Double Integration Change of Variables

Question

Evaluate $\int\int xy \, dx\,dy$ over the square with corners $(0,0), (1,1), (2,0),$ and $(1,−1)$ using

$x$=$\frac{u+v}{2}$

$y$=$\frac{u-v}{2}$

In my transformation, I get the limits for $v$ as $u-1$ to $u$ and $u$ as $0$ to $2$. But these limits are incorrect. The solution the book provides is zero but my evaluation leads to $\frac{1}{3}$

I computed the Jacobian and got the determinant equal to $\frac{1}{2}$

After transformation the integral I get is:

$\int_0^2 \int_{u-2}^{u} \frac{u^2-v^2}{8} \, dvdu$

But somehow the solution to the integral that is $\frac{1}{3}$ does not agree with the solution provided to me i.e. zero. Are my limits to the new integral incorrect or is my Jacobian wrong?

Answer 1

Your limiting values are incorrect.

If you convert the corners of the square to the new coordinate system, you see that:

• $(0, 0)$ maps to $(0, 0)$
• $(1, 1)$ maps to $(2, 0)$
• $(2, 0)$ maps to $(2, 2)$
• $(1, -1)$ maps to $(0, 2)$

So the boundaries are just $0$ to $2$, for both $u$ and $v$.