volume of solid generated by the regin bounded by curve $y=\sqrt{x},y=\frac{x-3}{2},y=0$ about $x$ axis 
Using sell method to find the volume of solid generated by revolving the region bounded by $$y=\sqrt{x},y=\frac{x-3}{2},y=0$$ about $x$ axis, is (using shell method)

What I try:

Solving two given curves $$\sqrt{x}=\frac{x-3}{2}\Longrightarrow x^2-10x+9=0$$
We have $x=1$ (Invalid) and $x=9$ (Valid).
Put $x=9$ in $y=\sqrt{x}$ we have $y=3$
Now Volume of solid form by rotation about $x$ axis is 
$$=\int^{9}_{0}2\pi y\bigg(y^2-2y-3\bigg)dy$$
Is my Volume Integral is right? If not then how do I solve it? Help me please.

So I would instead split this up into two integrals:
$$\pi\int_0^3{(\sqrt{x})^2}dx + \pi\int_3^9{(\sqrt{x})^2-\left(\frac{x-3}{2}\right)^2}dx$$.
Using the shell method:
$$\int_0^3{2\pi y(2y+3-y^2)}dy$$.
