# What is the volume of the solid produced by revolving f(x)=sec, x in [pi/8,pi/3] around the x-axis?

Volume $V = 4.140107808 \text{ }$cubic units

#### Explanation:

Given equation $y = \sec x$ and $x$ is from $\frac{\pi}{8}$ to $\frac{\pi}{3}$.

Differential volume $\mathrm{dv} = \pi \cdot {r}^{2} \cdot \mathrm{dh}$ for circular solid disk

In terms of our variable

$\mathrm{dv} = \pi \cdot f {\left(x\right)}^{2} \cdot \mathrm{dx}$

Volume $V$

$V = \int \mathrm{dv} = {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} \pi \cdot {\sec}^{2} x \cdot \mathrm{dx}$

$V = \pi \cdot {\left[\tan x\right]}_{\frac{\pi}{8}}^{\frac{\pi}{3}}$

V=pi*(tan pi/8-tan pi/3)#

$V = 4.140107808 \text{ }$cubic units

God bless....I hope the explanation is useful.