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

” option2=”\[\int\limits_{{\text{y}} = {{\text{x}}^2}}^{{\text{y}} = 1} {\int\limits_{{\text{x}} = {{\text{y}}^2}}^{{\text{x}} = 1} {{\text{f}}\left( {{\text{x,}}\,{\text{y}}} \right){\text{dx dy}}} } \]” option3=”\[\int\limits_{{\text{y}} = 0}^{{\text{y}} = 1} {\int\limits_{{\text{x}} = 0}^{{\text{x}} = 1} {{\text{f}}\left( {{\text{x,}}\,{\text{y}}} \right){\text{dx dy}}} } \]” option4=”\[\int\limits_{{\text{y}} = 0}^{{\text{y}} = \sqrt {\text{x}} } {\int\limits_{{\text{x}} = 0}^{{\text{x}} = \sqrt {\text{y}} } {{\text{f}}\left( {{\text{x,}}\,{\text{y}}} \right){\text{dx dy}}} } \]” correct=”option3″]

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

$$\iint_R f(x, y) \, dx dy = \int_0^1 \int_0^{\sqrt{y}} f(x, y) \, dx dy.$$