How should the change of variables be performed so that
How should the change of variables be performed so that the curvilinear quadrilateral bounded by the curves $xy=1$, $xy=2$, $x-y+1=0$, $x-y-1=0$ (where $x>0$, $y>0$) is mapped to a rectangle whose sides are parallel to the coordinate axes?
Explanation of Method
Each pair of boundary curves already has the form "some expression $=$ constant": the hyperbolas are level sets of $xy$, and the lines are level sets of $x-y$. Using these two expressions directly as the new coordinates turns all four curved/slanted boundaries into four straight, axis-aligned lines at once.
- Choose $u=xy$, $v=x-y$ as the new variables.
- Verify each boundary becomes a constant-$u$ or constant-$v$ line.
- Solve for $x,y$ in terms of $u,v$ and compute the Jacobian.
Step 1 · Identify the boundary curves
The region is bounded by $xy=1$, $xy=2$, $x-y=-1$, $x-y=1$.
Step 2 · Choose the new variables
Since the hyperbolas have the form $xy=\text{const}$ and the lines have the form $x-y=\text{const}$, the natural substitution is:
$$ u=xy, \qquad v=x-y. $$
Step 3 · Verify the transformed boundaries
$$ xy=1\Rightarrow u=1, \qquad xy=2\Rightarrow u=2, \qquad x-y=-1\Rightarrow v=-1, \qquad x-y=1\Rightarrow v=1. $$
The curvilinear quadrilateral maps to the rectangle:
$$ 1\leqslant u\leqslant2, \qquad -1\leqslant v\leqslant1. $$
Step 4 · Express $x,y$ in terms of $u,v$
From $v=x-y$: $y=x-v$. Substituting into $u=xy$: $u=x(x-v)=x^2-vx$, so $x^2-vx-u=0$. Taking the positive root (since $x>0$):
$$ x=\frac{v+\sqrt{v^2+4u}}{2}, \qquad y=x-v=\frac{\sqrt{v^2+4u}-v}{2}. $$
Step 5 · Compute the Jacobian
It is easier to compute $\dfrac{\partial(u,v)}{\partial(x,y)}$ first and invert:
$$ \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix}y&x\\1&-1\end{vmatrix} = -y-x = -(x+y). $$
$$ \left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \frac{1}{|{-(x+y)}|} = \frac{1}{x+y}. $$
Additional Notes
- Since $x+y=\sqrt{v^2+4u}$ (adding the expressions for $x$ and $y$ above), the Jacobian factor can also be written directly in terms of $u,v$ as $\dfrac{1}{\sqrt{v^2+4u}}$.
Result
The change of variables and transformed integral are: