"For shell method the equation is 2 pi times the integral from 1 to 3 of x=1+ (y-2)^2"

No, it is not. You are rotating around y= 2, not the x-axis. And it makes no sense to integrate a function of y from x= 1 to x= 3. Nor does it make sense to integrate from y= 1 when you are rotating around y= 2. You must mean to integrate from y= 2 to y= 3.

You don't give a right boundary but x= 2 is implied. For that I get neither 16/3 nor (32/3)pi. Doing it by both by shells and by disks, I get a volume of pi/4.