question_answer

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

Correct Answer: C

Solution :

[c] \[{{y}^{2}}=4[\sqrt{y}]x\] For \[y\in [1,4][\sqrt{y}]=1\]or \[{{y}^{2}}=4x.\] Similarly, for \[x\in [1,4),\left[ \sqrt{x} \right]=1\] and \[{{x}^{2}}=4\left\lfloor \sqrt{x} \right\rfloor y\]would Transform into \[{{x}^{2}}=4y.\] The required area is the shaded region. \[A=\int\limits_{1}^{2}{(2\sqrt{x}-1)dx+\int\limits_{2}^{4}{\left( 2\sqrt{x}-\frac{{{x}^{2}}}{4} \right)dx}}\] \[={{\left( \frac{4}{3}{{x}^{3/2}}-x \right)}_{1}}^{2}+{{\left( \frac{4}{3}{{x}^{3/2}}-\frac{{{x}^{3}}}{12} \right)}_{2}}^{4}=\frac{11}{3}\] Sq. units.