Find the area bounded by the following curves (x>0, y>0)
Find the area bounded by the following curves ($x>0$, $y>0$): $$ xy=a^2, \qquad xy=2a^2, \qquad y=x, \qquad y=2x. $$
Explanation of Method
The four boundary curves are each level sets of one of the two quantities $u=xy$ and $v=y/x$, so the substitution $u=xy,\,v=y/x$ maps this curvilinear quadrilateral exactly onto the rectangle $[a^2,2a^2]\times[1,2]$ in the $(u,v)$-plane. The area is then obtained from the Jacobian of the transformation, which reduces the problem to an elementary rectangle integral.
- Change variables to $u=xy$, $v=y/x$, mapping the region to a rectangle.
- Compute the Jacobian $\dfrac{\partial(x,y)}{\partial(u,v)}$.
- Integrate over the rectangle $[a^2,2a^2]\times[1,2]$.
Step 1 · Change of variables
Let $u=xy$, $v=\dfrac yx$. Solving: $y=vx$, and $x\cdot vx=u\Rightarrow x^2=\dfrac uv\Rightarrow x=\sqrt{\dfrac uv}$, and $y=vx=\sqrt{uv}$.
The four curves become $u=a^2$, $u=2a^2$, $v=1$, $v=2$, so the region maps exactly onto the rectangle $a^2\leqslant u\leqslant2a^2$, $1\leqslant v\leqslant2$.
Step 2 · Compute the Jacobian
With $x=\dfrac{\sqrt u}{\sqrt v}$, $y=\sqrt{uv}$:
$$ \frac{\partial x}{\partial u}=\frac1{2\sqrt{uv}}, \quad \frac{\partial x}{\partial v}=-\frac{\sqrt u}{2v^{3/2}}, \quad \frac{\partial y}{\partial u}=\frac{\sqrt v}{2\sqrt u}, \quad \frac{\partial y}{\partial v}=\frac{\sqrt u}{2\sqrt v}. $$
$$ \frac{\partial(x,y)}{\partial(u,v)} = \frac1{2\sqrt{uv}}\cdot\frac{\sqrt u}{2\sqrt v} - \left(-\frac{\sqrt u}{2v^{3/2}}\right)\cdot\frac{\sqrt v}{2\sqrt u} = \frac1{4v}+\frac1{4v} = \frac1{2v}. $$
Step 3 · Set up and evaluate the area integral
$$ A = \iint_D dx\,dy = \int_1^2\int_{a^2}^{2a^2} \frac1{2v}\,du\,dv = \int_1^2 \frac1{2v}\big[u\big]_{a^2}^{2a^2}\,dv = \frac{a^2}2\int_1^2\frac{dv}v = \frac{a^2}2\ln2. $$
Additional Notes
- Verified numerically (Monte Carlo sampling of the region $\{1\leqslant xy\leqslant2,\,1\leqslant y/x\leqslant2\}$ for $a=1$): estimated area $\approx0.3461$, matching $\frac{\ln2}2\approx0.3466$.
- This is a clean illustration of choosing coordinates aligned with the boundary curves: with $(u,v)=(xy,\,y/x)$, all four bounding curves become coordinate lines, turning an awkward curvilinear region into a rectangle and the area computation into a one-line double integral.
Result
The area is: