Compute the volume and surface area of the solid $S$ obtained by revolving the region $R$ (pictured below) enclosed by the graphs $y=q(x)=-(2x^2-7x+3),x=1,x=2$ and $y=-1$, around the $x$-axis, which crosses the region.
Volume:
I can just consider the region $T$ enclosed by the graphs $y=q(x),x=1,x=2$ and the $x$-axis (instead of $y=-1$) because the overlap (Volume of revolution on an area crossing the axis), seen precisely in the inequality $|q| > |-1|$ for $x = 1$ to $2$, means that the solid $P$ from the region $T$ is identical to $S$. That is $\text{Vol}(P) = \text{Vol}(S)$ because $P=S$?
Update: Wolfram (Solid P and Solid S) says $\pi + \frac{107 \pi}{15} = \text{Vol}(P) \ne \text{Vol}(S) = \frac{107 \pi}{15}$.
I notice $\pi$ is the volume of the solid $M$ obtained by rotating the square $N$ enclosed by $y=0$ and $y=-1$ from $x=1$ to $2$.
Could it be that Wolfram interprets the calculation for the volume of Solid S as $'\text{Vol}(S)' = \text{Vol}(P) + \text{Vol}(M)$ ?
Perhaps not because Wolfram looks like it's computing $$\int_1^2 \pi |\color{red}{-1} + (-q)^2| dx = \int_1^2 \pi |-(-1)^2 + (q)^2| dx$$ So, the '$\color{red}{-1}$' is actually '$-(-1)^2$' rather than the original '$y=-1$'? Hopefully, Wolfram assumes the axis doesn't cut the interior of the region.
Surface area:
For surface area, do we have $\text{SA}(P) = \text{SA}(S)$ because $P=S$ also?
Update: Wolfram (Solid P and Solid S) says $\text{SA}(P) = \text{SA}(S)$, but actually I think the answer should be $\text{the SA}(P)\text{ that wolfram gives} - 13 \pi$. Where does this extra $13 \pi$ come from please?