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Step-by-Step Solution
Step 1: Understand the Region and the Curves
We are given two parabolas:
1) $y^2 = 4x$, which can be rewritten as $x = \frac{y^2}{4}$.
2) $x^2 = 4y$, which can be rewritten as $y = \frac{x^2}{4}$.
The region of interest is the square bounded by the lines $x = 4$, $y = 4$, and the coordinate axes ($x = 0$ and $y = 0$). The two parabolas split this square into three parts, labeled $S_1$, $S_2$, and $S_3$ from top to bottom.
Step 2: Find the Points of Intersection
To locate the boundary of the region, note that these parabolas also pass through common points:
- At $(0, 0)$, substituting $x=0,\,y=0$ satisfies both $y^2=4x$ and $x^2=4y$.
- At $(4, 4)$, if $x=4$, then from $x^2=4y$, we get $16 = 4y \implies y=4$. Also, if $y=4$, from $y^2=4x$, we get $16=4x \implies x=4$.
Thus, the two parabolas intersect at $(0,0)$ and $(4,4)$.
Step 3: Visual Representation
A rough sketch (or the provided figure) shows how the two parabolas $y^2=4x$ and $x^2=4y$ divide the square.
Step 4: Use Symmetry to Identify Equal Areas
By symmetry, the topmost region ($S_1$) and the bottommost region ($S_3$) are mirror images of each other across the line $y=x$. Therefore, we have:
$S_1 = S_3.$
Step 5: Calculate One of the Equal Areas (e.g., $S_1$)
The region $S_1$ (topmost part) can be expressed as an integral. Here, $S_1$ lies between $x=0$ and $x=4$, and the boundary for $y$ is given by $y = \frac{x^2}{4}$. Thus,
$S_1 = \int_{0}^{4} \frac{x^2}{4}\,dx.$
Let us compute it step by step:
$
\displaystyle
S_1 = \int_{0}^{4} \frac{x^2}{4}\,dx
= \frac{1}{4}\int_{0}^{4} x^2 \,dx
= \frac{1}{4}\left[\frac{x^3}{3}\right]_{0}^{4}
= \frac{1}{4}\left(\frac{4^3}{3}\right)
= \frac{1}{4}\times \frac{64}{3}
= \frac{64}{12}
= \frac{16}{3}.
$
Step 6: Calculate the Middle Region Area $S_2$
The region $S_2$ (middle part) lies between the curves:
- $y = 2\sqrt{x}$ (since $y^2=4x \implies y = 2\sqrt{x}$ for $y \ge 0$),
- $y = \frac{x^2}{4}$,
from $x=0$ to $x=4$. Hence,
$
\displaystyle
S_2 = \int_{0}^{4} \Bigl(2\sqrt{x} - \frac{x^2}{4}\Bigr)\,dx.
$
Let us split the integral and compute:
$
\displaystyle
\int_{0}^{4} 2\sqrt{x}\,dx
= 2\int_{0}^{4} \sqrt{x}\,dx
= 2\int_{0}^{4} x^{1/2}\,dx
= 2\left[\frac{x^{3/2}}{\frac{3}{2}}\right]_{0}^{4}
= 2\left[\frac{2}{3}x^{3/2}\right]_{0}^{4}
= \frac{4}{3}\left[x^{3/2}\right]_{0}^{4}
= \frac{4}{3}\left[(4)^{3/2}\right]
= \frac{4}{3}\left[8\right]
= \frac{32}{3}.
$
$
\displaystyle
\int_{0}^{4} \frac{x^2}{4}\,dx
= \frac{1}{4}\int_{0}^{4} x^2\,dx
= \frac{1}{4}\left[\frac{x^3}{3}\right]_{0}^{4}
= \frac{1}{4}\times \frac{64}{3}
= \frac{16}{3}.
$
Therefore,
$
\displaystyle
S_2
= \left(\frac{32}{3}\right)
- \left(\frac{16}{3}\right)
= \frac{16}{3}.
$
Step 7: Conclude the Ratios
We have
$S_1 = \frac{16}{3}, \quad S_2 = \frac{16}{3}, \quad S_3 = S_1 = \frac{16}{3}.$
Thus,
$
\displaystyle
S_1 : S_2 : S_3 = 1 : 1 : 1.
$
Final Answer: $1 : 1 : 1$
