question_answer

The area included between the parabolas\[x=4y\] and \[y=4x\] is (in square units) :

A)  \[\frac{4}{3}\]                                  

B)  \[\frac{1}{3}\]

C)  \[\frac{16}{3}\]                               

D)  \[\frac{8}{3}\]

Correct Answer: C

Solution :

Here we have                 \[{{y}^{2}}=4x\]                                                ... (i)                 \[{{x}^{2}}=4y\]                                                ... (ii) from (i) and (ii) we get                 \[{{\left( \frac{{{x}^{2}}}{4} \right)}^{2}}=4x\] \[\Rightarrow \]               \[{{x}^{4}}-64x=0\]                 \[x({{x}^{3}}-{{4}^{3}})=0\] \[\therefore \,\,x=0,\,x=4\] \[\therefore \] required area \[\left| \int_{0}^{4}{\sqrt{4x}\,dx-}\int_{0}^{4}{\frac{{{x}^{2}}}{4}dx} \right|\] \[=\left| \left\{ 2\left( \frac{2}{3} \right){{x}^{3/2}} \right\}_{0}^{4}-\frac{1}{4}\left\{ \frac{{{x}^{3}}}{3} \right\}_{0}^{4} \right|\] \[=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}\] square units