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

Answer 1

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

Given
$x^2-y^2=a^2...\left(i\right)$
$x=a-h,(h>0,a>0)$
$\because$ The figure is bounded by $x=a,x=a+h,y=0$
$\therefore$ Volume of the solid revolution
$V=\pi {\int }_a^{a+h}{y}^{2}dx$
$=\pi {\int }_a^{a+h}{(x}^2-a^2)dx=\pi {[\frac{x^3}{3}-a^{2}x]}_a^a$
$=\pi [(\frac{{(a+h)}^3}{3}-\frac{a^3}{3})-\left(a^2\right(a+h)-a^3)]=\frac{\pi h^2}{3}(h+3a)$