Q1: R is the region between $f(x) = x^{2}$ and $g(x) = x + 2$. Find the volume $V_1$ of the region generated by revolving about the line $y = - 3$
Q2: R is the region between $f(x) = 3x$, and $g(x) = 3x$, Find the volume $V_2$ of the region generated by revolving about the line $y = - 2$
$$ \begin{split} V_1 &= \pi\int_{-1}^2 \left( (x^2 + 3)^2 - (x + 2 + 3)^2 \right) dx\\ V_2 &= π\int_0^3 \left( (x^2 + 2)^2 - (3x + 2)^2 \right) dx \end{split} $$
Thank you!
For $V_1$, you have the inner and outer radii reversed: on the interval $[-1,2]$, $g(x) \ge f(x)$, consequently your evaluation of $V_1$ will result in a negative number.
For $V_2$, I cannot verify your integral, because you have stated $f(x) = g(x) = 3x$, meaning there is no region enclosed by these two functions. However, if you mean that $f(x) = x^2$ and $g(x) = 3x$ as your integrand implies, then again you have the same problem as $V_1$: your inner and outer radii are reversed, since $3x \ge x^2$ on $[0,3]$.