The region under the curves #y=sinx/x, pi/2

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?

Charles Anctil · Accepted Answer

0.6372

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#

Emma Aldridge · Answer

undefined To find the volumes of the two solids of revolution formed by rotating the region under the curve $ y = \frac{\sin(x)}{x} $ from $ x = \frac{\pi}{2} $ to $ x = \pi $ about the x-axis:

1. **Setup**: Determine the axis of rotation and the limits of integration.
2. **Volume of Solid 1**: For the region above the x-axis, integrate $ \pi y^2 $ with respect to $ x $ from $ \frac{\pi}{2} $ to $ \pi $.
3. **Volume of Solid 2**: For the region below the x-axis, integrate $ -\pi y^2 $ with respect to $ x $ from $ \frac{\pi}{2} $ to $ \pi $.
4. **Combine Volumes**: Add the volumes of Solid 1 and Solid 2 to get the total volume of revolution.

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

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?