Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 4 + sec x, -pi/3 ≤ x≤ pi/3, y = 6;    about y = 4 V=______________?

Expert Answers

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(Note: the graphs do not intersect at -1, the way it kind of looks.  They intersect at `+- pi / 3`)

 

 

We will do this problem using the "slicing" or disk method.

Imagine you are putting a bunch of vertical cuts into the resulting solid. Each of the slices looks roughly like a washer (a cylinder with a smaller cylinder cut out of it).

The volume of a cylinder is `pi R^2 h`, so the volume of a washer is: `pi R^2 h - pi r^2 h = pi (R^2 - r^2)h`

When `-pi/3 <= x <= pi/3`, `6 >= 4 + "sec" x`

So our outer radius will be the distance from 6 to 4, so R=2

Our inner radius will be the distance from `4 + "sec" x` to 4, so `r="sec" x`

The height of our washer is `Delta x` (we made vertical slices).

As the number of slices goes to infinity, and the thickness of the slices goes to 0, we get the integral:

`int_(-pi / 3) ^(pi/3) pi (2^2-"sec" ^2 x) dx `

The solid is symmetrical about the y-axis, so we may instead calculate:

`2 int_(0) ^(pi/3) pi (2^2-"sec" ^2 x) dx `

This is:

`2 pi int_(0) ^(pi/3) (4-"sec" ^2 x) dx `

`2 pi [ 4x - "tan" x| _0 ^(pi/3) ]`

`2 pi [ 4(pi/3) - sqrt(3) ]`

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