$(1)$ \[ p\cdot dA-\left(p+\frac{\partial p}{\partial y} dy \right)dA-\rho \cdot dA\cdot dy\cdot g=0\\ \frac{\partial p}{\partial y}dy = -\rho \cdot dy \cdot g\\ \therefore \frac{\partial p}{\partial y} = - \rho g \] $(2)$ \[ p\cdot dA \cdot dr \cdot r\omega^2+p\cdot dA- \left(p+\frac{\partial p}{\partial r}dr \right)=0\\ \rho \cdot dr \cdot r\omega^2- \frac{\partial p}{\partial r}dr=0\\ \therefore \frac{\partial p}{\partial r}=\rho r \omega^2 \] $(3)$ \[ dp=\left(\frac{\partial p}{\partial r} \right)dr + \left(\frac{\partial p}{\partial y} \right)dy\\ dp=\rho r \omega^2\cdot dr+\rho g \cdot dy\\ \] 液面では$dp=0$だから，液面の微分方程式は \[ \rho g \cdot dy=\rho r \omega^2\cdot dr \] \[ \frac{\partial p}{\partial r}=\frac{r \omega^2}{g} \] 両辺を積分して \[ y=\frac{\omega^2 r^2}{2g}+C \] ここで$r=0$で$y=0$だから， \[ C=0 \] \[\therefore y=\frac{\omega^2 r^2}{2g} \]