Let $p>0$. Then express the area of the region bounded by the coordinate axes and the curve $x^p+y^p=1$ in the first quadrant in terms of the gamma function.
My thought is to consider polar coordinates. So I can set up an iterated integral like so $\int\limits_0^{\pi/2}\int\limits_0^\infty f(r,\theta) \, dr \, d\theta$ then I manipulate until I find a product of a gamma and beta function then write the beta in terms of gamma and I will be done. The trouble I am having is figuring out what $f(r,\theta)$ is.
The area is equal to $$ \int_0^1(1-x^p)^{\frac{1}{p}}\;dx $$ Setting $u=x^p$, this becomes $$ \frac{1}{p}\int_0^1u^{\frac{1}{p}-1}(1-u)^{\frac{1}{p}}\;du=B\Big(\frac{1}{p},\frac{1}{p}+1\Big)=\frac{1}{p}\frac{\Gamma(\frac{1}{p})\Gamma(\frac{1}{p}+1)}{\Gamma(\frac{2}{p}+1)}$$ $$ =\frac{1}{p^2}\frac{\Gamma(\frac{1}{p})^2}{\frac{2}{p}\Gamma(\frac{2}{p})}=\frac{1}{2p}\frac{\Gamma(\frac{1}{p})^2}{\Gamma(\frac{2}{p})}$$
Note that when $p=2$, this is equal to $\frac{\pi}{4}$, as we would expect.