The area enclosed between the curves y2 = 4x and x2 = 4y is A. \[\frac{{16}}{3}\] B. 8 C. \[\frac{{32}}{3}\] D. 16

” option2=”8″ option3=”\[\frac{{32}}{3}\]” option4=”16″ correct=”option1″]

The area enclosed between the curves $y^2 = 4x$ and $x^2 = 4y$ is $\frac{32}{3}$ square units.

To find the area, we can use the following steps:

1. Find the points of intersection of the two curves.
2. Set up a definite integral that represents the area enclosed by the curves.
3. Evaluate the definite integral.

The points of intersection of the two curves are $(0,0)$ and $(4,4)$.

The definite integral that represents the area enclosed by the curves is:

$$\int_0^4 \left( y^2 – x^2 \right) \, dx$$

Evaluating the definite integral, we get:

$$\left[ \frac{y^3}{3} – \frac{x^3}{3} \right]_0^4 = \frac{64}{3} – \frac{64}{3} = \boxed{\frac{32}{3}}$$

Here is a graph of the two curves:

[asy]
unitsize(1 cm);

real ticklen=3;
real tickspace=1;
real axisarrowsize=0.14inch;
real tickdown=-0.5;
real tickdownlength=-0.12inch;
real wholetickdown=-0.24inch;
real wholetickdownlength=-0.2inch;
real tickdownlengthshort=-0.08inch;
real wholetickdownlengthshort=-0.12inch;
real t=0;
real d=1;
real xleft=-4;
real xright=4;
real ybottom=-4;
real ytop=4;

draw((xleft,0)–(xright,0));
draw((0,ybottom)–(0,ytop));

real i;

for(i=xleft+tickspace; i<xright; i+=tickspace) {
draw((i,0.1)–(i,-0.1));
}

for(i=ybottom+tickspace; i<ytop; i+=tickspace) {
draw((0.1,i)–(-0.1,i));
}

label(“$x$”,(xright+0.2,0),SE);
label(“$y$”,(0,ytop+0.2),NW);

draw((0,-0.1)–(0,0.1));
draw((0.1,-0.1)–(0.1,0.1));

label(“$y^2=4x$”,(4,2),S);
label(“$x^2=4y$”,(2,4),E);

draw(graph(y^2=4x,-4,4),red);
draw(graph(x^2=4y,-4,4),blue);
[/asy]