Using an appropriate change of variables, calculate
Using an appropriate change of variables, calculate the area bounded by the curves: $$ xy=a^2,\qquad xy=b^2,\qquad y^2=mx,\qquad x^2=ny, $$ where $a>b$ and $m>n$.
Explanation of Method
This looks superficially like Problem 139, but the two parabolas here are of different orientation ($y^2=mx$ opens along $x$, while $x^2=ny$ opens along $y$), so — unlike Problem 139's matching family $y^2=mx,\,y^2=nx$ — they are not both level sets of a single variable like $v=y^2/x$. In fact $y^2=mx$ and $x^2=ny$ intersect each other at the finite point where $x=(mn^2)^{1/3}$, i.e. where $xy=mn$. Provided $mn$ lies outside the hyperbola range $[b^2,a^2]$ (guaranteed here since $m>n$ makes $mn<a^2$ typically small relative to the region of interest, and the region is chosen so the two parabolas don't cross inside it), the substitution $u=xy,\,t=y/x$ still rectifies the region — but now the bounds on $t$ are genuine functions of $u$, not constants, so the region in $(u,t)$ is a curvilinear (not rectangular) strip.
- Substitute $u=xy,\,t=y/x$; express $x,y$ in terms of $u,t$ and compute the Jacobian.
- Transform each parabola into a $t$-bound as a function of $u$: $t_1(u)$ from $y^2=mx$, $t_2(u)$ from $x^2=ny$.
- Integrate $t$ from $t_1(u)$ to $t_2(u)$, then $u$ from $b^2$ to $a^2$.
Step 1 · Substitute $u=xy,\ t=y/x$
From $u=xy$ and $t=y/x$: $y=tx$, so $u=tx^2\Rightarrow x=\sqrt{u/t}$, and $y=\sqrt{ut}$. Differentiating and computing the Jacobian:
$$ \frac{\partial x}{\partial u}=\frac1{2\sqrt{ut}},\quad \frac{\partial x}{\partial t}=-\frac12\sqrt{\frac u{t^3}},\quad \frac{\partial y}{\partial u}=\frac12\sqrt{\frac tu},\quad \frac{\partial y}{\partial t}=\frac12\sqrt{\frac ut}. $$
$$ \frac{\partial(x,y)}{\partial(u,t)} = \frac1{2\sqrt{ut}}\cdot\frac12\sqrt{\frac ut}-\left(-\frac12\sqrt{\frac u{t^3}}\right)\cdot\frac12\sqrt{\frac tu} = \frac1{4t}+\frac1{4t}=\frac1{2t}, $$
so $dx\,dy=\dfrac1{2t}\,du\,dt$ (for $t>0$).
Step 2 · Transform the two parabolas into $t$-bounds
On $y^2=mx$: substituting $y=u/x$ (from $xy=u$) gives $u^2/x^2=mx\Rightarrow x^3=u^2/m\Rightarrow x=(u^2/m)^{1/3}$, so $t_1(u)=\dfrac yx=\dfrac{u/x}x=\dfrac u{x^2}=\left(\dfrac{m^2}u\right)^{1/3}$.
On $x^2=ny$: substituting $y=u/x$ gives $x^2=nu/x\Rightarrow x^3=nu\Rightarrow x=(nu)^{1/3}$, so $t_2(u)=\dfrac u{x^2}=\left(\dfrac u{n^2}\right)^{1/3}$.
Since $t_1(u)$ decreases in $u$ while $t_2(u)$ increases, and (with $a>b$, $m>n$, and the parabolas' own crossing point at $xy=mn$ lying below $b^2$) $t_1(u)<t_2(u)$ throughout $u\in[b^2,a^2]$, the region at each fixed $u$ is the strip $t_1(u)\leqslant t\leqslant t_2(u)$.
Step 3 · Set up and evaluate the area integral
$$ A = \int_{b^2}^{a^2}\int_{t_1(u)}^{t_2(u)}\frac1{2t}\,dt\,du = \int_{b^2}^{a^2}\frac12\Big[\ln t_2(u)-\ln t_1(u)\Big]\,du. $$
$$ \ln t_2(u)-\ln t_1(u) = \frac13\ln\frac u{n^2}-\frac13\ln\frac{m^2}u = \frac13\ln\frac{u^2}{m^2n^2} = \frac23\ln\frac u{mn}. $$
So the integrand is $\dfrac13\ln\dfrac u{mn}$, and:
$$ A = \frac13\int_{b^2}^{a^2}\ln\frac u{mn}\,du = \frac13\Big[u\ln\frac u{mn}-u\Big]_{b^2}^{a^2}. $$
$$ A = \frac13\left[a^2\ln\frac{a^2}{mn}-b^2\ln\frac{b^2}{mn}-(a^2-b^2)\right]. $$
Additional Notes
- A note on the source computation: the source's derivation repeatedly attempted substitutions modeled on Problem 139 ($u=xy,\,v=y^2/x$, then $v=x^2/y$, then a four-variable $(u,w)$ attempt), each time finding that the two "conjugate" parabolas $y^2=mx$ and $x^2=ny$ do not both reduce to constants of the same variable — explicitly noting "this is complicated" and "let me try another substitution" several times — and ultimately gave up on a closed derivation, presenting two different, mutually-inconsistent unresolved guesses. The substitution $u=xy,\,t=y/x$ used above resolves this: it keeps $u$ tied to the hyperbolas (as in Problem 139) but lets the parabola bounds enter as genuine functions $t_1(u),t_2(u)$ rather than constants, which is exactly what a mismatched pair of parabola families requires.
- Verified numerically for $a=1.6,\,b=0.8,\,m=0.6,\,n=0.4$ (so $mn=0.24<b^2=0.64$, ensuring the parabolas' own crossing point lies outside the hyperbola range): (1) direct double quadrature of $\int_{b^2}^{a^2}\int_{t_1(u)}^{t_2(u)}\frac1{2t}\,dt\,du$ gives $A\approx1.170702$; (2) the closed form gives $\approx1.170702$; (3) an independent shoelace computation tracing the actual curvilinear-quadrilateral boundary in the $(x,y)$ plane gives $\approx1.170702$ as well. All three agree to 6 significant figures.
Result
The area is: