Finding region for Change of Variables and Double integral problem

Question

I'm running into some trouble on a problem in Vector Calc by Marsden and Tromba. I don't think I am correctly finding the region for my change of variables and the book doesn't have a similar example.

Question:

Let $D$ be the region $0 \leq y \leq x$ and $0 \leq x \leq 1$. Evaluate $$\iint_D (x + y) \,dx\,dy$$ by making the change of variables $x = u + v$, $y = u - v$.

Attempts:

$T(u,v) = A\left( \begin{array}{c} u \\ v \\\end{array} \right) = \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \\\end{array} \right)\left( \begin{array}{c} u \\ v \\\end{array} \right)$ Therefore $det A = -2$. From this we can find that $\left( \begin{array}{cc} 1/2 & 1/2 \\ 1/2 & -1/2 \\\end{array} \right)\left( \begin{array}{c} x \\ y \\\end{array} \right) = \left( \begin{array}{c} u \\ v \\\end{array} \right) $.

From here I am unsure how to proceed. By just solving the double integral in terms of $x,y$ I know that I should be getting $1/2$ but none of the regions $D^{*}$ that I've found have gotten me this (I run into the same issue on the next problem in the book.

Answer 1

If $x=u+v$ and $y=u-v$, you get the Jacobian $$J=\begin{vmatrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial y}{\partial u}\\[1ex]\dfrac{\partial x}{\partial v}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix}1&1\\1&-1\end{vmatrix}=-2~~\implies~~|J|=2$$ So, $$\begin{align*}\iint_D(x+y)\,\mathrm{d}x\,\mathrm{d}y&=2\iint_{D^*}(u+v+u-v)\,\mathrm{d}u\,\mathrm{d}v\\[1ex] &=4\int_0^{1/2}\int_v^{1-v}u\,\mathrm{d}u\,\mathrm{d}v\end{align*}$$ It seems to me that you're having trouble coming up with the right limits for integrating over $D^*$. What I basically did was transform each side of the triangular region $D$ via the given change of variable to make a corresponding plot of the region in the $u$-$v$ plane.

$D$ is bounded by three lines:

• $y=x$
• $y=0$
• $x=1$

Since $x=u+v$ and $y=u-v$, this means the first line translates to $u-v=u+v$, i.e. $v=0$. Similarly, $y=0$ gives $u-v=0$, or $u=v$; and $x=1$ gives $u+v=1$. If you plot each of

• $v=0$
• $v=u$
• $v=1-u$

in the $u$-$v$ plane, you'll make another triangle with its hypotenuse along the $v$ axis. (Essentially, the transformation rotates the triangle $45^\circ$ clock- or counter-clockwise, depending on the orientation of the $u$-$v$ plane's axes.)