I am trying to understand a proof where the following equality, without any further details, appears:
$$\int_{B_q}e^{-i\langle x,t\rangle}dx = c(k)\int_{-q}^q e^{-i|t|y}(q^2-y^2)^{k/2}dy,$$
where $B_q = \{x\in\mathbb{R}^{k+1};\,|x|\leq q\}$, $t\in\mathbb{R}^{k+1}$, $|x| = (x_1^2+\ldots + x_{k+1}^2)^{1/2}$, $\langle\cdot,\cdot\rangle$ is the standard scalar product and $c(k)$ is a constant depending only on $k$.
I am not able to see how can I pass from a $(k+1)$-dimensional integral to 1-dimensional integral. My intuition tells me that this constant $c(k)$ must be related with the surface area of $\mathbb{S}^k=\{x\in\mathbb{R}^{k+1};\,|x|=1\}$ and then the remaining integral will be related with the radius of $B_q$.
Let's make a linear variable change that is given by some orthogonal matrix that takes $t$ to $( 0, \ldots, 0, |t|)$. Since the Jacobian determinant is 1, and the domain of integration is invariant, we get that the integral we are interested is equal to $\int_{B_q} e^{-i|t|x_{k+1}} d x_1 \ldots d x_k d x_{k+1}$. Now let's call $x_{k+1}=y$ and integrate by iteration, with integration with respect with $d x_1 \ldots d x_k$ as the inner integral. That becomes $$\int_{B^{k}_{\sqrt{q^2-y^2}}} e^{-i|t|y} d x_1 \ldots d x_k=e^{-i|t|y}\int_{B^{k}_{\sqrt{q^2-y^2}}} 1 d x_1 \ldots d x_k=e^{-i|t|y} c(k)(q^2-y^2)^{k/2},$$ where $c(k)$ is the volume of the unit ball in $\mathbb{R}^k$. Now the outer integral is exactly as stated.