In working on an example of the Change of Variables Formula, I seem to have hit a wall in understanding how to set up an integral. For the following, I have found the correct Jacobian, but am unsure of where I am going wrong on the integration itself.
Let $I = \int\int_D(x^2-y^2)dxdy$
Where $D = {[(x,y):3\leq xy \leq 4, 0 \leq x-y \leq 4, x\geq0, y\geq0]}$
Show that the mapping $u = xy, v = x-y$ maps $D$ to the rectangle $R = [3, 4]$x$[0,4]$
I have found that $\frac {∂(x,y)}{∂(uv)}$ is $\frac1{-x-y}$.
It is when I try to show that $I$ is equal to the integral of $f(u,v) =v$ over $R$ that I get lost. So far, I have attempted to prove this through a number of trials, however, I cannot seem to quite get the right integral. The following is one of my attempts at it.
$$
\int_0^4\int_3^4x-v\left(\frac1{-x-y}\right)dxdy
$$
Why is it that this integral is incorrect?
Best Answer
If you perform the change $u = x-y, v = xy$ , the jacobian is $∂x\,∂y=\frac{1}{x+y}\,∂u\,∂v$.Your new variables $0<u<4\,\,,\,\,3<v<4$ and your integral
$$I = \int\int_D(x^2-y^2)dxdy=\int_0^4u\,du\int_3^4 dv =8$$