In a question I read Quarks in a hadron: where does the mass come from?
"The sum of the masses of the quarks in a proton is approximately $30~\text{MeV}/c^2$, whereas the mass of a proton is $931~\text{MeV}/c^2$. "
Is it correct, since a proton has zillions of partons.
Mass contributed by valence quarks is ~8-9MeV. Is OP correct about 30MeV.
What you are looking for is called the parton distribution function for the proton. It roughly describes the energy/momentum shared by the partons (quarks and gluons) inside a proton (or any other hadron, of course). It is deduced from experiments according to theoretical guidelines, and must be defined in reference to the energies involved in the experiment. You'll find more info on the web. Below, you can see the parton distribution functions for u, d, s, c quarks, their antiquarks and gluons, @ 4 GeV (the image is taken from Peskin and Schroeder's An Introduction to Quantum Field Theory, sec. 17.4, fig. 17.6). Beware of the plotted function ( $xf(x)$ instead of $f(x)$ ).
P.S.: It is not correct, protons do not contain zillions of mesons. Nuclei may contain them, but not hadrons.
The quark mass contribution to the mass of the proton can be determined in lattice QCD. The result is indeed about 30 MeV. More accurately, $\Delta M_p=37\pm 8\pm 6$ MeV, see Nucleon mass and sigma term from lattice QCD with two light fermion flavors.
This quantity is sometimes called the nucleon sigma term, and chiral symmetry implies that it is related to pion-nucleon scattering. This leads to independent determinations of $\Delta M_p$ (which, historically, precede the lattice results). These days, the results are consistent with lattice QCD.
Note that $$ \Delta M_p = \langle p|\sum_f m_f \bar\psi_f\psi_f |p\rangle $$ where $f$ labels quark flavor ($f$=up,down,..) and $m_f$ are quark masses. In a non-relativistic quark model we would assume that $\langle p|\bar{u}u|p\rangle=2$ and $\langle p|\bar{d}d|p\rangle=1$, and that $\Delta M_p=2m_u+m_d$.
This estimate is not right, for several reasons. Neither $m_f$ nor $\bar\psi_f\psi$ are renormalization group invariants, only the product is. Also, computed at a "reasonable" renormalization scale, the sum $2m_u+m_d$ is smaller than $\Delta M_p$. Roughly, there are more than three quarks in a proton (there are several quark-anti-quark pairs).
The number of quark-anti-quark pairs, or equivalently the number of mesons, is not sharply defined. Only $\Delta M_p$ is. Taking some phenomenologically reasonable cutoff to estimate the number gives values of a few, not "zillions".
The quark masses are not uniquely defined, they can be given in an $\bar{MS}(2GeV)$ scheme as $m_u \sim 2.2 MeV$, $m_d \sim 4.8 MeV$. The rest mass of a nucleon comes from dominantly the strong interaction that binds the whole nucleon together. (Only a tiny fraction of mass comes from the electromagnetic interactions and through the Higgs mechanism.) The precise determination of nucleon masses can be achieved by ab initio lattice calculations.
The net mass-energy of bare quarks is a miniscule fraction of the bound system. A key fact to be noticed is that the force due to strong interaction does not decrease with increasing distance as in the case of electromagnetism. Another fact is that Heisenberg uncertainity is there.
Now we have a bound state of three quarks. For an easy comparision with electromagnetism, lets take a bound state of two quarks at first and later on we can move onto the case of three quarks.
So firstly, if two unlike charges are bounded by electromagnetism, then the coupling between them is less and falls down as inverse square with distance. As a result, for a system bounded by electromagnetism like the hydrogen atom, the distance is of the order of $10^{-10} m$ and there is no appreciable correction to the net mass of the the system as compared to the proton. {This depends on the magnitude of the charge too and is only a representative example}
Now if we take 2 quarks, then since the attractive force increases with distance, they form a bound system having a very small radius (Of the order of 1 fermi). Now Heisenberg uncertainity says that
$$\Delta x \Delta p \approx \hbar$$
Putting the values you can see that the quarks have a momentum uncertainity in the relativistic regime and have speeds close to light. This gives rise to extra mass of the quarks inside the bound state and is called the constituent quark mass.
Now finally we put 3 quarks in the system and get the proton. We can easily see that in order to hold these quarks together, we require that the potential energy should be greater than their kinetic energy. The potential energy in this case is provided by the gluon field, which gives rise to the remaining extra mass of the proton. Thus the proton is more heavy than the sum of individual quark masses.
