Let $\mathcal S^2$ denote the unit sphere, i.e $\mathcal S^2=\{(x,y,z)\vert x^2+y^2+z^2=1 \}$ and consider a function $f \in C(\mathcal S^2)$. We set
$\begin{equation} g(x)=\frac{1}{2\pi\sqrt{1-x^2}}\int \limits_{\{y^2+z^2=1-x^2\}} f(x,y,z)\;ds\quad (*) \end{equation}$
Our professor said the other day: "The motivation behind $(*)$ is that in each slice of the sphere the function $f$ remains the same, hence it makes sense to average."
I did not quite get this statement.
- Since we're talking about average, shouldn't the denominator $2\pi\sqrt{1-x^2}$ be equal to $\int \limits_{\{y^2+z^2=1-x^2\}} 1\;ds\;$? To me, it seems that the term $2\pi\sqrt{1-x^2}$ is of the form $2\pi \;r(x)$ where $r(x)=1-x^2$ is the radius of the disk $\{y^2+z^2=1-x^2\}$. But this if I 'm not mistaken stands for the perimeter of the disk (not for the area= $\int \limits_{\{y^2+z^2=1-x^2\}} 1\;ds\;$). How does then the denominator appear?
- Why does the function remain the same on each slice of the sphere? What does this mean and how it is possible?
I hope my question is clear. I'm having a really hard time understanding this statement. Any help will be much appreciated.
UPDATE: I think I might (almost) found the answers I needed.
- After the comment of Ted Shifrin, I think that $\int \limits_{\{y^2+z^2=1-x^2\}} 1\;ds\;$ is indeed equal to $2\pi\sqrt{1-x^2}$ since line integrals estimate in general the length of the curve and in this particular case the perimeter of the circle.
- I think that what the professor was saying is that in each of these circles $\{y^2+z^2=1-x^2\}$ the function $f(x,\cdot,\cdot)$ remains constant. Thus it makes sense to take the average among all these constants.
Is my reasoning correct? I would be really glad if someone could verify this or explain instead what I'm missing here.
Many thanks in advance!