The region bounded by y= e^x, y=0, x= -1 and x=1 is rotated about the x-axis, find the volume generated.

To find the volume generated by rotating a region around the x-axis, we can use the method of cylindrical shells.

The formula for the volume of a cylindrical shell is V = 2πrhδx, where r is the distance from the axis of rotation to the shell, h is the height of the shell, and δx is the thickness of the shell.

In this case, the region bounded by y = e^x, y = 0, x = -1, and x = 1 is being rotated about the x-axis.

To find the volume, we need to integrate the volume of all the cylindrical shells that make up the region.

First, let's consider a small strip of width δx at a specific x-coordinate within the region. This strip will have a height y = e^x. The distance from the x-axis to this strip is simply y = 0. Therefore, the radius of the cylindrical shell is r = e^x.

Now, we need to find the height of the shell, h. Since we are rotating around the x-axis, the height will be the difference in y-coordinates between the upper and lower boundaries. In this case, the upper boundary is y = e^x and the lower boundary is y = 0. So, the height of the shell is h = e^x - 0 = e^x.

The thickness of the shell, δx, represents the width of the strip. As δx approaches 0, the number of cylindrical shells approaches infinity and the sum of their volumes gives the total volume.

Now, we can set up the integral to find the volume:

V = ∫(V) dx
= ∫(2πrhδx) dx
= 2π ∫(e^x)(e^x) dx
= 2π ∫(e^(2x)) dx
= 2π [(1/2) e^(2x)] + C

To evaluate this integral over the interval x = -1 to x = 1, substitute the limits of integration:

V = 2π [(1/2) e^(2(1))] - 2π [(1/2) e^(2(-1))]
= π [e^2 - (1/e^2)]

Therefore, the volume generated by rotating the region around the x-axis is π [e^2 - (1/e^2)].