The parabolas \(\mathrm{y}^{2}=4 \mathrm{x}\) and \(\mathrm{x}^{2}=4 \mathrm{y}\) divide the square region bounded the lines \(x=4, y=4\) and the coordinate axes. If \(S_{1}\), \(\mathrm{S}_{2}, \mathrm{~S}_{3}\) are respectively the areas of these parts numbered from top to bottom then \(S_{1}: S_{2}: S_{3}\) is........ (a) \(1: 2: 3\) (b) \(2: 1: 2\) (c) \(3: 2: 3\) (d) \(1: 1: 1\)

To find the points of intersection between the two parabolas and the square region, we need to set them equal to the boundary lines of the square and solve for x and y: Intersection Points with x-axis and the square's vertical boundary: (1) Set \(\mathrm{y}^2 = 4\mathrm{x}\) equal to 0 when \(\mathrm{x} = 4\): \(\mathrm{y}^2 = 4(4) = 16 \Rightarrow \mathrm{y} = 4\) (2) Set \(\mathrm{x}^2 = 4\mathrm{y}\) equal to 0 when \(\mathrm{x} = 4\): \(\mathrm{x} = \sqrt{4(4)} = 4 \Rightarrow \mathrm{y} = 4\) Intersection Points with y-axis and the square's horizontal boundary: (3) Set \(\mathrm{y}^2 = 4\mathrm{x}\) equal to 0 when \(\mathrm{y} = 4\): \(\mathrm{x}^2 = 4(4) = 16 \Rightarrow \mathrm{x} = 4\) (4) Set \(\mathrm{x}^2 = 4\mathrm{y}\) equal to 0 when \(\mathrm{y} = 4\): \(\mathrm{y} = \frac{4^2}{4} = 4 \Rightarrow \mathrm{x} = 4\) Intersection points between the parabolas: Set \(\mathrm{y}^2 = 4\mathrm{x}\) and \(\mathrm{x}^2 = 4\mathrm{y}\) equal to each other: \(\mathrm{x}^2 = \mathrm{y}^2 \Rightarrow \mathrm{y} = \pm\mathrm{x}\) Since we are interested in only the positive square region, \(\mathrm{y} = \mathrm{x}\): Substitute \(\mathrm{y} = \mathrm{x}\) into one of the parabolas' equations: \(\mathrm{x}^2 = 4\mathrm{x} \Rightarrow \mathrm{x} = 0, \mathrm{x} = 4\) The intersection points are: \((4,4)\), \((0,0)\), \((4,0)\), and \((0,4)\).

Now we can use integration to find the areas of the divided regions: (1) Area \(S_1\): \(\displaystyle\int_0^4 (\mathrm{y} - \frac{\mathrm{x}^2}{4})d\mathrm{x}\) (2) Area \(S_2\): \(\displaystyle\int_0^4 (\frac{\mathrm{y}^2}{4} - \frac{\mathrm{x}^2}{4})d\mathrm{x}\) (3) Area \(S_3\): \(\displaystyle\int_0^4 (\frac{\mathrm{x}^2}{4} - \mathrm{y})d\mathrm{x}\)

Write the integrals in terms of x and y and integrate: (1) Area \(S_1\): \(\displaystyle\int_0^4 (\mathrm{y} - \frac{\mathrm{x}^2}{4})d\mathrm{x}\) = \(\displaystyle\int_0^4 (\frac{4\sqrt{x} - x^2}{4})d\mathrm{x}\) = 8 (2) Area \(S_2\): \(\displaystyle\int_0^4 (\frac{\mathrm{y}^2}{4} - \frac{\mathrm{x}^2}{4})d\mathrm{x}\) = \(\displaystyle\int_0^4 (\frac{4x - x^2}{4})d\mathrm{x}\) = 4 (3) Area \(S_3\): \(\displaystyle\int_0^4 (\frac{\mathrm{x}^2}{4} - \mathrm{y})d\mathrm{x}\) = \(\displaystyle\int_0^4 (\frac{x^2 - 4\sqrt{x}}{4})d\mathrm{x}\) = 8 The areas of \(S_1\), \(S_2\), and \(S_3\) are 8, 4, and 8 units^2, respectively.

Since the areas are 8, 4, and 8, the ratio of \(S_1:S_2:S_3\) is: \(\frac{8}{4}:\frac{4}{4}:\frac{8}{4} = 2:1:2\) Therefore, the correct answer is option (b) \(2:1:2\).