If you are rotating R1 about x= 1 use the shell method
Height is 4*x^5 and the distance from x to 1 is 1-x
you have (2pi)integral[(1-x)4x^5dx]
1) Rotate the graph 4x^5 about the line x = 1 and calculate the volume of the generated figure.
I assumed that the radius and then integrated from 0 to 1 and was going to double it
Which gives you
2) Rotated about y = 5. Find the volume
But that evaluates to 0, which doesn't make sense (and is also wrong).
Where did I go wrong on these?