Find the area of the region bounded by the ellipses

Find the area of the region bounded by the ellipses: $$ \frac{x^2}{\cosh^2u}+\frac{y^2}{\sinh^2u}=c^2 \qquad(u=u_1,\,u_2) $$ and the hyperbolas: $$ \frac{x^2}{\cos^2v}-\frac{y^2}{\sin^2v}=c^2 \qquad(v=v_1,\,v_2), $$ where $0<u_1<u_2$, $0<v_1<v_2$, $x>0$, $y>0$.

Step 1 · Confocal elliptic coordinates

Let $x=c\cosh u\cos v,\ y=c\sinh u\sin v$ (with $u>0$, $0<v<\frac\pi2$ covering the first quadrant $x>0,y>0$). Then:

$$ \frac{x^2}{\cosh^2u}+\frac{y^2}{\sinh^2u} = c^2\cos^2v+c^2\sin^2v = c^2, $$

which is exactly the given ellipse equation, with $u$ fixed and $v$ the parameter along it. Likewise:

$$ \frac{x^2}{\cos^2v}-\frac{y^2}{\sin^2v} = c^2\cosh^2u-c^2\sinh^2u = c^2\big(\cosh^2u-\sinh^2u\big)=c^2, $$

which is exactly the given hyperbola equation, with $v$ fixed and $u$ the parameter along it. So the region bounded by the ellipses $u=u_1,u_2$ and hyperbolas $v=v_1,v_2$ is precisely the rectangle $u\in[u_1,u_2],\,v\in[v_1,v_2]$ in these coordinates.

Step 2 · Compute the Jacobian

$$ \frac{\partial x}{\partial u}=c\sinh u\cos v,\quad \frac{\partial x}{\partial v}=-c\cosh u\sin v,\quad \frac{\partial y}{\partial u}=c\cosh u\sin v,\quad \frac{\partial y}{\partial v}=c\sinh u\cos v. $$

$$ \frac{\partial(x,y)}{\partial(u,v)} = c^2\sinh u\cos v\cdot\sinh u\cos v - \big(-c^2\cosh u\sin v\cdot\cosh u\sin v\big) = c^2\big(\sinh^2u\cos^2v+\cosh^2u\sin^2v\big). $$

Using $\cosh^2u=1+\sinh^2u$: $\sinh^2u\cos^2v+\cosh^2u\sin^2v = \sinh^2u(\cos^2v+\sin^2v)+\sin^2v = \sinh^2u+\sin^2v$. So:

$$ \frac{\partial(x,y)}{\partial(u,v)} = c^2\big(\sinh^2u+\sin^2v\big) \geqslant0, $$

and $dx\,dy=c^2(\sinh^2u+\sin^2v)\,du\,dv$.

Step 3 · Set up and evaluate the area integral

$$ A = c^2\int_{v_1}^{v_2}\!\int_{u_1}^{u_2}\big(\sinh^2u+\sin^2v\big)\,du\,dv = c^2(v_2-v_1)\int_{u_1}^{u_2}\sinh^2u\,du + c^2(u_2-u_1)\int_{v_1}^{v_2}\sin^2v\,dv. $$

Using $\sinh^2u=\frac{\cosh2u-1}2$ and $\sin^2v=\frac{1-\cos2v}2$:

$$ \int_{u_1}^{u_2}\sinh^2u\,du = \frac{\sinh2u_2-\sinh2u_1}4-\frac{u_2-u_1}2, \qquad \int_{v_1}^{v_2}\sin^2v\,dv = \frac{v_2-v_1}2-\frac{\sin2v_2-\sin2v_1}4. $$

Substituting and collecting terms, the two "linear" pieces $-\frac{c^2}2(v_2-v_1)(u_2-u_1)$ and $+\frac{c^2}2(u_2-u_1)(v_2-v_1)$ cancel exactly, leaving:

$$ A = \frac{c^2}4\Big[(v_2-v_1)\big(\sinh2u_2-\sinh2u_1\big) - (u_2-u_1)\big(\sin2v_2-\sin2v_1\big)\Big]. $$

Additional Notes

• A note on this problem's source material: unlike every other problem in this collection, no worked solution was provided for this one — only the problem statement itself. The derivation above is an original, from-scratch solution, obtained by recognizing the two given curve families as the level curves of the standard confocal elliptic coordinate system and proceeding via the Jacobian, exactly as for the elliptical- and generalized-polar-coordinate substitutions used elsewhere in this collection.
• Verified numerically for $c=1.3,\,u_1=0.4,\,u_2=0.9,\,v_1=0.5,\,v_2=1.1$: direct double quadrature of $c^2\int_{v_1}^{v_2}\int_{u_1}^{u_2}(\sinh^2u+\sin^2v)\,du\,dv$ gives $A\approx0.527672$, matching the closed form $\approx0.527672$ to $10$ significant figures.