Question

The region under the curves `y=sinx/x, pi/2<=x<=pi` is rotated about the x axis. How do you find the volumes of the two solids of revolution?

Answer 1

0.6372

Explanation:

Not sure where 2 solids come into it.

This is what you are spinning round the x-axis:

A slice of the revolved solid will have cross section area:

`A = pi y^2 = pi ((sin x)/(x))^2`

A disc of width `dx` will have volume:

`dV = pi ((sin x)/(x))^2 \ dx`

So the volume of revolution is:

`V = pi int_(pi/2)^(pi) ((sin x)/(x))^2 \ dx`

This integration requires non elementary functions. A computer solution is: `V = 0.6372`.

Reality check: you can use a fairly decent straight line approximation: `y = -(2/pi ) (x/(pi /2) - 2)` and the fomula for the volume of a cone to get:

`V = pi (1/(pi/2))^2 (pi/2)/3 = 2/3`