Consider the first eq i.e

$qxy = 4$ i.e $qx = 4/y$

Let $x = 0$ then $y = \infty$

There $(0, \infty )$

Let $x = \infty$ then $y = 0$

$( \infty , 0)$

When $ x = 1, y = 4/q$

When $y = 1, x = 4/q$

Plotting through diagram

We get,

Now consider 2nd equation

i.e $2x + y = z --------- (2)$

Let $x = 0 y = 2$

Let $y = 0 x = 1$

Equation (2) is a straight line passing from $(0, 2)$ and $(1, 0)$

This line is intersecting curve at solving eq (2)

$Xy = 4/9, x = 4/9y$

Substituting value of x in eq (2)

$\therefore y=\dfrac 43 $ & $\dfrac 23$

Corresponding x is

$x=\dfrac 4{9\times \dfrac43}=\dfrac 13$

$x=\dfrac 4{9\times \dfrac43}=\dfrac 23$

Hence the points of intersection of the hyperbola and the line are $(2/3, 2/3)$ and $(1/3, 4/3)$

Now considering the horizontal strip which will slide along x axis i.e from $1/3$ to $22/3 $ which is outer limit.

Upper limit is equation of line $2x + y = 2$ in terms of y we get $y = 2 – 2x$

Lower limit is equation of curves

$qxy = 4$ in terms of y

we get $y = 4/qx$

$A=\int\limits_{1/3}^{2/3}\int\limits_{4/9x}^{2-2x} dydx$

Integrating w.r.t y we get

$A=\int\limits_{1/3}^{2/3}[y]^{2-2x}_{4/9x}dx$

$A=\int\limits_{1/3}^{2/3}2-2x-\dfrac 4{9x}dx$

$A=\Bigg[2x-\dfrac {2x^2}2-\dfrac 49\log x\Bigg]^{2/3}_{1/3}$

$A=\dfrac 43-\dfrac 49-\dfrac 49\log 2/3$

$=-\dfrac 23+\dfrac 19+\dfrac 49\log 1/3$

$=\dfrac 13-\dfrac 49\Big[\log 2-\log 3-\log 1+\log 3\Big]$

$=\dfrac 13 -\dfrac 49\log 2$