[amp_mcq option1=”\\[\\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}}} } \\]” 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″]<\/p>\n
The correct answer is $\\boxed{\\int_0^1 \\int_0^{\\sqrt{y}} f(x, y) \\, dx dy}$.<\/p>\n
The volume under a surface $z=f(x, y)$ over the region $R$ is given by the double integral<\/p>\n
$$\\iint_R f(x, y) \\, dx dy.$$<\/p>\n
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 $y > x^2$.<\/p>\n
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 $y=1$.<\/p>\n
The shaded region in the following figure shows the region $R$:<\/p>\n
[asy]
\nunitsize(1 cm);<\/p>\n
draw((0,0)–(1,0)–(1,1)–(0,1)–cycle);
\ndraw((0,0)–(0,1));
\ndraw((1,0)–(1,1));
\ndraw((0.25,0)–(0.25,1));
\ndraw((0.5,0)–(0.5,1));
\ndraw((0.75,0)–(0.75,1));<\/p>\n
label(“$x$”, (1,0), E);
\nlabel(“$y$”, (0,1), N);
\nlabel(“$y=x^2$”, (0.5,0.25), S);
\nlabel(“$y=1$”, (1,0.5), W);
\nlabel(“$x=y^2$”, (0.25,0.5), W);
\nlabel(“$x=1$”, (0.75,0.5), W);
\n[\/asy]<\/p>\n
The double integral that gives the volume under $f(x, y)$ over $R$ is then<\/p>\n
$$\\iint_R f(x, y) \\, dx dy = \\int_0^1 \\int_0^{\\sqrt{y}} f(x, y) \\, dx dy.$$<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”\\[\\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}}} } \\]” 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}} = … <\/p>\n