For the sketch,

-----x +y = 3 is a straight line passing through (0,3) and (3,0).

-----x = 4 -(y-1)^2 is a horizontal parabola that opens to the left whose vertex is at (4,1), and whose y-intercepts are (0,3) and (0,-1), and whose x-intercept is (3,0).

If by shell method, rotating about the x-axis,

dV = [2pi(y)(x2 -x1)](dy)-----where x2 is from the parabola and x1 is from the line.

dV = 2pi(y)[(4 -(y-1)^2) -(3-y)]dy

dV = 2pi(y)[(4 -(y^2 -2y +1) -3 +y]dy

dV = 2pi(y)[4 -y^2 +2y -1 -3 +y]dy

dV = 2pi(y)[-y^2 +3y]dy

dV = 2pi[3y^2 -y^3]dy ---------------**

The boundaries of dy are from y=0 up to y=3.

So,

V = (2pi)INT.(0-->3)[3y^2 -y^3]dy

V = (2pi)[y^3 -(1/4)y^4] |(0-->3)

V = (2pi){[(3^3) -(1/4)(3^4)] -[0]}

V = (2pi){(27) -(81/4)]

V = (2pi)(27)[1 -3/4]

V = (54pi)[1/4]

V = 13.5pi cu.units ----------------answer.