question_answer

The area bounded by the parabolas\[{{y}^{2}}=4ax\]and\[{{x}^{2}}=4ay\]is:

A)  \[\frac{8{{a}^{3}}}{3}sq\ unit\] 

B)                         \[\frac{16{{a}^{2}}}{3}sq\ unit\]

C)         \[\frac{32{{a}^{2}}}{3}sq\ unit\]              

D)                         \[\frac{64{{a}^{2}}}{3}sq\ unit\]

E)         \[\frac{128{{a}^{2}}}{3}sq\ unit\]

Correct Answer: B

Solution :

Required area \[=\int_{0}^{4a}{2\sqrt{a}\sqrt{x}}dx-\int_{0}^{4a}{\frac{{{x}^{2}}}{4a}}dx\] \[=2\sqrt{a}.\frac{2}{3}[{{x}^{3/2}}]_{0}^{4a}-\frac{1}{4a}\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{4a}\] \[=\frac{4\sqrt{a}}{3}8.{{a}^{3/2}}-\frac{1}{12a}64{{a}^{3}}\] \[=\frac{32}{3}{{a}^{2}}-\frac{16}{3}{{a}^{2}}\] \[=\frac{16}{3}{{a}^{2}}sq\,unit\]