2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $$ \frac{1}{8}\int_{0}^{1}a^3\sqrt{1-a}\,da \int_{-1}^{1}(1-b^2)\,db = \frac{16}{945}.$$

$\endgroup$
  • $\begingroup$ Can you please explain, how you determined the limits of the integral? $\endgroup$ – Mohammed Arshaan Apr 22 '18 at 21:40
  • $\begingroup$ And how did you derive $x = \frac{u+v}{\sqrt{2}}$ and $x = \frac{u-v}{\sqrt{2}}$ $\endgroup$ – Mohammed Arshaan Apr 22 '18 at 21:41
  • $\begingroup$ @MohammedArshaan: a $45^\circ$ rotation matrix has a well-known structure. $\endgroup$ – Jack D'Aurizio Apr 22 '18 at 22:08
  • $\begingroup$ I am sorry but we haven't been taught that. Can you give hint as you how you get the idea that is it a $45^{\circ}$ rotation matrix? Like how did the equationa $u=x+y$ and $y=uv$ help you? $\endgroup$ – Mohammed Arshaan Apr 22 '18 at 22:09
  • 1
    $\begingroup$ Basic knowledge of linear algebra is essential in understanding the change of variables in multiple integrals. A $45^\circ$ clockwise rotation around the origin is a linear map sending $(1,0)$ and $(0,1)$ into... that is straightforward. $\endgroup$ – Jack D'Aurizio Apr 22 '18 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.