The parabolas `y^2=4xa n dx^2=4y` divide the square region bounded by the lines `x=4,y=4` and the coordinate axes. If `S_1,S_2,S_3` are the areas of these parts numbered from top to bottom, respectively, then `S_1: S_2-=1:1` (b) `S_2: S_3-=1:2` `S_1: S_3-=1:1` (d) `S_1:(S_1+S_2)=1:2`

Options

A. `1:1:1`
B. `2:1:2`
C. `1:2:3`
D. `1:3:2`

Explanation

To solve the problem, we need to find the areas \( S_1, S_2, \) and \( S_3 \) formed by the parabolas \( y^2 = 4x \) and \( x^2 = 4y \) within the square region bounded by the lines \( x = 4 \), \( y = 4 \), and the coordinate axes.

### Step 1: Identify the curves and their intersections
The equations of the parabolas are:
1. \( y^2 = 4x \) (which opens to the right)
2. \( x^2 = 4y \) (which opens upwards)

To find the points of intersection, we can set \( y^2 = 4x \) and \( x^2 = 4y \).

Substituting \( y = \frac{x^2}{4} \) into \( y^2 = 4x \):
\[ \left(\frac{x^2}{4}\right)^2 = 4x \]
\[ \frac{x^4}{16} = 4x \]
\[ x^4 - 64x = 0 \]
Factoring gives:
\[ x(x^3 - 64) = 0 \]
Thus, \( x = 0 \) or \( x = 4 \). Therefore, the points of intersection are \( (0, 0) \) and \( (4, 4) \).

### Step 2: Calculate \( S_3 \)
The area \( S_3 \) is bounded by the curve \( x^2 = 4y \) from \( x = 0 \) to \( x = 4 \):
\[ S_3 = \int_0^4 \frac{x^2}{4} \, dx \]
Calculating the integral:
\[ S_3 = \frac{1}{4} \int_0^4 x^2 \, dx = \frac{1}{4} \left[ \frac{x^3}{3} \right]_0^4 = \frac{1}{4} \left[ \frac{64}{3} \right] = \frac{16}{3} \]

### Step 3: Calculate \( S_1 \)
The area \( S_1 \) is bounded by the curve \( y^2 = 4x \) from \( y = 0 \) to \( y = 4 \):
\[ S_1 = \int_0^4 \frac{y^2}{4} \, dy \]
Calculating the integral:
\[ S_1 = \frac{1}{4} \int_0^4 y^2 \, dy = \frac{1}{4} \left[ \frac{y^3}{3} \right]_0^4 = \frac{1}{4} \left[ \frac{64}{3} \right] = \frac{16}{3} \]

### Step 4: Calculate \( S_2 \)
The area \( S_2 \) is bounded between the curves \( y^2 = 4x \) and \( x^2 = 4y \):
\[ S_2 = \int_0^4 \left( \frac{y^2}{4} - \frac{x^2}{4} \right) \, dx \]
Using the relationship \( y = 2\sqrt{x} \) from \( y^2 = 4x \):
\[ S_2 = \int_0^4 \left( 2\sqrt{x} - \frac{x^2}{4} \right) \, dx \]
Calculating the integral:
\[ S_2 = \left[ \frac{4}{3} x^{3/2} - \frac{x^3}{12} \right]_0^4 \]
Calculating at the bounds:
\[ = \left[ \frac{4}{3} \cdot 8 - \frac{64}{12} \right] = \left[ \frac{32}{3} - \frac{16}{3} \right] = \frac{16}{3} \]

### Step 5: Ratios of Areas
Now we have:
- \( S_1 = \frac{16}{3} \)
- \( S_2 = \frac{16}{3} \)
- \( S_3 = \frac{16}{3} \)

Thus, the ratios are:
\[ S_1 : S_2 : S_3 = 1 : 1 : 1 \]

### Conclusion
The correct option is:
(a) \( S_1 : S_2 = 1 : 1 \)