Find the area bounded by the following curves: y=x², x=y²

Find the area bounded by the following curves: $$ y=x^2, \qquad x=y^2. $$

Explanation of Method

These two parabolas are reflections of each other across the line $y=x$, and both pass through the origin and through $(1,1)$. Between these two intersection points, one curve lies above the other, so the enclosed area is a single elementary integral of the vertical (or, by symmetry, horizontal) gap between them.

Find the intersection points.
Determine which curve is above the other between them.
Integrate the gap between the curves.

Step 1 · Find the intersection points

Substituting $y=x^2$ into $x=y^2$:

$$ x=(x^2)^2=x^4 \quad\Longrightarrow\quad x^4-x=0 \quad\Longrightarrow\quad x(x^3-1)=0 \quad\Longrightarrow\quad x=0 \text{ or } x=1. $$

Intersection points: $(0,0)$ and $(1,1)$.

Step 2 · Determine which curve is above

For $0<x<1$, write $x=y^2$ as $y=\sqrt x$ (first-quadrant branch). At $x=\dfrac14$: $y=x^2=\dfrac1{16}$ from the first curve, and $y=\sqrt x=\dfrac12$ from the second. Since $\dfrac12>\dfrac1{16}$, the curve $x=y^2$ (i.e. $y=\sqrt x$) lies above $y=x^2$ throughout $(0,1)$.

Step 3 · Set up and evaluate the area integral

$$ A = \int_0^1\left(\sqrt x-x^2\right)dx = \left[\frac{2x^{3/2}}3-\frac{x^3}3\right]_0^1 = \frac23-\frac13 = \frac13. $$

Additional Notes

Confirmed by symmetry: since the two curves are reflections of one another across $y=x$, integrating with horizontal strips instead gives the identical computation $\displaystyle\int_0^1(\sqrt y-y^2)\,dy=\frac13$.
The bounded region is a symmetric lens shape straddling the line $y=x$, consistent with the two curves being mirror images of each other.

Result

The area is:

$$ A = \frac13. $$