The only problem finding the limits of integration. Thats why i want to convert to polar. I want to make a substitution so that the limits of r can be easily evaluated. I was thinking of $$ x=r^2 cos^2 \phi $$ $$ y=r^2 sin^2 \phi $$
The absolute values scare me a little. I am not sure if they even change anything, r is positive, but i dont know what to do with absolute values of trig functions.
The shape $$B:\quad\bigl(\sqrt{x}+\sqrt{y}\bigr)^{12}\leq xy,\quad x\geq0,\quad y\geq0\tag{1}$$ in the first quadrant makes up half the required area (see the figure in Cesareo's answer). We first set up a parametrization of $B$ of the following form: $$\psi:\quad\left\{\eqalign{x= u\cos^4 t\cr y=u\sin^4 t\cr}\right.\qquad\left(0\leq u\leq \bar u(t), \ 0\leq t\leq{\pi\over2}\right)\ .\tag{2}$$ Here the exponent $4$ of $\cos$ and $\sin$ was chosen such that $\sqrt{x}+\sqrt{y}$ comes out particularly simple. Note that the $u$-lines $t={\rm const.}$ are rays emanating from $(0,0)$ out into the first quadrant. In order to determine the upper $u$-limit $\bar u(t)$ for given $t$ we plug $(2)$ into $(1)$ and obtain $u^6\leq u^2\cos^4 t\sin^4 t$. It follows that $$0\leq u\leq \bar u(t)=\cos t\sin t\ .$$ From $(2)$ one computes $$d\psi(u,t)=\left[\matrix{\cos^4 t&-4u\cos^3 t\sin t\cr \sin^4 t&4u\sin^3 t\cos t\cr}\right]\ ,$$ leading to the Jacobian $$J_\psi(u,t)=4u\cos^3 t\sin^3 t\ .$$ In this way we obtain $$\eqalign{{\rm area}(B)&=\int_B 1\>{\rm d}(x,y)=\int_0^{\pi/2}\int_0^{\bar u(t)} J_\psi(u,t)\>du\>dt\cr &=\int_0^{\pi/2}\cos^3 t\sin^3 t\> \bigl(2\bar u^2(t)\bigr)\>dt\cr &=2\int_0^{\pi/2}\cos^5 t\>\sin^5 t\>dt={1\over30}\ .\cr}$$