$\int_{B}{f(x.z)dx}=\int_B{f(x_n|z|)dx}$ for $z \in \mathbb{R^n}$ fixed and $f$ continuous

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function, $z \in \mathbb{R}^n$ and $B \subset \mathbb{R}^n$ the unit closed ball.

Show $\int_{B}{f(x \cdot z)dx}=\int_B{f(x_n|z|)dx}$ ($x_n$ is the last coordinate of $x$).

I know that $x \cdot z \leq |z|$ and $x_n|z| \leq |z|$, I tried to define two functions

$h_1(x)=x \cdot z$, $h_2(x)=x_n|z|$, but couldn't get any further with this idea.

Any advices?

Answer 1

The inner product and the integration set $B^n$ are both invariant under rotations, so you can simply choose the $x$ coordinates so that $z$ is parallel to the $n$th axis, $z = \lvert z \rvert e_n$. Then $x \cdot z = \lvert z \rvert x \cdot e_n = x_n \lvert z\rvert$.