Question

A closed figure $S$ is bounded by the hyperbola $x^2-y^2=a^2$ and the straight line $x=a+h;(h>0,a>0)$. This closed figure is rotated about the x-axis. Then the volume of the solid of revolution is :

• A. $\frac{\pi h^2}{2}(3a+h)$
• B. $\pi h^2(3a+h)$
• C. $\frac{\pi h^2}{6}(3a+h)$
• D. $\frac{\pi h^2}{3}(3a+h)$

Answer 1

The correct option is D $\frac{\pi h^2}{3}(3a+h)$


Volume of solid of revolution of the closed figure $S$ is
$V=\pi \underset{a}{\overset{a+h}{\int }}{y}^{2}dx\Rightarrow V=\pi \underset{a}{\overset{a+h}{\int }}(x^2-a^2)dx$
$\Rightarrow V=\frac{\pi h^2}{3}[3a+h]$