Compute the following double integral: $$ \iint\limits_{G} \sqrt{x^{2} - y^{2}} \, dx\, dy, $$ where $$ G=\{(x,y) \in \mathbb{R_{+}^{2}}: 0 \leq x + y \leq 2, 0 \leq x - y \leq 2\} $$ In order to solve this question, I wanted to use the following substitutions $$ u = x + y, v = x-y $$ This resulted in an answer of $16/9$. However, according to the answer in my book the correct value is $8/9$. This seems to be the result of $(x,y) \in \mathbb{R_{+}^{2}}$ which cuts the rectangle in half but I don't know how to write this down using proper mathematical notation. I was hoping somebody could help me with this double integral.
Exact Steps: $$ \iint\limits_{G} \sqrt{x^{2} - y^{2}} \, dx\, dy \\ \iint\limits_{G} \sqrt{(x-y)(x+y)} \, dx\, dy \\ \iint\limits_{G} \frac{1}{2}\sqrt{(x-y)(x+y)} \, 2 \, dx\, dy, \\ \iint\limits_{\widetilde{G}} \frac{1}{2}\sqrt{vu} \, du\, dv, \\ \frac{1}{2} \iint\limits_{\widetilde{G}} \sqrt{vu} \, du\, dv, \\ \frac{1}{2} \int_{0}^{2} \sqrt{v} \left( \int_{0}^{2} \sqrt{u} \, du \right) \,dv \\ \frac{1}{2} \int_{0}^{2} \sqrt{v} \left(\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{0}^{2} \right) \,dv \\ \frac{1}{3}2^{\frac{3}{2}} \int_{0}^{2} \sqrt{v} \,dv \\ \frac{1}{3}2^{\frac{3}{2}} \left[\frac{2}{3}v^{\frac{3}{2}}\right]_{0}^{2} \\ \frac{2}{9}2^3 \\ \frac{16}{9} $$
The bounds of the integral should be different:
$$\frac{1}{2}\int_0^2\int_v^2\sqrt{uv}\ du\ dv$$
If we just look at the region $\mathbb{R}^2_+$, we have
Adding in the bounds $0 \leq x + y \leq 2$ and $0 \leq x - y \leq 2$, we have
The region in the original problem would be the triangle, whereas the region $$0 \leq u = x + y\leq 2 \\0 \leq v = x-y\leq 2$$ would be the square where the blue and green regions overlap.
To remove the half of the triangle below the $x$-axis, we need to add the bound $$y \geq 0 \iff 2y \geq 0 \iff u - v \geq 0 \iff u \geq v$$ which gives the correct region,
$$v \leq u = x + y\leq 2 \\0 \leq v = x-y\leq 2$$ or equivalently $$0 \leq u = x + y\leq 2 \\0 \leq v = x-y\leq u$$ (depending on the order of integration).