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I am trying to show the following thing.

Let $\Omega_{q}$ be the surface of unit sphere in $\mathbb{R}^q$ and let $x$ be the variable and $y$ an element of $\Omega_{q}.$ Then denote by $t = x^{T}y$ their scalar product and then we may write

$$ x = ty + (1 - t^2)^{1/2}\xi,$$

where $\xi$ is the unit vector, perpendicular to $x$. So far, it is clear to me. But then what I do not know is how to show the derivation of surface area element on this sphere, with the use of the previous formula and differentials $dt, d\xi,$ in the following form

$$\omega_{q}(dx) = (1 - t^2)^{p}dt\, \omega_{q-1}(d\xi),$$
where $p = (q - 3) / 2.$

Thank in advance for any comments on my problem.

stanly
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    This is a very useful formula that is often explained hastily. I like how it is treated in the books of Claus Müller on spherical harmonics (he has two, the latest one is more recent but it is also basically a remake of the first one from the 1960s). I have written something about this formula on Math.SE, but not a full proof. – Giuseppe Negro Oct 13 '21 at 15:02
  • Thank you very much. I will check the book as well as your post. Have a nice day! – stanly Oct 13 '21 at 15:09
  • @GiuseppeNegro I have realised that, the formula for expressing elements of sphere $$ x = ty + (1 - t^2)^{1/2}\xi,$$ is not so clear to me as i thought. Could i ask you for some hints to derive this formula. Thanks! – stanly Oct 20 '21 at 14:13
  • Because you have some mistakes there. Your y should be fixed once and for all and your xi should be orthogonal to y, not to x. – Giuseppe Negro Oct 20 '21 at 14:52
  • @GiuseppeNegro And, why exactly coefficients $t$ and $(1−t^2)^{1/2}$ ? I don't see the purpose or how i would derive it. – stanly Oct 20 '21 at 15:03
  • Draw a picture. Recall that $\lvert x \rvert=1$. That formula is nothing more than Pythagoras' theorem. – Giuseppe Negro Oct 20 '21 at 15:48

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