Water pressure on a dam

The water pressure acts on a triangular area, whose cross-sectional width $\Delta x = \dfrac{d}{h} y$ increases linearly with the height $y$ and varies between the value $\Delta x = 0$ at the bottom and $\Delta x = d$ at the top of the lake. On the other hand, the pressure $p = \rho g(h - y)$ decreases linearly with $y$ , so that we have $p = \rho g h$ at the bottom and $p = 0$ at the top. The total force can be estimated by the surface integral $\begin{aligned} F &= \int p \, dA = \int_0^h p \cdot \Delta x \, dy = \int_0^h \rho g(h - y) \cdot \dfrac{d}{h} y \, dy = \frac{\rho g d}{h} \int_0^h y (h - y) dy \\ &= \frac{\rho g d}{h} \left[ \frac{h^3}{2} - \frac{h^3}{3} \right] = \frac{\rho g d h^2}{6} = 5 \cdot 10^9 \,\text{N} \end{aligned}$