Let $A_{1}, A_{2}$ and $A_{3}$ be the regions on $R^{2}$ defined by $$ \begin{array}{l} A_{1}=\left\{(x, y) ; x \geq 0, y \geq 0,2 x+2 y-x^{2}-y^{2}>1>x\right. \\ +y\} \\ A_{2}=\left\{(x, y) ; x \geq 0, y \geq 0, x+y>1>x^{2}+y^{2}\right\} \\ A_{3}=\left\{(x, y) ; x \geq 0, y \geq 0, x+y>1>x^{3}+y^{3}\right\} \end{array} $$ Denote by $\left|A_{1}\right|,\left|A_{2}\right|$ and $\left|A_{3}\right|$ the areas of the regions $A_{1}, A_{2}$, and $A_{3}$ respectively. Then $$(a) $\left|A_{1}\right|>\left|A_{2}\right|>\left|A_{3}\right|$$ $$(b) $\left|A_{1}\right|>\left|A_{3}\right|>\left|A_{2}\right|$$ $$(c) $\left|A_{1}\right|=\left|A_{2}\right|<\left|A_{3}\right|$$ $$(d) $\left|A_{1}\right|=\left|A_{3}\right|>\left|A_{2}\right|$$.
My method was that x+y >1 will represent points above the x+y = 1 line as not containing origin as it doesnt satify it. The curve x^2 +y^2 <1 will represent the inner section of the circle x^2 + y^2 = 1 , simalrily with the 2x+2y -x^2 -y^2 one circle , but how do we check whether x^3 + y^3 <1 locus as such for that we need x^3 +y^3 =1 locus and the set would be inside that locus . It will be somewhat circle but not similar but for sure the symmetric points would be a bit closer to y=x axis so curve at that point will be closer than circle x^2 +y^2 =1 , but how to get the shape roughly so as to get the area which might be intersections ?