$\iint (x+y)e^{xy} dA$ . where A is region enclosed by $x-y = 1$, $x-y=4$, $xy = 1$, $xy=2$

Question

Find the integral $\iint (x+y)e^{xy} dA$ . where A is region enclosed by $x-y = 1$, $x-y=4$, $xy = 1$, $xy=2$

So I tried the following substitution: $$u = xy$$ $$v = x-y$$

And want to find the partial derivatives of $x$ and $y$ w.r.t. $u$ and $v$.

However I am unable to isolate the partials as only functions of $u$ and $v$ For example for $dx/dv$ I get $1/(1+ u/x^2)$

I am unsure of how to proceed from this point.

Any help would be much appreciated.

Answer 1

The transformation you chose is easy to invert:

$y=u/x$ and $v=x-u/x$ or $x^2-vx-u=0$ that can be solved for $x$: $x=\dfrac{v\pm\sqrt{v^2+4u}}{2}$. The for $y$: $y=\dfrac{-v\pm\sqrt{v^2+4u}}{2}$

From here, the partial derivatives can be explicitly calculated for you to compute the surface element.