Volume of rotation about x-axis with part of the area under the x-axis

First find $x \in (-1,3)$ such that 
\[ |x^2-3x|=|3-x|.\]
Then 
\[ |x^2-3x|=3x-x^2=3-x=|3-x|    \Rightarrow  x^2-4x+3=0.\]
\[ \Rightarrow  (x-1)(x-3)=0   \Rightarrow x=1, x=3.\]
So $x=1$ is the only solution in $(-1,3)$. Note that for $x\in (1,3)$ we have $|x^2-3x|>|3-x|$ and for $x\in (0,1)$ we have $|x^2-3x|<|3-x|$. When rotating about $x$-axis, the volume will be created by the curve that has the larger distance from the $x-$axis, i.e. by $y=x^2-3x$ on $(1,3)$ and by $y=3-x$ on $(0,1)$. So we need to break up this into three integrals:
\[\text{Volume}=\int_0^1 \pi (3-x)^2dx+\int_1^3 \pi (3x-x^2)^2dx+\int_{-1}^{0}\pi [(3-x)^2-(x^2-3x)^2]dx.\]
Note that on $(0,1)$ and $(1,3)$ we should use the disk method, but on $(-1,0)$ we use the washer method.