Let \(z(y)\) be the curve in the \(yz\)-plane. The direction of the tangent, as given by the unit vector \(\hat t\text{,}\) will be
\begin{equation*}
\hat t = b \left( \hat j + \left( \frac{dz}{dy} \right)\hat k \right),
\end{equation*}
where
\begin{equation*}
b = 1/\sqrt{1 + (dz/dy)^2}.
\end{equation*}
From this we get the direction of the force by pressure, to be denited by \(\hat u\text{,}\) will be
\begin{equation*}
\hat u = b\left( - \left( \frac{dz}{dy}\right)\hat j + \hat k \right).
\end{equation*}
Therefore, force by pressure behind the element is
\begin{equation*}
\vec F = PdA\;\hat u = b PdA \left[ - \left( \frac{dz}{dy}\right)\hat j + \hat k\right].
\end{equation*}
The other force is the weight
\begin{equation*}
\vec W = -dm\; g \;\hat k.
\end{equation*}
Net force, \(\vec F_\text{net} = \vec F + \vec W = m\vec a\text{,}\) where \(\vec a\) is the centripetal acceleration, which for radius \(y\) and angular speed \(\omega\) will be
\begin{equation*}
\vec a = - \omega^2 y \hat j.
\end{equation*}
Therefore, we will get equations along \(y\) and \(z\) axis will be
\begin{align*}
- b\amp P dA \left( \frac{dz}{dy}\right) = -dm\; \omega^2 y.\\
b\amp PdA - dm\; g = 0.
\end{align*}
Eliminating \(PdA\) from these two gives
\begin{equation*}
\frac{dz}{dy} = \frac{\omega^2}{g} y.
\end{equation*}
This differential equation has a simple solution.
\begin{equation*}
z = \frac{\omega^2}{2g}\; y^2 + c,
\end{equation*}
where \(c\) is constant of integration. Since, \(y=0,z=0\) is on the curve, we have \(c=0\text{.}\) Thus,
\begin{equation*}
z = \frac{\omega^2}{2g}\; y^2.
\end{equation*}
Replacing \(|y|\) by \(r\) we get
\begin{equation*}
z = \frac{\omega^2}{2g}\; r^2.
\end{equation*}
