Question

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.

Answer 1

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.