$d K=\frac{1}{2} d m \cdot v_{x}^{2} \Rightarrow d K=\frac{1}{2} 4 \pi x^{2} \cdot d x \cdot \rho \cdot v_{x}^{2}$

or $d K=2 \pi R^{2} \cdot \rho \cdot v_{x}^{2} \cdot d x \quad \ldots$ $(i)$

Considering laminar flow, for section $1$ and $2$ shown in figure, by equation of continuity

$A_{1} v_{1}=A_{2} v_{2} \Rightarrow 4 \pi x^{2} v_{x}=4 \pi R^{2} v$

$\Rightarrow \quad v_{x} =\frac{R^{2}}{x^{2}} \cdot v\quad \ldots$ $(ii)$

From Eqs. $(i)$ and $(ii)$, we have

$d K=2 \pi x^{2} \rho\left(\frac{R^{2}}{x^{2}} \cdot v \cdot\right)^{2} d x$

$\Rightarrow d K=2 \pi R^{4} \rho v^{2} \cdot\left(\frac{d x}{x^{2}}\right)$

So, kinetic energy of complete volume of water,

$K=\int \limits_{R}^{\infty} d K=2 \pi R^{4} \rho v^{2} \int \limits_{R}^{\infty} \frac{d x}{x^{2}}$

$\Rightarrow \quad K=2 \pi R^{4} \cdot \rho \cdot v\left[-\frac{1}{x}\right]_{R}^{\infty}$

$=2 \pi \rho R^{3} v^{2}$