To do this problem, we're going to set up an integral for the area of the surface generated by revolving the curve XY equal 1/5 less than or equal to Y less than or equal to 8 about the Y axis. So if it's about the Y axis, we know we're going to use DY, so I need to get X to be in terms of DY. So X = 1 / y. If we think about rotating around the Y axis, our radius each time is going to be the distance of X, wherever that happens to fall. So we're going to get 2π. The radius is the X, which is really 1 / y in terms of Y, the square root 1 plus our DXDY is -1 / y ^2, and we're going to square it DY. Our bounds were given from 5:00 to 8:00. So we're going to get 2π five to eight of 1 / y square root 1 + 1 / y to the 4th dy. When we put that into our calculator, we're going to get 2.95. We do not know how to evaluate that currently.