Proving $\int_0^r{(r^m-x^m)^{1/m}dx}=\frac{\Gamma\left(\frac{1}{m}+1\right)\Gamma\left(\frac{1}{m}+1\right)}{\Gamma\left(\frac{2}{m}+1\right)}r^2$

Under $x\to rx$ and $x^m\to x$, one has \begin{eqnarray} &&\int_0^r{(r^m-x^m)^{1/m}dx}\\ &=&\int_0^1{(r^m-r^mx^m)^{1/m}rdx}\\ &=&r^2\int_0^1(1-x^m)^{1/m}dx\\ &=&r^2\frac1m\int_0^1(1-x)^{1/m}x^{\frac1m-1}dx\\ &=&r^2\frac1m\frac{\Gamma(\frac1m+1)\Gamma(\frac1m)}{\Gamma(\frac2m+1)}\\ &=&\frac{\Gamma^2(\frac{1}{m}+1)}{\Gamma(\frac{2}{m}+1)}r^2. \end{eqnarray} Here $$ \int_0^1x^{p-1}(1-x)^{q-1}dx=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)},\Gamma(x+1)=x\Gamma(x) $$ are used.