Calculus: Early Transcendentals, Chapter 6, 6.3, Section 6.3, Problem 17

y=4x-x^2, y=3
The point of intersection of the curves will be ,
4x-x^2=3
4x-x^2-3=0
-x^2+4x-3=0
-(x^2-4x+3)=0
x^2-4x+3=0
factorizing the above equation,
(x-3)(x-1)=0
x=3 , x=1
The shell has radius (x-1) , circumference 2pi(x-1) and height (4x-x^2)-3
Volume generated by rotating the region bounded by the given curves about x=1 (V)=int_1^3(2pi)(x-1)(4x-x^2-3)dx
V=int_1^3(2pi)(4x^2-x^3-3x-4x+x^2+3)dx
V=(2pi)int_1^3(-x^3+5x^2-7x+3)dx
V=2pi[-x^4/4+5x^3/3-7x^2/2+3x]_1^3
V=2pi[-3^4/4+5/3*3^3-7/2*3^2+3*3]-2pi[-1^4/4+5/3*1^3-7/2*1^2+3*1]
V=2pi((-81/4+45-63/2+9)-(-1/4+5/3-7/2+3))
V=2pi(9/4-11/12)
V=2pi(16/12)
V=(8pi)/3