# Evaluate double integral by making change of variables

$\iint_D \frac{\sqrt{x+y}}{\sqrt{x-2y}}~dA$ where D is the region in R2 enclosed by the lines y =$\frac{x}{2}$, y = 0, and x+y = 1. I set $u=x+y$ and $v=x-2y$. When I take the Jacobian I get $J=\frac{1}{3}$. Through change of variables I get $\iint_D \frac{\sqrt{x+y}}{\sqrt{x-2y}}~dA$ = $\iint_D \frac{1}{3}\sqrt{\frac{u}{v}}~dudv$. Am I correct so far?

The integral of a positive function cannot be negative: this is a reminder that the determinant of the Jacobian has to be taken in absolute value. Additionally, the integration range is a triangle, with a parametrization given by $$(x,y) = a\left(\frac{2}{3},\frac{1}{3}\right)+b(1,0),\qquad a,b\in[0,1],a+b\leq 1.$$ By enforcing the substitution $$\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}\tfrac{2}{3}&1 \\ \tfrac{1}{3}&0\end{pmatrix}\begin{pmatrix}a \\ b\end{pmatrix}$$ the original integral is converted into $$\begin{eqnarray*} \frac{1}{3}\iint_{\substack{a,b\in(0,1)\\a+b\leq 1}}\sqrt{\frac{a+b}{b}}\,da\,db&=&\frac{1}{3}\int_{0}^{1}\frac{1}{\sqrt{b}}\int_{0}^{1-b}\sqrt{a+b}\,da\,db\\&=&\frac{2}{9}\int_{0}^{1}\frac{1}{\sqrt{b}}-b\,db=\color{red}{\frac{1}{3}}.\end{eqnarray*}$$
• @mr_geoo: the usual way: compute the new integration range (a right triangle), the jacobian of the transformation, the new integrand function, perform a bunch of computations and voilà $\frac{1}{3}$. – Jack D'Aurizio Dec 18 '17 at 8:34
• @mr_geoo: I set $x=\frac{2}{3}a+b$ and $y=\frac{1}{3}a$ as stated. As a consequence, $x+y=a+b$ and $x-2y=b$. – Jack D'Aurizio Dec 18 '17 at 8:48