A) \[\pi {{a}^{2}}sq\,\,unit\]
B) \[2\pi {{a}^{2}}sq\,\,unit\]
C) \[3\pi {{a}^{2}}sq\,\,unit\]
D) None of these
Correct Answer: A
Solution :
We have, the equation of the curve \[y=\frac{{{a}^{3}}}{{{x}^{2}}+{{a}^{2}}}=f(x)\] (say) Here, \[f(-x)=\frac{{{a}^{3}}}{{{(-x)}^{2}}+{{a}^{2}}}=\frac{{{a}^{3}}}{{{x}^{2}}+{{a}^{2}}}=f(x)\] \[\therefore \]The function is symmetrical about y-axis. Now, the required area \[=\int_{-\infty }^{\infty }{y}\,\,dx=2\int_{0}^{\infty }{y\,\,dy}\] \[=2\int_{0}^{\infty }{\frac{{{a}^{3}}}{{{x}^{2}}+{{a}^{2}}}dx}\] \[=2{{a}^{3}}\left[ \frac{1}{a}{{\tan }^{-1}}\frac{x}{a} \right]_{0}^{\infty }\] \[=2{{a}^{2}}({{\tan }^{-1}}\infty -{{\tan }^{-1}}0)\] \[=2{{a}^{2}}\cdot \left( \frac{\pi }{2}-0 \right)\] \[=\pi {{a}^{2}}sq\,\,unit\]You need to login to perform this action.
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