n$. At this point, one might want to use polar coordinates parametrized by $r$ and a point in $\mathbb{S}^{n-1}$. In other words, this integral equals $$ \int_{\mathbb{S}^{n-1}}\int_{0}^1r^{n-1}drdA. $$ This is $\frac{1}{n}$ times the volume of $\mathbb{S}^{n-1}$.

Similarly, $$ \int_{\mathbb{S}^n_+}(x_0)^3dA=\int_{D^n}(x_0)^2dx_1\cdots dx_n=\int_{D^n}(1-x_1^2-\dots-x_n^2)dx_1\cdots dx_n. $$ At this point, one might want to use polar coordinates parametrized by $r$ and a point in $\mathbb{S}^{n-1}$. In other words, this integral equals $$ \int_{\mathbb{S}^{n-1}}\int_{0}^1(1-r^2)r^{n-1}drdA. $$ This is the volume of $\mathbb{S}^{n-1}$ times $$ \int_0^1r^{n-1}(1-r^2)dr=\frac{1}{n}-\frac{1}{n+2}=\frac{2}{n(n+2)}. $$

Answer 2

This isn't an answer directly, but explains why the OP's formula fails. The problem is in the computation of the Jacobian. It looks like the OP uses the map $$ f:[0,2\pi]\times\mathbb{S}^{n-1}\rightarrow\mathbb{S}^n $$ and the formula $$ (\theta,(x_1,\dots,x_n))\mapsto (\sin(\theta),\cos(\theta)x_1,\dots,\cos(\theta)x_n). $$ It looks like the Jacobian would be $\cos(\theta)^{n+1}$, but the problem is that $x_1,\dots,x_n$ are not independent coordinates since they lie on an $(n-1)$-dimensional sphere. Similarly, the coordinates for the output are not independent since they lie on an $n$-dimensional sphere. Therefore, you can't take the Jacobian by just computing derivatives with respect to $\theta$ and $x_i$'s.

Answer 3

The intersection of the unit sphere with the hyperplane $x_0=x$ has $n-1$ dimensional measure $\omega_{n-1}\left(1-x^2\right)^{\large\frac{n-1}2}$ and the strip between the hyperplanes $x_0=x$ and $x_0=x+\mathrm{d}x$ has width $\left(1-x^2\right)^{\large-\frac12}\mathrm{d}x$. Thus, the area of this strip is $\omega_{n-1}\left(1-x^2\right)^{\large\frac{n-2}2}\mathrm{d}x$. Therefore, the integral of $x_0^k$ over the portion of the sphere where $x_0\gt0$ is $$ \begin{align} \omega_{n-1}\int_0^1x^k\left(1-x^2\right)^{\large\frac{n-2}2}\,\mathrm{d}x &=\omega_{n-1}\frac{\Gamma\!\left(\frac{k+1}2\right)\Gamma\!\left(\frac{n}2\right)}{2\,\Gamma\!\left(\frac{n+k+1}2\right)}\tag1\\ &=\pi^{n/2}\frac{\Gamma\!\left(\frac{k+1}2\right)}{\Gamma\!\left(\frac{n+k+1}2\right)}\tag2 \end{align} $$ Explanation:
$(1)$: Beta Integral
$(2)$: $\omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$ (see here)

For $k=1$, $(1)$ gives $$ \omega_{n-1}\frac{\Gamma(1)\,\Gamma\left(\frac{n}2\right)}{2\,\Gamma\left(\frac{n}2+1\right)}=\frac{\omega_{n-1}}n\tag3 $$ For $k=3$, $(1)$ gives $$ \omega_{n-1}\frac{\Gamma(2)\,\Gamma\left(\frac{n}2\right)}{2\,\Gamma\left(\frac{n}2+2\right)}=\frac{2\omega_{n-1}}{n(n+2)}\tag4 $$