Find the volume using disk/washer method of region bound by curves y=e^-x/2, y=ln(9)? The solid is generated when R is revolves around x-axis. The boundaries are ln(9) and 0. I just need help setting up the integral. Thank you.

1 Answer
Mar 24, 2018

See below.

Explanation:

If you look at the graph of #y=ln9# and #y=e^[-x/2# you will notice that the area bounded by #y=ln9#, from #0# to #ln9# is a square, and has an area greater than #y=e^[-x/2# from #0# to #ln9#.

So we would need to find #piint_[0]^ln9 ##[ln9]^2#dx -#piint_[0]^ln9##[e^[-x/2]]^2]#dx

Volume of solid of revolution=#piint_a^b[y^2]dx#, where #y=f[x]#. Sorry I am unable to show the graphs[ don't know how!]

Leave it to you to evaluate the integrals. Hope this was helpful.