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I'm looking for a proof for the formula

$$\int_{B(0,1)}f(|x|)dx = |S^{n-1}|\int_0^1f(r)r^{n-1}dr,$$ where $B(0,1)$ is the unit ball in $\mathbb{R}^n$ and $x\mapsto f(|x|) $ is some integrable function on $B(0,1).$

Thank you in advanced

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Change variables to polar coordinates and use Fubini's theorem, you get

$$\int_{B(0,1)}f(|x|)\,dx=\int_0^1\left(\int_{S^{n-1}}f(r)r^{n-1}\,d\sigma\right)\,dr$$

where $d\sigma$ is the surface area measure on $S^{n-1}$. Then just carry out the inner integration and you get your result.

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    @A.MONNET no, it's not a function, it's a set. – 2017-02-05
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    Yes it is a set but in a previous step shouldn't we integrate on the surface of the ball of radius $r$ ? I mean $I = \int_0^1 f(r)\left( \int_{S^{n-1}(0,r) }d\sigma \right)dr$ – 2017-02-05
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    No, the space $B(0,1)\setminus\{0\}\cong (0,1]\times S^{n-1}$ because the first factor is the radial one. Omitting a single point doesn't affect anything because the integral over that point is $0$, so you (essentially, up to removing that point) have a product space and each factor is independent of the other. – 2017-02-05
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    so the main step is a change of coordinate ? – 2017-02-05
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    @a.monnett yes that's accurate – 2017-02-05
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    Thank you. Do we have an explicit formula for such change of coordinate (I mean $x= \phi(\sigma, r)$) ? – 2017-02-05
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    @a.monnett Yes, this is the standard spherical coordinates. If you want all the sines and cosines check out Wikipedia for the full formula – 2017-02-05