As long as you're only asking for an expression as an integral, and not an actual number, we can calculate the area as follows:
Let
- $a$ be the positive real solution of $x^5-11x^3-1=0$
- $b$ be the positive real solution of $x^{7/2}-11x^{3/2}-3\sqrt{3}=0$
- $c$ be the positive real solution of $x^4-26x^2-1=0$
- $d$ be the positive real solution of $x^3-26x-3=0$
We have $a
$$\color{red}{xy=1},\quad \color{green}{xy^2=3},\quad \color{blue}{x^2-y^2=26},\quad \color{black}{x^2-y^3=11}$$

As you can see, the equations for $a,b,c,d$ were obtained by solving for the $x$-coordinate of the relevant intersections of the curves.
In the upper right quadrant, we can re-express our four curves as $$\color{red}{y=\tfrac{1}{x}},\quad \color{green}{y=\sqrt{\tfrac{3}{x}}},\quad \color{blue}{y=\sqrt{x^2-26}},\quad \color{black}{y=(x^2-11)^{1/3}}$$
The area below the black curve and above the red curve, from $a$ to $b$, is $$\int_a^b\left((x^2-11)^{1/3}-\tfrac{1}{x}\right)dx$$ The area below the green curve and above the red curve, from $b$ to $c$, is $$\int_b^c\left(\sqrt{\tfrac{3}{x}}-\tfrac{1}{x}\right)dx$$ The area below the green curve and above the blue curve, from $c$ to $d$, is $$\int_c^d\left(\sqrt{\tfrac{3}{x}}-\sqrt{x^2-26}\right)dx$$ Thus the area of the upper region is $$\int_a^b\left((x^2-11)^{1/3}-\tfrac{1}{x}\right)dx+\int_b^c\left(\sqrt{\tfrac{3}{x}}-\tfrac{1}{x}\right)dx+\int_c^d\left(\sqrt{\tfrac{3}{x}}-\sqrt{x^2-26}\right)dx$$
We can do a similar computation for the lower region.
Mathematica code:
NSolve[x^5 - 11x^3 - 1 == 0, x] NSolve[x^(7/2) - 11x^(3/2) - 3*Sqrt[3] == 0, x] NSolve[x^4 - 26x^2 - 1 == 0, x] NSolve[x^3 - 26x - 3 == 0, x] a = 3.320739129529704 b = 3.437347103656831 c = 5.102784025451723 d = 5.155761179910075 ContourPlot[{x*y == 1, x*y^2 == 3, x^2 - y^2 == 26, x^2 - y^3 == 11, x == a, x == b, x == c, x == d}, {x, 2.5, 6}, {y, -2, 2}, ContourStyle -> {{Red, Thick}, {Green, Thick}, {Blue, Thick}, {Black, Thick}, {Black, Dashed}, {Black, Dashed}, {Black, Dashed}, {Black, Dashed}}]