Double integral over a specific region

Question

I was given an exercise:

Calculate the integral of $\begin{gather*} \iint_D (\sqrt{x/a}+\sqrt{y/b})^2 \,dx\,dy \end{gather*}$, where D is the region bounded by the curve $\sqrt{x/a}+\sqrt{y/b}=1$ and two axes. $a$ and $b$ are both positive real numbers.

I tried to draw the region of D but doesn't find it useful to compute directly or use polar/spherical coordinates, but I feel like there's some connection between the region and the function inside the integral.Any help will be appreciated!

Answer 1

Take $u=\sqrt{\frac{x}{a}}$ and $v=\sqrt{\frac{y}{b}}$. Then $\frac{\partial(x,y)}{\partial(u,v)}=4abuv$ and the transformation $(x,y)\mapsto\left(\sqrt{\frac{x}{a}},\sqrt{\frac{y}{b}}\right)$ maps $D$ bijectively to the triangular shaped region beneath the curve $v=1-u$ in the first quadrant of the $uv-$plane. We get $$\iint_{D}\left(\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}\right)^2\mathrm{d}A=\int_0^1\int_0^{1-u}4abuv(u+v)^2\mathrm{d}v\mathrm{d}u=\frac{ab}{9}$$

Answer 2

Have you tried these coordinates? $x=a r \cos^4 (t)$, $y=b r\cos^4 (t)$, $r \in [0,1]$, $t \in [0,2\pi]$.