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$B_n$ is the unit ball in $n$ dimensions and $v_n$ is its volume.

$f:\mathbb R^n \to \mathbb R$ is a function that is only dependent on its first variable, meaning $f(x_1,x_2, \dots, x_n) = g(x_1)$ where $g: \mathbb R \to \mathbb R$.

We wish to show that $\int_{B_n}f(x)dx = v_{n-1}\int_{-1}^{1}g(t)(1-t^2)^{\frac{n-1}{2}}dt$

And we are given a hint: If $B \subset \mathbb R^n$ is a ball with radius $R$, then its volume is $R^nv_n$.

I tried calculating the integral using hyperspherical transformation but I failed, I also tried induction but that doesn't seem to be the way.

I don't see how to use the hint.

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I may have found a simple answer. If you are experienced in this field, please review this if you can spare the time.

The intersection of $B_n$ with the plane $x_1 = t$ is well defined for all $t \in (-1,1)$, and we have that $B_n \bigcap \{x \in \mathbb R^n| x_1 = t\} = \{x \in \mathbb R^n| x_2^2+x_3^2 + \dots + x_n^2 \leq 1-t^2\}$ which is an $n-1$ dimensional ball with radius $\sqrt{1-t^2}$.

Now from Cavalieri's principle:

$\int_{B_n}f(x)dx = \int_{B_n}g(x_1)dx = \int_{-1}^{1}\int_{B^*}g(t)dx_2dx_3\dots dt = \int_{-1}^{1}g(t)\int_{B^*}1dx_2dx_3\dots dt=\int_{-1}^{1}g(t)v_{n-1}(\sqrt{1-t^2})^{n-1}dt = v_{n-1}\int_{-1}^{1}g(t)(1-t^2)^{\frac{n-1}{2}}$

Where $B^*$ is an $n-1$ dimensional ball with radius $\sqrt{1-t^2}$ centered at zero.

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