-1
$\begingroup$

$\iint_D \frac{\sqrt{x+y}}{\sqrt{x-2y}}~dA$ 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?

$\endgroup$
0
$\begingroup$

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 $$\begin{eqnarray*} \frac{1}{3}\iint_{\substack{a,b\in(0,1)\\a+b\leq 1}}\sqrt{\frac{a+b}{b}}\,da\,db&=&\frac{1}{3}\int_{0}^{1}\frac{1}{\sqrt{b}}\int_{0}^{1-b}\sqrt{a+b}\,da\,db\\&=&\frac{2}{9}\int_{0}^{1}\frac{1}{\sqrt{b}}-b\,db=\color{red}{\frac{1}{3}}.\end{eqnarray*}$$

$\endgroup$
  • $\begingroup$ I get that the integration range is a triangle but I'm not sure I understand how you got the converted integral $\endgroup$ – mr_geoo Dec 18 '17 at 8:07
  • $\begingroup$ @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 $\frac{1}{3}$. $\endgroup$ – Jack D'Aurizio Dec 18 '17 at 8:34
  • $\begingroup$ I get that the jacobian is 1/3. What I don't get is how you changed the top and bottom functions. It looks like you set a+b=x+y and b=x-2y? $\endgroup$ – mr_geoo Dec 18 '17 at 8:40
  • $\begingroup$ @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 $x-2y=b$. $\endgroup$ – Jack D'Aurizio Dec 18 '17 at 8:48
  • $\begingroup$ could you help me take a look at this problem? math.stackexchange.com/questions/2571239/… $\endgroup$ – mr_geoo Dec 18 '17 at 18:13

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.