Question

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.? y = 2 + sec(x), −π/3 ≤ x ≤ π/3, y = 4; about y = 2

Answer 1

Answer:

The volume of the solid is: $\mathbf{V = \pi [ \dfrac{8 \pi}{3} - 2\sqrt{3}]}$

Step-by-step explanation:

GIven that :

$y = 2 + sec \ x , -\dfrac{\pi}{3} \leq x \leq \dfrac{\pi}{3} \\ \\ y = 4\\ \\ about \ y \ = 2$

This implies that the distance between the x-axis and the axis of the rotation = 2 units

The distance between the x-axis and the inner ring is r = (2+sec x) -2

Let R be the outer radius and r be the inner radius

By integration; the volume of the of the solid  can be calculated as follows:

$V = \pi \int\limits^{\dfrac{\pi}{3}}_{\dfrac{-\pi}{3}} [(4-2)^2 - (2+ sec \ x -2)^2]dx \\ \\ \\ V = \pi \int\limits^{\dfrac{\pi}{3}}_{\dfrac{-\pi}{3}} [(2)^2 - (sec \ x )^2]dx \\ \\ \\ V = \pi \int\limits^{\dfrac{\pi}{3}}_{\dfrac{-\pi}{3}} [4 - sec^2 \ x ]dx$

$V = \pi [4x - tan \ x]^{\dfrac{\pi}{3}}_{\dfrac{-\pi}{3}} \\ \\ \\ V = \pi [4(\dfrac{\pi}{3}) - tan (\dfrac{\pi}{3}) - 4(-\dfrac{\pi}{3})+ tan (-\dfrac{\pi}{3})] \\ \\ \\ V = \pi [4(\dfrac{\pi}{3}) - tan (\dfrac{\pi}{3}) + 4(\dfrac{\pi}{3})- tan (\dfrac{\pi}{3})] \\ \\ \\ V = \pi [8(\dfrac{\pi}{3}) - 2 \ tan (\dfrac{\pi}{3}) ]$

$\mathbf{V = \pi [ \dfrac{8 \pi}{3} - 2\sqrt{3}]}$