Answer
$V = \frac{\pi \times D^{2}}{4} \times V \times t $
Work Step by Step
Assuming the pool was empty:
Since the hose's cross-section is a circle, its area would be:
$ A(m^{2}) = \frac{\pi \times D^{2}}{4} $
The hose's flowrate would be:
$ v(\frac{m^{3}}{s}) = A(m^{2}) \times V(\frac{m}{s}) $
$ v(\frac{m^{3}}{s}) = \frac{\pi \times D^{2}}{4} \times V $
With a filling time t(s), the volume of the pool would be:
$ V(m^{3}) = v(\frac{m^{3}}{s}) \times t(s) $
$ V(m^{3}) = \frac{\pi \times D^{2}}{4} \times V \times t $
P.S. If the pool has an initial volume V_{0} (m^{3}), its total volume would be given by:
$ V = \frac{\pi \times D^{2}}{4} \times V \times t + V_{0} $
