For example, the answer to this question:
"Let $W$ be the solid cone bounded by $z=\sqrt{x^2 + y^2}$ and $ z=2$. Decide (without calculating) whether $\int_W x \ dV$ is positive, negative, or zero."
Is zero, because the integrand, $x$, is symmetrical for the region (if we imagine taking slices perpendicular to the $xy$-axis, then each slice is a circle with the top half above the $x$-axis and the bottom half below it).
However, if we modified the question to say:
"Let $W$ be the solid cone bounded by $z=\sqrt{x^2 + y^2}, z=2,$ and the $yz$-plane with $x \geq 0$. Decide (without calculating) whether $\int_W x \ dV$ is positive, negative, or zero."
Then the integral is positive, because we've restricted the region to only consist of points above the $x$-axis ("above" if we look at the $xy$-plane head on).
So is that all there is to it? Does just knowing the sign of the integrand over the region tell us with certainty the sign of the entire integral?
Your examples are correct and your argument is valid for double and triple integrals.
For integrals over an interval such that $$\int _a^b f(x)dx,$$ it is also important to know whether $a<b$ or $b<a$
As you know if $a<b$ and $f(x)\ge 0$ then $\int _a^b f(x)dx \ge 0$ and $\int _b^a f(x)dx \le 0$
So there may be more factors than just the sign of integrand and the symmetry of the region.