$\displaystyle y=6-2x-x^{2}$
$\displaystyle y=x+6$
Rotate around the x-axis.
First find the points of intersection.
They intersect when $\displaystyle x+6=6-2x-x^2\implies x^2+3x=0\implies x(x+3)=0\implies x=0,-3$.
Thus, by the washers, you integral will take on the form of $\displaystyle \pi\int_a^b\left(\left[f\!\left(x\right)\right]^2-\left[g\!\left(x\right)\right]^2\right)\,dx$
In this case $\displaystyle f\!\left(x\right)=6-2x-x^2$ and $\displaystyle g\!\left(x\right)=x+6$
Thus, the volume of the solid about the x axis is $\displaystyle \pi\int_{-3}^0\left(6-2x-x^2\right)^2-\left(x+6\right)^2\,dx=\pi\int_{-3}^0 x^4+4x^3-9x^2-36x\,dx$
Can you take it from here?