Question

Find the volume of the solid that is generated when the given region is revolved as described. The region bounded by ​f(x)equalse Superscript negative x​, xequalsln 16​, and the coordinate axes is revolved around the​ y-axis.

Answer 1

The answer is "
`= (\pi)/(8)(15-41n^2 )\\`".

Explanation:

The radius = x

the value of height is=
`e^(-x)`

The Formula for the volume by the shell method:

`\bold{V= \int\limits^b_a {(2\pi\ rad)(height)} \, dx }`

`= 2\pi \int\limits^(In 16)_0 {xe^(-x)} \, dx\\\\\\= 2\pi {(e^(-x)[-x-1])}_(0)^(In 16)\\\\ = 2\pi {((1)/(16) * (15-41n^2 ))}\\\\ = (\pi)/(8)(15-41n^2 )\\`