Using an appropriate change of variables, calculate

Using an appropriate change of variables, calculate the area bounded by the curves: $$ xy=a^2,\qquad xy=b^2,\qquad y^2=mx,\qquad y^2=nx $$ (assuming $a^2<b^2$ and $m<n$).

Step 1 · Substitute and identify the region

Let $u=xy,\,v=\dfrac{y^2}x$. Then $xy=a^2\Rightarrow u=a^2$, $xy=b^2\Rightarrow u=b^2$, $y^2=mx\Rightarrow v=m$, and $y^2=nx\Rightarrow v=n$. The region maps to the rectangle $[a^2,b^2]\times[m,n]$.

Step 2 · Invert and compute the Jacobian

From $y^2=vx$ and $y=\dfrac ux$: $\dfrac{u^2}{x^2}=vx\Rightarrow x^3=\dfrac{u^2}v\Rightarrow x=u^{2/3}v^{-1/3}$, and $y=\dfrac ux=u^{1/3}v^{1/3}$. Differentiating:

$$ \frac{\partial x}{\partial u}=\frac23u^{-1/3}v^{-1/3},\quad \frac{\partial x}{\partial v}=-\frac{u^{2/3}}{3v^{4/3}},\quad \frac{\partial y}{\partial u}=\frac13u^{-2/3}v^{1/3},\quad \frac{\partial y}{\partial v}=\frac{u^{1/3}}{3v^{2/3}}. $$

$$ \frac{\partial(x,y)}{\partial(u,v)} = \frac23u^{-1/3}v^{-1/3}\cdot\frac{u^{1/3}}{3v^{2/3}} - \left(-\frac{u^{2/3}}{3v^{4/3}}\right)\cdot\frac13u^{-2/3}v^{1/3} = \frac2{9v}+\frac1{9v} = \frac1{3v}, $$

so $dx\,dy=\dfrac1{3v}\,du\,dv$.

Step 3 · Set up and evaluate the area integral

$$ A = \int_{a^2}^{b^2}\!\int_m^n\frac1{3v}\,dv\,du = \frac13\int_{a^2}^{b^2}\ln\frac nm\,du = \frac13\ln\frac nm\cdot(b^2-a^2) = \frac{b^2-a^2}3\ln\frac nm. $$

Additional Notes

• This derivation matches the source exactly.
• Verified numerically for $a=0.9,\,b=1.6,\,m=0.7,\,n=1.5$: direct double quadrature of $\int_{a^2}^{b^2}\int_m^n\frac1{3v}\,dv\,du$ gives $\approx0.444582$, matching $\dfrac{b^2-a^2}3\ln\dfrac nm\approx0.444582$.