Prove that the curves $y^{2}=4x$ and $x^{2}=4y$ divide the area of the square bounded by $x=0, x=4, y=4$ and $y=0$ into three equal parts.

(N/A) The points of intersection of the parabolas $y^{2}=4x$ and $x^{2}=4y$ are found by substituting $y = \frac{x^{2}}{4}$ into $y^{2}=4x$,which gives $(\frac{x^{2}}{4})^{2} = 4x$,so $x^{4} = 64x$. This implies $x(x^{3}-64) = 0$,so $x=0$ or $x=4$. Thus,the intersection points are $(0,0)$ and $(4,4)$.
$1$. The area of the region bounded by the curves $y^{2}=4x$ and $x^{2}=4y$ is:
$\int_{0}^{4} (2\sqrt{x} - \frac{x^{2}}{4}) dx = [2 \times \frac{2}{3} x^{\frac{3}{2}} - \frac{x^{3}}{12}]_{0}^{4} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$ square units.
$2$. The area of the region bounded by the curve $x^{2}=4y$,the $x$-axis,and the lines $x=0$ and $x=4$ is:
$\int_{0}^{4} \frac{x^{2}}{4} dx = \frac{1}{12} [x^{3}]_{0}^{4} = \frac{64}{12} = \frac{16}{3}$ square units.
$3$. The area of the region bounded by the curve $y^{2}=4x$,the $y$-axis,and the lines $y=0$ and $y=4$ is:
$\int_{0}^{4} \frac{y^{2}}{4} dy = \frac{1}{12} [y^{3}]_{0}^{4} = \frac{64}{12} = \frac{16}{3}$ square units.
Since the total area of the square is $4 \times 4 = 16$ square units and each of the three regions has an area of $\frac{16}{3}$ square units,the curves divide the square into three equal parts.