Exactly. You have a little mass below you, and far more above you. But the mass above you is also in far more distance.
The result is the - for the first spot, surprising - fact, that these two exactly balance eachother.
Thus, if somehow you don't feel the gravity of the Sun, then there is a weightlessness in the whole sphere.
Actually, you can imagine that so, that consider looking into any direction and a small $d\omega$ solid angle. In that direction, the wall of the sphere is in $r_1$ distance. In the exactly opposite direction (with the same solid angle), you see the wall in $r_2$ distance.
It doesn't matter, what is $r_1$ and $r_2$, because the wall size you see on the depends quadratically on them, thus the wall mass depends also qudratically. But their gravity to you is proportional to the quadratical recipe of the distances.
The result is that these opposite areas off the wall have exactly the same force on you, but into the opposite direction.
On the language of the formulas:
- You see $d\omega$ solid angle in both direction
- The distance of the wall is $r_1$ and $r_2$
- Let the "surface density" (i.e. the mass of a single $m^2$ wall) $\rho$
- Then you see $d\omega\cdot\rho\cdot r_1^2$ and $d\omega\rho\cdot r_2^2$ masses in the both directions,
- Their gravititational acceleration ($G\frac{m}{r^2}$) are thus $G\frac{d\omega\cdot\rho\cdot r_1^2}{r_1^2} = Gd\omega\rho$. It doesn't depend on $r$!
