Consider this sector S what's area is the following: $$ A = \frac{1}{2}r^2\theta $$ Where theta is in radians. I would like to create a solid out of this sector by rotating it about $\phi$ with the following: $$ V =\int_{0}^{2\pi} \Big[\frac{1}{2}r^2\theta\Big]\phi \:\:d\phi $$ which evaluates to: $$ V =\frac{r^2\theta \phi^2}{4} $$ and then let $\phi = 2\pi$ (upper bound of the integral): $$ V =\frac{r^2 \pi^2\theta}{2} $$ Is this correct?
Edit: Add picture, red is S, light green is surface, dark green is inside cross section. This is a bad picture, the dotted lines should be straight not curved. Image of the solid
The first thing to notice about your solution is that it is dimensionally incorrect. Volume should be proportional to $r^3$.
A solution can had very quickly with Pappus's (2nd) Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. The bottom line is that the volume is given simply by $V=2πRA$.
Without any further ado, the properties of the sector are
$$ R=\frac{2r\sin\alpha}{3\alpha}\\ A=\alpha r^2\\ V=2\pi RA=\frac{4}{3}\pi r^3 \sin\alpha $$
where $\alpha$ is the half-angle of the sector.