I am trying to solve this exercise from a textbook:
$y = x^4, y = 0, x = 1;$ rotated about $x=2$
This is my attempt at solving the problem:
Shell radius: $2 - x$
Shell height: $x^4$
$a = 1$
$b = 2$
$\int_1^2 2π(2-x)x^4dx$
$= 2π(2x^5/5 - x^6/6)_1^2$
$= 2π(2*2^5/5 - 2^6/6 - 2/5 + 1/6)_1^2$
$= 2π(19/10)$
$= 19π/5$
However, the answer is supposed to be 7π/15 according to the solutions manual. What am I doing wrong?