Evaluate
$$\iint_R(x+y)\,dA$$
Where $R$ is the trapezoidal region with vertices at $(0,0),(5,0),(5/2,5/2)$ and $(5/2,−5/2)$, using the transformation $x=2u+3v$ and $y=2u−3v$.
Approach:
The Jacobian comes out to be $12$, and the region in the $u−v$ plane consists of the points $(0,0),(5/2,0),(5/2,5/6),(0,5/6)$ which is a rectangle, and therefore, the double integral $\iint_{R'}(4u)[J\,du\,dv]$ is simply $$48\int_{0}^{5/6}\int_{0}^{5/2}u\,du\,dv= 125.$$
However, the answer given is $125/4$.
To kind of "check" where my mistake is, I came up with a small test:
By putting $f=1$ in the variable transformation formula for double integrals, we get:
$$Area(R)=J\times Area(R')$$
However, in my case, area of the original region is $25/2$, whereas $12$*(area of the transformed region) is $25$.
So have I messed up in drawing the transformed region? Or somewhere else?
Best Answer
Your work is correct except for the points of the new region:
$$\begin{cases} 2u + 3v = 5/2\\ 2u - 3v = 5/2 \end{cases}$$ yields the point $(5/4, 0)$ and $$\begin{cases} 2u + 3v = 5\\ 2u - 3v = 0 \end{cases} $$ gives the point $(5/4, 5/6)$.
With these new points:
$$48\int_{0}^{5/6}\int_{0}^{5/4}u \ dudv=\frac{125}{4}$$