I have a question: Evaluate $\iint_{D} xy(\sqrt{1-x-y})\,dx\,dy$, where $D$ is the region bounded by $x=0$, $y=0$, $x+y=1$ using the transformations $x+y=u$ and $y=uv$.
I can see that the region $D$ is a triangle with coordinates $(0,0)$, $(1,0)$ and $(0,1)$. However in my book when transformed the new region is a rectangle with the vertices $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1)$.
My question: I am unable to understand how the new region would turn out to be rectangle. How does the new region turn out as a rectangle?
For $(0,0)$, we have $u=0$ and $v=\frac{0}{0}$ (what now?).
For $(0,1)$, we have $u=1$ and $v=1$.
For $(1,0)$, we have $u=1$ and $v=0$.
What am I doing wrong here?
Let $T$ be the triangle $T=\{(x,y)\in\mathbb{R}^2: x,y\geq 0, x+y\leq 1\}$. A $45^\circ$ clockwise rotation around the origin, induced by $x=\frac{u+v}{\sqrt{2}}, y=\frac{u-v}{\sqrt{2}}$, maps $T$ into the triangle $$T'=\{(u,v)\in\mathbb{R}^2:0\leq u\leq\tfrac{1}{\sqrt{2}}, |v|\leq u\}$$ and $$ \iint_{T}xy\sqrt{1-(x+y)}\,dx\,dy = \iint_{T'}\frac{u^2-v^2}{2}\sqrt{1-u\sqrt{2}}\,du\,dv\\=\int_{0}^{1/\sqrt{2}}\int_{-u}^{u}\frac{u^2-v^2}{2}\sqrt{1-u\sqrt{2}}\,du\,dv $$ Now by setting $u=\frac{a}{\sqrt{2}}, du=\frac{1}{\sqrt{2}}da$ and $v=u b,dv = d\,db$ the last integral becomes $$ \frac{1}{8}\int_{0}^{1}\int_{-1}^{1}a^3(1-b^2)\sqrt{1-a}\,db\,da $$ and by Fubini's theorem this integral equals $$ \frac{1}{8}\int_{0}^{1}a^3\sqrt{1-a}\,da \int_{-1}^{1}(1-b^2)\,db = \frac{16}{945}.$$