Question

The volume generated when the plane region bounded by $y=secx,x=-\frac{\pi }{4},x=\frac{\pi }{3}$ and $y=0$ revolves about x-axis is?

• A. $\pi (1+\sqrt{3})$ cu unit
• B. $\pi +\sqrt{3}$ cu unit
• C. $2\pi (1+\sqrt{5})$ cu unit
• D. $\frac{\pi +\sqrt{3}}{2}$ cu unit

Answer 1

The correct option is D $\frac{\pi +\sqrt{3}}{2}$ cu unit

$\because$ The required volume is

$V=\pi {\int }_{\frac{-\pi }{4}}^{\frac{\pi }{3}}{y}^{2}dx$

$\therefore V=\pi {\int }_{\frac{-\pi }{4}}^{\frac{\pi }{3}}{\sec }^{2}xdx$

$\therefore V=\pi .[\tan x{]}_{\frac{-\pi }{4}}^{\frac{\pi }{3}}$

$\therefore V=\pi .(\tan \frac{\pi }{3}+\tan \frac{\pi }{4})$

$V=\pi .(\sqrt{3}+1)$ cu unit.