Using an appropriate change of variables, calculate

Using an appropriate change of variables, calculate the area bounded by the curves: $$ x+y=a,\qquad x+y=b,\qquad y=\alpha x,\qquad y=\beta x, $$ where $a<b$ and $\alpha<\beta$.

Step 1 · Substitute and identify the region

Let $u=x+y,\,v=\dfrac yx$. Then $x+y=a\Rightarrow u=a$, $x+y=b\Rightarrow u=b$, $y=\alpha x\Rightarrow v=\alpha$, and $y=\beta x\Rightarrow v=\beta$. The region maps to the rectangle $[a,b]\times[\alpha,\beta]$ in the $(u,v)$ plane.

Step 2 · Invert and compute the Jacobian

From $y=vx$ and $u=x+y=x(1+v)$: $x=\dfrac u{1+v}$, $y=\dfrac{uv}{1+v}$. Differentiating:

$$ \frac{\partial x}{\partial u}=\frac1{1+v},\quad \frac{\partial x}{\partial v}=-\frac u{(1+v)^2},\quad \frac{\partial y}{\partial u}=\frac v{1+v},\quad \frac{\partial y}{\partial v}=\frac u{(1+v)^2}. $$

$$ \frac{\partial(x,y)}{\partial(u,v)} = \frac1{1+v}\cdot\frac u{(1+v)^2} - \left(-\frac u{(1+v)^2}\right)\cdot\frac v{1+v} = \frac{u(1+v)}{(1+v)^3} = \frac u{(1+v)^2}, $$

so $dx\,dy=\dfrac u{(1+v)^2}\,du\,dv$.

Step 3 · Set up and evaluate the area integral

$$ A = \int_a^b\!\int_\alpha^\beta\frac u{(1+v)^2}\,dv\,du = \int_a^b u\left[\int_\alpha^\beta\frac{dv}{(1+v)^2}\right]du. $$

The inner integral: $\displaystyle\int_\alpha^\beta\frac{dv}{(1+v)^2}=\left[-\frac1{1+v}\right]_\alpha^\beta=\frac1{1+\alpha}-\frac1{1+\beta}=\frac{\beta-\alpha}{(1+\alpha)(1+\beta)}$. So:

$$ A = \frac{\beta-\alpha}{(1+\alpha)(1+\beta)}\int_a^b u\,du = \frac{\beta-\alpha}{(1+\alpha)(1+\beta)}\cdot\frac{b^2-a^2}2 = \frac{(b^2-a^2)(\beta-\alpha)}{2(1+\alpha)(1+\beta)}. $$

Additional Notes

• This derivation matches the source exactly.
• Verified numerically for $a=1.2,\,b=2.3,\,\alpha=0.4,\,\beta=1.1$: direct double quadrature of $\int_a^b\int_\alpha^\beta\frac u{(1+v)^2}\,dv\,du$ gives $\approx0.458333$, matching $\dfrac{(b^2-a^2)(\beta-\alpha)}{2(1+\alpha)(1+\beta)}\approx0.458333$.