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

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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));
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