The parabolas y^2 = 4x and x^2 = 4y divide th | vedclass

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(B) The total area of the square bounded by $x=0, x=4, y=0, y=4$ is $4 	imes 4 = 16$ square units.The area $S_2$ is the area enclosed between the two

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$1:1:1$

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(B) The total area of the square bounded by $x=0, x=4, y=0, y=4$ is $4 	imes 4 = 16$ square units.The area $S_2$ is the area enclosed between the two parabolas $y^2 = 4x$ and $x^2 = 4y$. The intersection points are $(0,0)$ and $(4,4)$.$S_2 = \int_0^4 (\sqrt{4x} - \frac{x^2}{4}) dx = \int_0^4 (2x^{1/2} - \frac{x^2}{4}) dx = [2 \cdot \frac{2}{3} x^{3/2} - \frac{x^3}{12}]_0^4 = \frac{4}{3}(8) - \frac{64}{12} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$.The area $S_1$ is the area of the region bounded by $x=0, y=4$ and the parabola $y^2=4x$. This is the area of the square minus the area under the parabola $y^2=4x$ from $x=0$ to $4$.Area under $y^2=4x$ is $\int_0^4 2\sqrt{x} dx = [\frac{4}{3} x^{3/2}]_0^4 = \frac{32}{3}$.So,$S_1 = 16 - \frac{32}{3} - S_2 = 16 - \frac{32}{3} - \frac{16}{3} = 16 - 16 = 0$ is incorrect logic. Let's re-evaluate.$S_1$ is the area bounded by $x=0, y=4$ and $y^2=4x$. $S_1 = \int_0^4 (4 - 2\sqrt{x}) dx = [4x - \frac{4}{3}x^{3/2}]_0^4 = 16 - \frac{32}{3} = \frac{16}{3}$.Similarly,$S_3$ is the area bounded by $y=0, x=4$ and $x^2=4y$. $S_3 = \int_0^4 (4 - \frac{x^2}{4}) dx = [4x - \frac{x^3}{12}]_0^4 = 16 - \frac{64}{12} = 16 - \frac{16}{3} = \frac{32}{3}$. Wait,the diagram shows $S_1$ and $S_3$ are symmetric.Actually,$S_1 = \int_0^4 (4 - 2\sqrt{x}) dx = \frac{16}{3}$ and $S_3 = \int_0^4 (4 - \frac{x^2}{4}) dx = \frac{32}{3}$ is wrong. By symmetry,$S_1 = S_3 = \frac{16}{3}$.Thus,$S_1:S_2:S_3 = \frac{16}{3} : \frac{16}{3} : \frac{16}{3} = 1:1:1$.

The parabolas $y^2 = 4x$ and $x^2 = 4y$ divide the square region bounded by the lines $x = 4$,$y = 4$ and the coordinate axes. If $S_1, S_2, S_3$ are respectively the areas of these parts numbered from top to bottom,then $S_1:S_2:S_3$ is

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