Actually, the question asks us to find the area of this figure (and divide it by `pi `). The inequalities may seem scary, but actually they are not. Each of them is an equation of inner or outer part of some circle:

`1 / 11 (x^2 + y^2 + 1 ) lt= x ` is the same as `x^2 - 11x + y^2 + 1 lt= 0 , ` which is the same as `( x - 11 / 2 )^2 + y^2 lt= ( 11 / 2 )^2 - 1 ,`

which is the equation of the inner part of the circle with the center `( 11 / 2 , 0 ) ` and the radius squared of `R^2 = ( 11 / 2 )^2 - 1 .`

The second equation `1 / 7 (x^2 + y^2 + 1 ) gt= x ` becomes `x^2 -7x + y^2 + 1 gt= 0 , ` or `( x - 7 / 2 )^2 + y^2 gt= ( 7 / 2 )^2 - 1 ,`

which is the equation of the outer part of the circle with the center `( 7 / 2 , 0 ) ` and the radius squared of `r^2 = ( 7 / 2 )^2 - 1 .`

It is simple to prove that the smaller circle lies entirely in the greater, so the area of the figure (divided by `pi`) is simply

`R^2 - r^2 = ( 11 / 2 )^2 - (7/2)^2 = ((11-7)(11+7))/4 = 18 .`

The graph is made with Desmos.com.

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