Assume $ k < n$. Consider the Lebesgue measure $v $ in $\mathbb{C}^n$ (normalized so the unit ball $B_n$ of $\mathbb{C}^n$ has measure 1). I want to integrate a function $f\circ P$, where $f$ is defined $\mathbb{C}^n$ but depends only the values on $\mathbb{S}_k = \partial B_k$, and $P$ is the projection of $\mathbb{C}^n$ onto $\mathbb{C}^k$.
I want to use fubini theorem on $$I(r) = \int_{rB_n} f\circ P \ dv$$ to show that
$$I(r) = C \int_{rB_k}(r^2 - |w|^2)^{n-k}f(w) \ dv_k$$
where C is a constant that depends on the normalization of the measure $v_{n}$ and $v_{k}$. I'm running into problems on how exactly make so that the constant C shows up. Here's what I did, if $(w,z)\in\mathbb{C}^k \times \mathbb{C}^{n-k}$, then $$ I(r) = \int_{rB_k} \bigg[\int_{(r^2 - |w|^2)B_{n-k}} f\circ P \ dv_{n-k}(z) \bigg] dv_k(w) = \int_{rB_k} f(w) \bigg[\int_{(r^2 - |w|^2)B_{n-k}} \ dv_{n-k}(z) \bigg] dv_k(w) $$
where the second equality is valid due the definition of $f$. Now, on my mind, evaluation of the integral $\int_{(r^2 - |w|^2)B_{n-k}} \ dv_{n-k}(z)$ would result in $$v_{n-k}((r^2 - |w|^2)B_{n-k}) = (r^2 - |w|^2)^{n-k}v_{n-k}(B_{n-k}) = (r^2 - |w|^2)^{n-k} $$
because I was assuming the measure $v_{k-n}$ would automatically satisfy $v_{n-k}(B_{n-k}) = 1$. In this case, the constant would be 1, which is false.
I want to know why I'm wrong on this assumption (I've spent way more time than I would like to admit on this, and made no progress), and if not, did I made a mistake on the previous step, that is, are the regions I'm integrating correct?