Answer doesn't make sense when doing substitution in multiple integrals

Question

The following transformation is given: u= x-y and v= x+y. I was then told to evaluate $\int \int \frac{x-y}{(x+y)^2}dA$ where R is the square with vertices $(0,2), (1,1), (2,2), (1,3)$. When I converted the coordinates to (u,v), I got :$(-2,2), (0,2), (0,4), (-2,4)$ respectively. I found my jacobian as 1/2. I then setup my integral as $\frac{1}{2}\int_{-2}^{0}\int_{2}^{4}\frac{u}{v^2}dvdu$. When I evaluated it, I got $-\frac{1}{4}$. I am confused because I thought area could not be negative. Is this correct?

Answer 1

Your answer is correct. Area (or volume) is positive, but an integral can be thought of as a signed area (or volume) and thus negative. In fact, the integrand $(x-y)/(x+y)^2\le0$ in $R$, since $x\le y$ and $(x+y)^2\ge0$.

The given integral could also be computed without a transformation of coordinates as follows:

$$\int_0^1\int_{2-x}^{2+x} \dfrac {x-y}{(x+y)^2} dydx + \int_1^2\int_{x}^{4-x} \dfrac {x-y}{(x+y)^2} dydx=\left(\frac12-\ln2\right)+\left(\ln2-\frac34\right)=-\frac14.$$