Calculate the value of the integral: double integral over
Calculate the value of the integral: $$ \iint_D (y-x)\,dx\,dy, $$ where the region $D$ is bounded by the lines $$ y=x+1,\quad y=x-3,\quad y=-\frac13x+\frac79,\quad y=-\frac13x+5, $$ using the substitution $u=y-x$, $v=y+\frac13x$.
Explanation of Method
Each pair of parallel boundary lines becomes a constant value of $u$ or $v$, so the parallelogram $D$ transforms into a plain rectangle in the $(u,v)$-plane. Solving for $x,y$ in terms of $u,v$ gives a constant Jacobian, and since the integrand $y-x$ is exactly $u$, the transformed integral is a simple product of a linear integral in $u$ and a constant-length integral in $v$.
- Identify the rectangle $D'$ in $(u,v)$-coordinates from the four boundary lines.
- Solve for $x,y$ in terms of $u,v$ and compute the Jacobian.
- Integrate $u$ over the rectangle, scaled by the Jacobian.
Step 1 · Identify the boundaries in $(u,v)$
$$ y-x=1 \Rightarrow u=1, \qquad y-x=-3 \Rightarrow u=-3, $$
$$ y+\tfrac13x=\tfrac79 \Rightarrow v=\tfrac79, \qquad y+\tfrac13x=5 \Rightarrow v=5. $$
So $D'$ is the rectangle $-3\leqslant u\leqslant1$, $\frac79\leqslant v\leqslant5$.
Step 2 · Solve for $x,y$ and find the Jacobian
Subtracting $u=y-x$ from $v=y+\frac13x$: $v-u=\frac43x \Rightarrow x=\frac{3(v-u)}{4}$. Then $y=u+x=\frac{u+3v}{4}$.
$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}-\frac34&\frac34\\[2pt]\frac14&\frac34\end{vmatrix} = -\frac34\cdot\frac34-\frac34\cdot\frac14 = -\frac34, \qquad \left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac34. $$
Step 3 · Transform the integral
Since $y-x=u$ by definition:
$$ I = \iint_D (y-x)\,dx\,dy = \frac34\int_{-3}^1\int_{7/9}^5 u\,dv\,du. $$
Step 4 · Evaluate
$$ I = \frac34\int_{-3}^1 u\left(5-\frac79\right)du = \frac34\cdot\frac{38}{9}\int_{-3}^1 u\,du = \frac{19}{6}\left[\frac{u^2}{2}\right]_{-3}^1 = \frac{19}{6}\cdot\frac{1-9}{2} = \frac{19}{6}\cdot(-4) = -\frac{38}{3}. $$
Additional Notes
- Verified numerically: direct integration of $y-x$ over the parallelogram (vertices computed as $(-\frac16,\frac56)$, $(3,4)$, $(6,3)$, $(\frac{17}6,-\frac16)$ from the four line intersections) gives $I\approx-12.667$, matching $-\frac{38}{3}\approx-12.667$.
Result
The solution to the integral is: