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 :

Answer:

The answer is "[tex]= \frac{\pi}{8}(15-41n^2 )\\[/tex]".

Step-by-step explanation:

The radius = x

the value of height is= [tex]e^{-x}[/tex]

The Formula for  the volume by the shell method:

[tex]\bold{V= \int\limits^b_a {(2\pi\ rad)(height)} \, dx }[/tex]

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

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