question_answer

Area included between the two curves \[{{y}^{2}}=4ax\] and \[{{x}^{2}}=4ay,\] is        [SCRA 1986; Roorkee 1984; RPET 1999; Kerala (Engg.) 2002, 05]

A)            \[\frac{32}{3}\,{{a}^{2}}\] sq. unit                                  

B)            \[\frac{16}{3}\] sq. unit

C)            \[\frac{32}{3}\] sq. unit   

D)            \[\frac{16}{3}\,{{a}^{2}}\] sq. unit

Correct Answer: D

Solution :

           Solving the two equations, we have\[{{x}^{4}}=64{{a}^{3}}x\]                    \[\Rightarrow x=0,\,\,4a\]        Required area =\[\int_{0}^{4a}{2{{a}^{1/2}}{{x}^{1/2}}dx-\int_{0}^{4a}{\frac{{{x}^{2}}}{4a}dx}}\]                                                                         \[=\frac{32}{3}{{a}^{2}}-\frac{16}{3}{{a}^{2}}=\frac{16}{3}{{a}^{2}}\]sq. unit.