question_answer

The parabolas\[{{y}^{2}}=4x\]and\[{{x}^{2}}=4y\]divide the square region bounded by the lines\[x=4,\text{ }y=4\]and the coordinates 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

Correct Answer: D

Solution :

 \[{{S}_{1}}={{S}_{3}}=\int_{0}^{4}{\frac{{{x}^{2}}}{4}}dx\] \[=\frac{1}{4}\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{4}=\frac{1}{12}\times 64=\frac{16}{3}sq\,units\] \[\therefore \] \[{{S}_{2}}+{{S}_{3}}=\int_{0}^{4}{\sqrt{4x}}dx\] \[=2\times 2\left[ \frac{{{x}^{3/2}}}{3} \right]_{0}^{4}=\frac{4}{3}\times 8=\frac{32}{3}sq\,units\] \[\Rightarrow \] \[{{S}_{2}}=\frac{16}{3}sq\,units\] \[\therefore \] \[{{S}_{1}}:{{S}_{2}}:{{S}_{3}}=\frac{16}{3}:\frac{16}{3}:\frac{16}{3}\] \[=1:1:1\]