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Consider fluid in a cylindrical vessel of radius a which is undergoing solid-body rotation so that its velocity field is given by $v = (−\sigma y, \sigma x, 0)$. (The z-axis is along the axis of the vessel and vertical.)

(a) Show that to within an additive constant, the pressure distribution is given by $p=\frac{1}{2} \rho \sigma^{2} (x^{2}+y^{2})-\rho g z$

Hence find the shape of the free surface of the fluid (i.e. the interface between the rotating fluid and the atmosphere.) And show that the fluid at the center of the vessel is lower than that at the edges by $\frac{\sigma^{2}a^{2}}{2g}$.

(b) Estimate the magnitude of this difference in heights for a coffee cup of radius 5cm which is stirred rapidly at a rate of 2 revolutions per second.

(c) If the height of the vessel is equal to its radius a. And if it is filled up to a level 3a/4 before it is set to rotate. What is the maximum angular velocity for which the liquid stays in the vessel? What is the minimum level of the water at this maximum angular velocity?

for the(a) question I noticed that you could draw the vessel with the axes in the middle. You can write down $a^{2}=x^{2}+y^{2}$. But now I don't see how you can becom the solution of the pressure. I thought easy to use $p=\rho g h$ but then I'm stuck with how to get a $\sigma$ in the solution.

for the (b) question I can't really describe it.

For the (c) question I thought i needed to calculate the volume of the rotating fluid. The maximum height that the fluid reaches on the edge must be at most equal to the height of the vessel. But because I'm stuck with (a) I can't use the result here.

questmath
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I can get you started with part (a) and we can take it from there if needed. This problem lends itself naturally to cylindrical coordinates such that "radial" direction is perpendicular to the axis of symmetry, and the $z$-axis does its usual job.

Now let me look at a small piece of fluid somewhere in the vessel. It feels a force from two effects: (i) the fluid rotation, and (ii) gravity. The effect of rotation acts purely radially, and naturally, gravity pulls along $-z$. This lets me write Newton's second law separately for each component: $$m a_r = m \frac{v^2}{r} \hspace{2.54cm} ma_z = -mg$$

In the above, we have $m$ as the (differential) mass of the small piece of fluid, $v$ being its instantaneous velocity, and $a_r$, $a_z$ being the respective acceleration components. In the left equation, note that we may replace $m$ by its equivalent expression $$m = \rho \cdot dA \cdot dr \:,$$ where $dA \cdot dr$ is the differential volume that $m$ occupies. We can also replace $v^2$ by its equivalent expression, $\sigma^2 \left(x^2+y^2\right)$ to write $$ma_r = \rho \cdot dA \cdot dr \cdot \sigma^2 \frac{x^2+y^2}{\sqrt{x^2+y^2}}$$ Divide through by $dA$, and we have $$\frac{ma_r}{dA} = \rho \cdot dr \cdot \sigma^2 \frac{x^2+y^2}{\sqrt{x^2+y^2}} \:.$$ The form on the left is, by definition, the differential "radial" contribution to the pressure (just force over area). On the right, we can substitute $x^2 + y^2 = r^2$ to simplify things greatly: $$dp_{rad} = \rho \cdot \sigma^2 r \: dr$$ Integrate both sides to get $$p_{rad} = \int dp_{rad} = \int \rho \cdot \sigma^2 r \: dr = \frac{1}{2}\rho \sigma^2 r^2 + p_0 \:,$$ where $p_0$ is an arbitrary constant (of the proper units). Restoring the $xy$-variables, we are halfway done: $$p_{rad} = \frac{1}{2}\rho \sigma^2 \left(x^2 + y^2\right) + p_0$$

That takes care of the rotational part, now for the gravitational. Start with the same setup: take a very very small mass $m$ occupying a differential volume $dA \cdot dz$ such that $$m a_z = -mg = - \rho \cdot dA \cdot dz \cdot g \:.$$ Divide through by $dA$ again to write $$dp_{z} = \frac{m a_z}{dA} = - \rho g \: dz \:.$$ Once again, the form on the left is just force over area, a differential contribution to pressure $p_z$. Integrate both sides, and we have $$p_z = \int dp_z = -\rho g \int dz = -\rho g z + \tilde{p}_0 \:,$$ where $\tilde{p}_0$ is another integration constant.

Finally, sum the two pressure values to crank out the answer: $$p = p_{rad} + p_z = \frac{1}{2}\rho \sigma^2 \left(x^2 + y^2\right) - \rho g z + p_0 + \tilde{p}_0$$ Of course, $p_0 + \tilde{p}_0$ is just a constant too, so lets leave it at: $$p = \frac{1}{2}\rho \sigma^2 \left(x^2 + y^2\right) - \rho g z + C$$

... Hope that helps things along. Also - we "should" replace all masses $m$ with the differential notation $dm$, but hopefully the context makes everything clear enough to not litter the page with more $d$'s.