f(x, y) is a continuous function defined over (x, y) [ in ] [0, 1] Ã [0, 1]. Given the two constraints, x > y2 and y > x2, the volume under f(x, y) is A. [intlimits_{{text{y}} = 0}^{{text{y}} = 1} {intlimits_{{text{x}} = {{text{y}}^2}}^{{text{x}} = sqrt {text{y}} } {{text{f}}left( {{text{x,}},{text{y}}} right){text{dx dy}}} } ] B. [intlimits_{{text{y}} = {{text{x}}^2}}^{{text{y}} = 1} {intlimits_{{text{x}} = {{text{y}}^2}}^{{text{x}} = 1} {{text{f}}left( {{text{x,}},{text{y}}} right){text{dx dy}}} } ] C. [intlimits_{{text{y}} = 0}^{{text{y}} = 1} {intlimits_{{text{x}} = 0}^{{text{x}} = 1} {{text{f}}left( {{text{x,}},{text{y}}} right){text{dx dy}}} } ] D. [intlimits_{{text{y}} = 0}^{{text{y}} = sqrt {text{x}} } {intlimits_{{text{x}} = 0}^{{text{x}} = sqrt {text{y}} } {{text{f}}left( {{text{x,}},{text{y}}} right){text{dx dy}}} } ]
The correct answer is $boxed{int_0^1 int_0^{sqrt{y}} f(x, y) , dx dy}$.
The volume under a surface $z=f(x, y)$ over the region $R$ is given by the double integral
$$iint_R f(x, y) , dx dy.$$
In this case, we are given that $f(x, y)$ is a continuous function defined over the region $R = [0, 1] times [0, 1]$. We are also given the two constraints $x > y^2$ and $y > x^2$.
The first constraint means that the region $R$ is bounded by the lines $x=y^2$ and $x=1$. The second constraint means that the region $R$ is also bounded by the lines $y=x^2$ and $y=1$.
The shaded region in the following figure shows the region $R$:
[asy]
unitsize(1 cm);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((0,0)--(0,1));
draw((1,0)--(1,1));
draw((0.25,0)--(0.25,1));
draw((0.5,0)--(0.5,1));
draw((0.75,0)--(0.75,1));
label("$x$", (1,0), E);
label("$y$", (0,1), N);
label("$y=x^2$", (0.5,0.25), S);
label("$y=1$", (1,0.5), W);
label("$x=y^2$", (0.25,0.5), W);
label("$x=1$", (0.75,0.5), W);
[/asy]
The double integral that gives the volume under $f(x, y)$ over $R$ is then
$$iint_R f(x, y) , dx dy = int_0^1 int_0^{sqrt{y}} f(x, y) , dx dy.$$