Answer
$65333333.33 \ N$
Work Step by Step
The width of the dam at height $y$ is given by $y=\dfrac{x^2}{16} \implies x=\pm 4 \sqrt y$
So, $w(y)=8 \sqrt y$
We need to use the formula for finding the total force on the dam such as: $F=\int_0^a \rho g (a-y) w(y) \ dy$.
Plug in the above formula the given values to obtain:
$F=\int_0^a \rho g (a-y) w(y) \ dy \\= 8 \times \int_{0}^{25} \rho g (25-y) \sqrt y \ dy\\= 8 \times \rho g \times \dfrac{2500}{3} \\= 8 \times \dfrac{2500}{3} \times (9.81) \times (1000) \\=65333333.33 \ N$
