question_answer

    The volume of the solid generated by the revolution of the curve\[y=\frac{{{a}^{3}}}{{{a}^{2}}+{{x}^{2}}}\]about\[x-\]axis is

A)  \[\frac{1}{2}{{\pi }^{3}}{{a}^{2}}\]                           

B)  \[{{\pi }^{3}}{{a}^{2}}\]

C)  \[\frac{1}{2}{{\pi }^{2}}{{a}^{3}}\]                           

D)  \[{{\pi }^{2}}{{a}^{3}}\]

Correct Answer: C

Solution :

                Key Idea: Volume\[=2\int_{0}^{\infty }{\pi {{y}^{2}}}dx\] The figure of the given curve \[y=\frac{{{a}^{3}}}{{{a}^{2}}+{{x}^{2}}}\]is \[\therefore \]  Required volume                 \[=2\int_{0}^{\infty }{x\,{{y}^{2}}}dx\]                 \[=2\pi {{a}^{6}}\int_{0}^{\infty }{\frac{1}{{{({{a}^{2}}+{{x}^{2}})}^{2}}}}dx\]                 \[=\frac{{{\pi }^{2}}{{a}^{3}}}{2}\]cu unit