f(x, y) is a continuous function defined over (x, y) \[ \in \] [0, 1] × [0, 1]. Given the two constraints, x > y² and y > x², the volume under f(x, y) is

A. \[\int\limits_{{\text{y}} = 0}^{{\text{y}} = 1} {\int\limits_{{\text{x}} = {{\text{y}}^2}}^{{\text{x}} = \sqrt {\text{y}} } {{\text{f}}\left( {{\text{x,}}\,{\text{y}}} \right){\text{dx dy}}} } \]

B. \[\int\limits_{{\text{y}} = {{\text{x}}^2}}^{{\text{y}} = 1} {\int\limits_{{\text{x}} = {{\text{y}}^2}}^{{\text{x}} = 1} {{\text{f}}\left( {{\text{x,}}\,{\text{y}}} \right){\text{dx dy}}} } \]

C. \[\int\limits_{{\text{y}} = 0}^{{\text{y}} = 1} {\int\limits_{{\text{x}} = 0}^{{\text{x}} = 1} {{\text{f}}\left( {{\text{x,}}\,{\text{y}}} \right){\text{dx dy}}} } \]

D. \[\int\limits_{{\text{y}} = 0}^{{\text{y}} = \sqrt {\text{x}} } {\int\limits_{{\text{x}} = 0}^{{\text{x}} = \sqrt {\text{y}} } {{\text{f}}\left( {{\text{x,}}\,{\text{y}}} \right){\text{dx dy}}} } \]

Answer: Option A