1. ## Volume by revolution

$y=x$
$y=0$
$x=5$
$x=6$

Rotate the area bounded by the given curves around $x=1$

The easy numbers are throwing me off I think. My outside radius would just be 6-1 and my inner radius would just be 5 -1

Correct?

I have my integral set up as $\int_0^6 (5^2-4^2)$

Did i do this problem correctly? Thanks

2. method of cylindrical shells.

rotation radius ... $(x-1)$

shell height ... $x$

limits ... $x = 5$ to $x = 6$

$V = 2\pi \int_5^6 (x-1)x \, dx$