Question

Consider two curves $C_1:y^2=4[\sqrt{y}]x$ and $C_2:x^2=4[\sqrt{x}]y$, where [.] denotes the greatest integer function. Then the area of region enclosed by these two curves within the square formed by the lines $x=1,y=1,x=4,y=4$ is

• A. $8/3$ sq. units
• B. $10/3$ sq. units
• C. $11/3$ sq. units
• D. $11/4$ sq. units

Answer 1

Answer: (C)

$y^2=4\left[\sqrt{y}\right]x$
For $y\in [1,4),\left[\sqrt{y}\right]=1$ or $y^2=4x$
Similarly, for $x\in [1,4),\lfloor \sqrt{x}\rfloor =1$ and $x^2=4\lfloor \sqrt{x}\rfloor y$ would transform into $x^2=4y$

The required area is the shaded region
$A=\underset{1}{\overset{2}{\int }}\left(2\sqrt{x}-1\right)dx+\underset{2}{\overset{4}{\int }}\left(2\sqrt{x}-\frac{x^2}{4}\right)dx$
$={\left(\frac{4}{3}{x}^{3/2}-x\right)}_1^2+{\left(\frac{4}{3}{x}^{3/2}-\frac{x^3}{12}\right)}_2^4$
$=\frac{11}{3}$ sq. units