Explanation

(c)
$The\ figure\ of\ the\ given\ curve\ y = \frac{a^{3}}{a^{2} + x^{2}}\ is$
$\therefore\$Required volume
$V = 2\int_{0}^{\infty}{\pi y^{2}\ dx}$
$= 2\pi a^{6}\int_{0}^{\infty}{\frac{1}{(a^{2} + x^{2})^{2}}\ dx}$
[image:8_image54.emf]
$Put\ x = a\tan\theta$
$\Longrightarrow \ \ dx = {a\ \sec}^{2}\theta\ d\theta$
$\therefore V = {2\pi a}^{6}\int_{0}^{\pi/2}{\frac{{a\sec}^{2}\theta}{(a^{2} + a^{2}\ \tan^{2}\theta)^{2}}d\theta}$
$= \frac{{2\pi a}^{6}}{a^{3}}\int_{0}^{\pi/2}{\cos^{2}\theta d\theta = {2\pi a}^{3}\left\lbrack \frac{1}{2} \bullet \frac{\pi}{2} \right\rbrack = \frac{\pi^{2}a^{3}}{2}\ cu\ units}$