1. Tardigrade
  2. Question
  3. Mathematics
  4. The parabolas y2 = 4x and x2 = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If S1 , S2 , S3 are respectively the areas of these parts numbered from top to bottom; then S1: S2: S3 is

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Q. 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

Application of Integrals

Solution:

Intersection points of $x^2 = 4y$ and $y^2 = 4x$ are (0, 0) and (4, 4). The graph is as shown in the figure.
image
By symmetry, we observe
$S_{1} = S_{3} = \int\limits^{4}_{0 }ydx$
$ = \int\limits^{4}_{0} \frac{x^{2}}{4} dx = \left[\frac{x^{3}}{12}\right]^{4}_{0} = \frac{16}{3}$ sq. units
Also
$ S_{2} = \int\limits^{4}_{0} \left(2 \sqrt{x } - \frac{x^{2}}{4} \right)dx = \left[\frac{2x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{3}}{12}\right] ^{4}_{0} $
$ = \frac{4}{3} \times8- \frac{16}{3} = \frac{16}{3} $ sq. units
$\therefore S_{1} : S_{2} : S_{3} = 1 : 1 : 1$