Help me find a transfor for the followin inequality to be a rectangle which can be solved by using one single double integral

Question

fixxer · Answer

$$0\le x+y \le \pi  $$$$3\le 2x.y \le 7$$

fixxer · Answer

2x-y sorry for that typo

irishboy123 · Answer

do the Jacobian :)

i think, as you have typo's, that you mean a co-ordinate shift like this:

$u = x + y, \qquad v = 2x -y $ 


with $u \in [0, \pi], v \in [3,7]$

and you have the transform $dx~ dy \to  |J| ~ du ~ dv$

fixxer · Answer

Hi,
Thank you very much for the transform.

irishboy123 · Answer

don't thank me mate, thank whoever invented partial differentiation :)

fixxer · Answer

lol :p they aint alive anymore I guess, anyway since we are at it is the jacobian the determinant of this matrix??$$\left[\begin{matrix}\frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } \ \frac{ 2 }{ 3 } & -\frac{ 1 }{ 3 }\end{matrix}\right]$$

Which will give me -1/3 and trapped area of -4?

irishboy123 · Answer

i'm getting -1/3 too

for $\det \left[\begin{matrix}1 & 1 \ 2 & -1\end{matrix}\right]$

fixxer · Answer

Great! means im doing at least a small part right lols