Storage change from inflow equation I = 2t

The water balance gives $I - O = \Delta S$. Losses are neglected, so outflow $O = 0$ and the change in storage equals the total inflow volume.

Given: $I = 2t$ m³/s, with $t$ in seconds; duration $T = 3$ h $= 3 \times 3600 = 10800$ s.

$$\Delta S = \int_0^{T} I\,dt = \int_0^{10800} 2t\,dt = \left[t^2\right]_0^{10800}$$ $$\Delta S = (10800)^2 = 1.1664 \times 10^{8}\ \text{m}^3$$

Answer: Change in storage = 1.1664 × 10⁸ m³ (116.64 million m³) increase in 3 hours.

(If $t$ were taken in hours, the answer would be different; $t$ in seconds is consistent with the unit m³/s.)