Question

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

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

Answer 1

Answer: (D)

Intersection points of $x^2=4y$ and $y^2=4x$ are (0, 0) and (4, 4). The graph is as shown in the figure.

By symmetry, we observe
$S_1=S_3=\underset{0}{\overset{4}{\int }}ydx$
$=\underset{0}{\overset{4}{\int }}\frac{x^2}{4}dx={\left[\frac{x^3}{12}\right]}_0^4=\frac{16}{3}$ sq. units
Also
$S_2=\underset{0}{\overset{4}{\int }}\left(2\sqrt{x}-\frac{x^2}{4}\right)dx={\left[\frac{2x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^3}{12}\right]}_0^4$
$=\frac{4}{3}\times 8-\frac{16}{3}=\frac{16}{3}$ sq. units
$\therefore S_1:S_2:S_3=1:1:1$