The force on an elemental area \(\triangle A\), at depth \(x\) on the side of such a cone is given by
\({ F }_{ s }=\rho g.x\triangle A\),
where \(g\) is gravity, \(\rho\) is density of the fluid.
\(\triangle A\) is an elemental ring of radius \(\triangle y\), height \(\triangle x\),
\(\triangle A=2\pi \triangle y. \triangle x\)
\(\therefore { F }_{ s }=2\pi \rho g. x.\triangle y \triangle x\)
The work done against such a force in moving the fluid out to the perimeter \(y\), at a given depth \(x\) is,
\(\triangle W_r = F_s . \triangle y = 2\pi \rho g. x. \triangle y^2 \triangle x\)
\(W_r = 2\pi \rho g. x. \triangle x \int_{0}^{y} (dy)^2\)
\(W_r = 2\pi \rho g. x. \triangle x |_{0}^{y} \frac {y^2}{2}\)
\(W_r = \pi \rho g. x.y^2 \triangle x \)
And so, total work done in establishing such a cone of height, \(h\) in the fliud is,
\(W = \int_0^h dW_r = \pi \rho g. \int_0^h x.y^2 dx\), since \(y=\frac r h x\)
\(W = \pi \rho g. \frac {r^2}{ h^2} \int _{0}^{h} x^3 dx\)
\(W = \frac{1}{4} \pi \rho g r^2h^2\)
If we model a vortex as a inverted cone, than this expression is the Potential Energy of the vortex.
